Should I use printf in my C++ code? - c++

I generally use cout and cerr to write text to the console. However sometimes I find it easier to use the good old printf statement. I use it when I need to format the output.
One example of where I would use this is:
// Lets assume that I'm printing coordinates...
printf("(%d,%d)\n", x, y);
// To do the same thing as above using cout....
cout << "(" << x << "," << y << ")" << endl;
I know I can format output using cout but I already know how to use the printf. Is there any reason I shouldn't use the printf statement?

My students, who learn cin and cout first, then learn printf later, overwhelmingly prefer printf (or more usually fprintf). I myself have found the printf model sufficiently readable that I have ported it to other programming languages. So has Olivier Danvy, who has even made it type-safe.
Provided you have a compiler that is capable of type-checking calls to printf, I see no reason not to use fprintf and friends in C++.
Disclaimer: I am a terrible C++ programmer.

If you ever hope to i18n your program, stay away from iostreams. The problem is that it can be impossible to properly localize your strings if the sentence is composed of multiple fragments as is done with iostream.
Besides the issue of message fragments, you also have an issue of ordering. Consider a report that prints a student's name and their grade point average:
std::cout << name << " has a GPA of " << gpa << std::endl;
When you translate that to another language, the other language's grammar may need you to show the GPA before the name. AFAIK, iostreams has not way to reorder the interpolated values.
If you want the best of both worlds (type safety and being able to i18n), use Boost.Format.

I use printf because I hate the ugly <<cout<< syntax.

Adaptability
Any attempt to printf a non-POD results in undefined behaviour:
struct Foo {
virtual ~Foo() {}
operator float() const { return 0.f; }
};
printf ("%f", Foo());
std::string foo;
printf ("%s", foo);
The above printf-calls yield undefined behaviour. Your compiler may warn you indeed, but those warnings are not required by the standards and not possible for format strings only known at runtime.
IO-Streams:
std::cout << Foo();
std::string foo;
std::cout << foo;
Judge yourself.
Extensibility
struct Person {
string first_name;
string second_name;
};
std::ostream& operator<< (std::ostream &os, Person const& p) {
return os << p.first_name << ", " << p.second_name;
}
cout << p;
cout << p;
some_file << p;
C:
// inline everywhere
printf ("%s, %s", p.first_name, p.second_name);
printf ("%s, %s", p.first_name, p.second_name);
fprintf (some_file, "%s, %s", p.first_name, p.second_name);
or:
// re-usable (not common in my experience)
int person_fprint(FILE *f, const Person *p) {
return fprintf(f, "%s, %s", p->first_name, p->second_name);
}
int person_print(const Person *p) {
return person_fprint(stdout, p);
}
Person p;
....
person_print(&p);
Note how you have to take care of using the proper call arguments/signatures in C (e.g. person_fprint(stderr, ..., person_fprint(myfile, ...), where in C++, the "FILE-argument" is automatically "derived" from the expression. A more exact equivalent of this derivation is actually more like this:
FILE *fout = stdout;
...
fprintf(fout, "Hello World!\n");
person_fprint(fout, ...);
fprintf(fout, "\n");
I18N
We reuse our Person definition:
cout << boost::format("Hello %1%") % p;
cout << boost::format("Na %1%, sei gegrüßt!") % p;
printf ("Hello %1$s, %2$s", p.first_name.c_str(), p.second_name.c_str());
printf ("Na %1$s, %2$s, sei gegrüßt!",
p.first_name.c_str(), p.second_name.c_str());
Judge yourself.
I find this less relevant as of today (2017). Maybe just a gut feeling, but I18N is not something that is done on a daily basis by your average C or C++ programmer. Plus, it's a pain in the a...natomy anyways.
Performance
Have you measured the actual significance of printf performance? Are your bottleneck applications seriously so lazy that the output of computation results is a bottleneck? Are you sure you need C++ at all?
The dreaded performance penalty is to satisfy those of you who want to use a mix of printf and cout. It is a feature, not a bug!
If you use iostreams consistently, you can
std::ios::sync_with_stdio(false);
and reap equal runtime with a good compiler:
#include <cstdio>
#include <iostream>
#include <ctime>
#include <fstream>
void ios_test (int n) {
for (int i=0; i<n; ++i) {
std::cout << "foobarfrob" << i;
}
}
void c_test (int n) {
for (int i=0; i<n; ++i) {
printf ("foobarfrob%d", i);
}
}
int main () {
const clock_t a_start = clock();
ios_test (10024*1024);
const double a = (clock() - a_start) / double(CLOCKS_PER_SEC);
const clock_t p_start = clock();
c_test (10024*1024);
const double p = (clock() - p_start) / double(CLOCKS_PER_SEC);
std::ios::sync_with_stdio(false);
const clock_t b_start = clock();
ios_test (10024*1024);
const double b = (clock() - b_start) / double(CLOCKS_PER_SEC);
std::ofstream res ("RESULTS");
res << "C ..............: " << p << " sec\n"
<< "C++, sync with C: " << a << " sec\n"
<< "C++, non-sync ..: " << b << " sec\n";
}
Results (g++ -O3 synced-unsynced-printf.cc, ./a.out > /dev/null, cat RESULTS):
C ..............: 1.1 sec
C++, sync with C: 1.76 sec
C++, non-sync ..: 1.01 sec
Judge ... yourself.
No. You won't forbid me my printf.
You can haz a typesafe, I18N friendly printf in C++11, thanks to variadic templates. And you will be able to have them very, very performant using user-defined literals, i.e. it will be possible to write a fully static incarnation.
I have a proof of concept. Back then, support for C++11 was not as mature as it is now, but you get an idea.
Temporal Adaptability
// foo.h
...
struct Frob {
unsigned int x;
};
...
// alpha.cpp
... printf ("%u", frob.x); ...
// bravo.cpp
... printf ("%u", frob.x); ...
// charlie.cpp
... printf ("%u", frob.x); ...
// delta.cpp
... printf ("%u", frob.x); ...
Later, your data grows so big you must do
// foo.h
...
unsigned long long x;
...
It is an interesting exercise maintaining that and doing it bug-free. Especially when other, non-coupled projects use foo.h.
Other.
Bug Potential: There's a lot of space to commit mistakes with printf, especially when you throw user input bases strings in the mix (think of your I18N team). You must take care to properly escape every such format string, you must be sure to pass the right arguments, etc. etc..
IO-Streams make my binary bigger: If this is a more important issue than maintainability, code-quality, reuseability, then (after verifying the issue!) use printf.

Use boost::format. You get type safety, std::string support, printf like interface, ability to use cout, and lots of other good stuff. You won't go back.

Use printf. Do not use C++ streams. printf gives you much better control (such as float precision etc.). The code is also usually shorter and more readable.
Google C++ style guide agrees.
Do not use streams, except where
required by a logging interface. Use
printf-like routines instead.
There are various pros and cons to
using streams, but in this case, as in
many other cases, consistency trumps
the debate. Do not use streams in your
code.

No reason at all. I think it's just some strange ideology that drives people towards using only C++ libraries even though good old C libs are still valid. I'm a C++ guy and I use C functions a lot too. Never had any problems with them.

On the whole I agree (hate the << syntax especially if you need complex formatting)
But I should point out the safety aspects.
printf("%x",2.0f)
printf("%x %x",2)
printf("%x",2,2)
Probably won't be noticed by the compiler but could crash your app.

Streams are the canonical way. Try making this code work with printf:
template <typename T>
void output(const T& pX)
{
std::cout << pX << std::endl;
}
Good luck.
What I mean is, you can make operators to allow your types to be outputted to ostream's, and without hassle use it just like any other type. printf doesn't fit the the generality of C++, or more specifically templates.
There's more than usability. There's also consistency. In all my projects, I have cout (and cerr and clog) tee'd to also output to a file. If you use printf, you skip all of that. Additionally, consistency itself is a good thing; mixing cout and printf, while perfectly valid, is ugly.
If you have an object, and you want to make it output-able, the cleanest way to do this is overload operator<< for that class. How are you going to use printf then? You're going to end up with code jumbled with cout's and printf's.
If you really want formatting, use Boost.Format while maintaining the stream interface. Consistency and formatting.

Use whatever fits your needs and preferences. If you're comfortable with printf then by all means use it. If you're happier with iostreams stick to 'em. Mix and match as best fits your requirements. This is software, after all - there's better ways and worse ways, but seldom is there only ONE way.
Share and enjoy.

You can get the best of both worlds with the {fmt} library which combines safety and extensibility of iostreams with usability and performance of (s)printf. Example:
fmt::print("The answer is {}.", 42);
The library supports Python-like and printf format string syntax.
Disclaimer: I'm the author of this library.

I do not like printf. Its lack of type-safety makes it dangerous to use, plus the need to remember format specifiers is a pain. The templated operators that smartly do the right thing are much better. So I always use the C++ streams in C++.
Granted, many people prefer printf, for other reasons, enumerated elsewhere.

I often "drop back" to using printf(), but more often snprintf() for easier formatted output. When programming in C++ I use this wrapper I wrote a while back, called like this (to use your example as above): cout << format("(%d,%d)\n", x, y);
Here's the header (stdiomm.h):
#pragma once
#include <cstdarg>
#include <string>
template <typename T>
std::basic_string<T> format(T const *format, ...);
template <typename T>
std::basic_string<T> vformat(T const *format, va_list args);
And the source (stdiomm.cpp):
#include "stdiomm.h"
#include <boost/scoped_array.hpp>
#include <cstdio>
template <>
std::wstring vformat(wchar_t const *format, va_list arguments)
{
#if defined(_WIN32)
int required(_vscwprintf(format, arguments));
assert(required >= 0);
boost::scoped_array<wchar_t> buffer(new wchar_t[required + 1]);
int written(vswprintf(buffer.get(), required + 1, format, arguments));
assert(written == required);
return std::wstring(buffer.get(), written);
#else
# error "No implementation yet"
#endif
}
template <>
std::string vformat(char const *format, va_list arguments)
{
#if defined(_WIN32)
int required(_vscprintf(format, arguments));
assert(required >= 0);
boost::scoped_array<char> buffer(new char[required + 1]);
int written(vsnprintf(buffer.get(), required + 1, format, arguments));
assert(written == required);
return std::string(buffer.get(), written);
#else
char *buffer;
int printed = vasprintf(&buffer, format, arguments);
assert(printed != -1);
std::string retval(buffer, printed);
free(buffer);
return retval;
#endif
}
template <typename T>
std::basic_string<T> format(T const *format, ...)
{
va_list ap;
va_start(ap, format);
std::basic_string<T> retval(vformat(format, ap));
va_end(ap);
return retval;
}
template std::wstring format(wchar_t const *format, ...);
template std::string format(char const *format, ...);
Update
After reading some of the other answers, I might have to make a switch to boost::format() myself!

I almost always use printf for temporary debugging statements. For more permanent code, I prefer the 'c' streams as they are The C++ Way. Although boost::format looks promising and might replace my stream usage (especially for complexly formatted output), probably nothing will replace printf for me for a long time.

C++ streams are overrated, after all they're in fact just classes with an overloaded operator <<.
I've read many times that streams are the C++ way as printf is the C way, but they are both library features available in C++, so you should use what suits best.
I mostly prefer printf, but I've also used streams, which provide cleaner code and prevent you from having to match % placeholders to arguments.

Even though the question is rather old, I want to add my two cents.
Printing user-created objects with printf()
This is rather simple if you think about it - you can stringify your type and sent the string to printf:
std::string to_string(const MyClass &x)
{
return to_string(x.first)+" "+to_string(x.second);
}
//...
printf("%s is awesome", to_string(my_object).c_str()); //more or less
A shame there wasn't (there is C++11 to_string()) standarized C++ interface to stringify objects...
printf() pitfall
A single flag - %n
The only one that is an output parameter - it expects pointer to int. It writes number of succesfully written characters to location pointed by this pointer. Skillful use of it can trigger overrun, which is security vulnerability (see printf() format string attack).

I have read warnings saying that cout and cerr are unsafe for multithreading. If true, this is a good reason to avoid using them. Note: I use GNU g++ with openMP.

(See the fmt library homepage)
In C++20, the fmt library is standardized for the formatting part:
std::format("({},{})\n", x, y) // returns a std::string
You can avoid the dynamic allocation overhead by using format_to:
std::format_to(/* output iterator */, "({},{})\n", x, y);
This should be considered the canonical way of formatting because it combines the benefits of streams:
Safety: the library is fully type safe. Automatic memory management prevents buffer overflow. Errors in format strings are reported using exceptions or at compile time.
Extensibility: overloading operator<< is easy, whereas extending printf is ... not so easy.
and that of printf:
Ease to use: the % syntax is supported rather than the verbose manipulators. The {} syntax is also introduced to eliminate the specifiers.
Performance: measurement has shown that the fmt library is so far the fastest output method in C++. Faster than printf and streams.

It depends on the situation. Nothing is perfect. I use both. Streams are good for custom types as you can overload the >> operator in ostream. But when it comes to spacing and etc it's better to use printf(). stringstream and like are better than the C style strcat(). So use one that's appropriate for the situation.

streams are preferred in cpp as they adhere to the object oriented paradigm of cpp,
beside being type safe.
printf , on the other hand is more of a functional approach.
only reason for not using printf in cpp code that i can think of is not being object oriented.
its more of a personal choice.

Related

C++ equivalent of printf() or formatted output

I am relatively new to the C++ world.
I know std::cout is used for console output in C++. But consider the following code in C :
#include<stdio.h>
int main(){
double dNum=99.6678;
printf("%.02lf",dNum);
//output: 99.67
return 0;
}
How do I achieve the similar formatting of a double type value upto 2 decimal places using cout in C++ ?
I know C++ is backward compatible with C. But is there a printf() equivalent in C++ if so, then where is it defined ?
This is what you want:
std::cout << std::fixed << std::setprecision(2);
std::cout << dNum;
and don't forget to :
#include <iostream>
#include <iomanip>
There is no equivalent. It is a pain using cout for formatted output.
All suggested solutions calling setprecision and similar are awfull when using long formats.
boost::format does not support some very nice features. Incompatibilities with printf. I still use printf because it is unbeatable.
If you want to use printf like formatting you should probably use snprintf (or build an allocating variant of that on top of that). Note that sprintf requires you to be able to guarantee that the result will not overrun the buffer you have to keep defined behaviour. With snprintf on the other hand can guarantee that it will not overrun the buffer since you specifiy the maximal number of characters that will be written to the string (it will instead truncate the output).
You could even build something that can directly be fed to an ostream on top of snprintf by automatically allocate the buffer and place in an object that on destruction free that memory. This in addition with a method to feed the object to an ostream would finish it off. Something like (with room for improvements):
struct Format {
char buf[999];
Format(fmt, ...) {
va_list ap;
va_start(fmt, ap);
vsnprintf(buf, sizeof(buf), fmt, ap);
va_end(ap);
}
}
ostream& operator<< (ostream& os, Format const& str);
then you use this as:
cout << Format("The answer is %d", 42) << endl;
If you're using the GNU libraries you could of course use printf directly since cout and stdout are the same object then. Otherwise you should probably avoid mixing stdio and iostreams as there is no guarantee that these are synchronized with each other.
If you really want to reuse the same formatting techniques as in C, you may use Boost::format, which does exactly that:
cout << boost::format("%.02lf") % dNum;
But is there a printf() equivalent in C++ if so, then where is it defined?
There is a standards proposal P0645 to add a similar formatting facility to C++. In the meantime you can use the {fmt} library that implements this proposal and more:
#include <fmt/core.h>
int main()
fmt::print("{:.02f}", 99.6678);
}
P0645 and {fmt} use Python-like format string syntax which is similar to printf's but uses {} as delimiters instead of %.
Also the type information is preserved so you don't need l or other noisy specifiers.
Save your program as CPP and run it.
It runs and prints the answer.
Because C++ also has the printf() and scanf() like C.
You can also use sprintf in C++ to 'print' into a string and then cout that string. This strategy leverages your experience with printf-style formatting.
To output a value to the console using C++, you need the global ostream object cout and the << operator. endl is another global ostream object used as line break.
All are defined in the <iostream> header file. You can also use various formatting flags to control the presentation of the output...
#include<iostream>
using namespace std;
int main() {
double dNum = 99.6678;
cout << dNum;
cout.setf(ios::scientific, ios::floatfield); // format into scientific notation
cout << dNum;
cout.precision(8); // change precision
cout << dNum;
system("pause");
return 0;
}
The functional equivalent of your printf() call, using std::cout, is
std::cout << fixed << setprecision(2) << dNum;
It is necessary to #include <iomanip> and <iostream>.
The printf() equivalent in C++ is std::printf(), declared in <cstdio>.
Also, thanks to backward compatibility to C - specifically C++98 was required to maximise backward compatiblility to C89 - C's printf() declared in <stdio.h> is also available in C++. Note, however, that <stdio.h> is deprecated (tagged for removal from a future version of the C++ standard). Also, not all features of printf() introduced in C99 (the 1999 C standard) or later are necessarily supported in C++.

Extracting and Creating separate txt files for each line of the source file [duplicate]

This question already has answers here:
Easiest way to convert int to string in C++
(30 answers)
Closed 7 years ago.
I was wondering if there was an alternative to itoa() for converting an integer to a string because when I run it in visual Studio I get warnings, and when I try to build my program under Linux, I get a compilation error.
In C++11 you can use std::to_string:
#include <string>
std::string s = std::to_string(5);
If you're working with prior to C++11, you could use C++ streams:
#include <sstream>
int i = 5;
std::string s;
std::stringstream out;
out << i;
s = out.str();
Taken from http://notfaq.wordpress.com/2006/08/30/c-convert-int-to-string/
boost::lexical_cast works pretty well.
#include <boost/lexical_cast.hpp>
int main(int argc, char** argv) {
std::string foo = boost::lexical_cast<std::string>(argc);
}
Archeology
itoa was a non-standard helper function designed to complement the atoi standard function, and probably hiding a sprintf (Most its features can be implemented in terms of sprintf): http://www.cplusplus.com/reference/clibrary/cstdlib/itoa.html
The C Way
Use sprintf. Or snprintf. Or whatever tool you find.
Despite the fact some functions are not in the standard, as rightly mentioned by "onebyone" in one of his comments, most compiler will offer you an alternative (e.g. Visual C++ has its own _snprintf you can typedef to snprintf if you need it).
The C++ way.
Use the C++ streams (in the current case std::stringstream (or even the deprecated std::strstream, as proposed by Herb Sutter in one of his books, because it's somewhat faster).
Conclusion
You're in C++, which means that you can choose the way you want it:
The faster way (i.e. the C way), but you should be sure the code is a bottleneck in your application (premature optimizations are evil, etc.) and that your code is safely encapsulated to avoid risking buffer overruns.
The safer way (i.e., the C++ way), if you know this part of the code is not critical, so better be sure this part of the code won't break at random moments because someone mistook a size or a pointer (which happens in real life, like... yesterday, on my computer, because someone thought it "cool" to use the faster way without really needing it).
Try sprintf():
char str[12];
int num = 3;
sprintf(str, "%d", num); // str now contains "3"
sprintf() is like printf() but outputs to a string.
Also, as Parappa mentioned in the comments, you might want to use snprintf() to stop a buffer overflow from occuring (where the number you're converting doesn't fit the size of your string.) It works like this:
snprintf(str, sizeof(str), "%d", num);
Behind the scenes, lexical_cast does this:
std::stringstream str;
str << myint;
std::string result;
str >> result;
If you don't want to "drag in" boost for this, then using the above is a good solution.
We can define our own iota function in c++ as:
string itoa(int a)
{
string ss=""; //create empty string
while(a)
{
int x=a%10;
a/=10;
char i='0';
i=i+x;
ss=i+ss; //append new character at the front of the string!
}
return ss;
}
Don't forget to #include <string>.
С++11 finally resolves this providing std::to_string.
Also boost::lexical_cast is handy tool for older compilers.
I use these templates
template <typename T> string toStr(T tmp)
{
ostringstream out;
out << tmp;
return out.str();
}
template <typename T> T strTo(string tmp)
{
T output;
istringstream in(tmp);
in >> output;
return output;
}
Try Boost.Format or FastFormat, both high-quality C++ libraries:
int i = 10;
std::string result;
WIth Boost.Format
result = str(boost::format("%1%", i));
or FastFormat
fastformat::fmt(result, "{0}", i);
fastformat::write(result, i);
Obviously they both do a lot more than a simple conversion of a single integer
You can actually convert anything to a string with one cleverly written template function. This code example uses a loop to create subdirectories in a Win-32 system. The string concatenation operator, operator+, is used to concatenate a root with a suffix to generate directory names. The suffix is created by converting the loop control variable, i, to a C++ string, using the template function, and concatenating that with another string.
//Mark Renslow, Globe University, Minnesota School of Business, Utah Career College
//C++ instructor and Network Dean of Information Technology
#include <cstdlib>
#include <iostream>
#include <string>
#include <sstream> // string stream
#include <direct.h>
using namespace std;
string intToString(int x)
{
/**************************************/
/* This function is similar to itoa() */
/* "integer to alpha", a non-standard */
/* C language function. It takes an */
/* integer as input and as output, */
/* returns a C++ string. */
/* itoa() returned a C-string (null- */
/* terminated) */
/* This function is not needed because*/
/* the following template function */
/* does it all */
/**************************************/
string r;
stringstream s;
s << x;
r = s.str();
return r;
}
template <class T>
string toString( T argument)
{
/**************************************/
/* This template shows the power of */
/* C++ templates. This function will */
/* convert anything to a string! */
/* Precondition: */
/* operator<< is defined for type T */
/**************************************/
string r;
stringstream s;
s << argument;
r = s.str();
return r;
}
int main( )
{
string s;
cout << "What directory would you like me to make?";
cin >> s;
try
{
mkdir(s.c_str());
}
catch (exception& e)
{
cerr << e.what( ) << endl;
}
chdir(s.c_str());
//Using a loop and string concatenation to make several sub-directories
for(int i = 0; i < 10; i++)
{
s = "Dir_";
s = s + toString(i);
mkdir(s.c_str());
}
system("PAUSE");
return EXIT_SUCCESS;
}
Allocate a string of sufficient length, then use snprintf.
int number = 123;
stringstream = s;
s << number;
cout << ss.str() << endl;
I wrote this thread-safe function some time ago, and am very happy with the results and feel the algorithm is lightweight and lean, with performance that is about 3X the standard MSVC _itoa() function.
Here's the link. Optimal Base-10 only itoa() function? Performance is at least 10X that of sprintf(). The benchmark is also the function's QA test, as follows.
start = clock();
for (int i = LONG_MIN; i < LONG_MAX; i++) {
if (i != atoi(_i32toa(buff, (int32_t)i))) {
printf("\nError for %i", i);
}
if (!i) printf("\nAt zero");
}
printf("\nElapsed time was %f milliseconds", (double)clock() - (double)(start));
There are some silly suggestions made about using the caller's storage that would leave the result floating somewhere in a buffer in the caller's address space. Ignore them. The code I listed works perfectly, as the benchmark/QA code demonstrates.
I believe this code is lean enough to use in an embedded environment. YMMV, of course.
The best answer, IMO, is the function provided here:
http://www.jb.man.ac.uk/~slowe/cpp/itoa.html
It mimics the non-ANSI function provided by many libs.
char* itoa(int value, char* result, int base);
It's also lightning fast and optimizes well under -O3, and the reason you're not using c++ string_format() ... or sprintf is that they are too slow, right?
If you are interested in fast as well as safe integer to string conversion method and not limited to the standard library, I can recommend the format_int method from the {fmt} library:
fmt::format_int(42).str(); // convert to std::string
fmt::format_int(42).c_str(); // convert and get as a C string
// (mind the lifetime, same as std::string::c_str())
According to the integer to string conversion benchmarks from Boost Karma, this method several times faster than glibc's sprintf or std::stringstream. It is even faster than Boost Karma's own int_generator as was confirm by an independent benchmark.
Disclaimer: I'm the author of this library.
Note that all of the stringstream methods may involve locking around the use of the locale object for formatting. This may be something to be wary of if you're using this conversion from multiple threads...
See here for more. Convert a number to a string with specified length in C++
On Windows CE derived platforms, there are no iostreams by default. The way to go there is preferaby with the _itoa<> family, usually _itow<> (since most string stuff are Unicode there anyway).
Most of the above suggestions technically aren't C++, they're C solutions.
Look into the use of std::stringstream.

Why doesn't the string class have a << operator (operator<<) predefined so that strings work like ostringstreams?

It seems to me that defining the << operator (operator<<) to work directly with strings is more elegant than having to work with ostringstreams and then converting back to strings. Is there a reason why c++ doesn't do this out of the box?
#include <string>
#include <sstream>
#include <iostream>
using namespace std;
template <class T>
string& operator<<(string& s, T a) {
ostringstream ss;
ss << a;
s.append(ss.str());
return s;
}
int main() {
string s;
// this prints out: "inserting text and a number(1)"
cout << (s << "inserting text and a number (" << 1 << ")\n");
// normal way
ostringstream os;
os << "inserting text and a number(" << 1 << ")\n";
cout << os.str();
}
Streams contain additional state. Imagine if this were possible:
std::string str;
int n = 1234;
str << std::hex;
str << n;
return str; // returns "0x4d2" (or something, I forget)
In order to maintain this additional state, strings would have to have storage for this state. The C++ standards committee (and C++ programmers in general) have generally frowned upon superfluous resource consumption, under the motto "pay only for what you use". So, no extra fields in the string class.
The subjective answer: is that I think the std::string class was quite poorly designed to begin with, especially compared to other parts of C++'s excellent standard library, and adding features to std::string is just going to make things worse. This is a very subjective opinion and feel free to dismiss me as a raving lunatic.
The problem with the idea of strings being output streams is that they would become too heavy.
Strings are intended to "hold string data", not to format some output. Output streams have a heavy "state" which can be manipulated (see <iomanip>) and thus has to be stored. This means that, of course, this has to be stored for every string in every program, but almost none of them are used as an output stream; so it's a huge waste of resources.
C++ follows the "zero overhead" design principle (or at least no more overhead than totally necessary). Not having a string class which doesn't add any unnecessary overhead would be a huge violation of this design principle. If this was the case: what would people do in overhead-critical cases? Use C-strings... ouch!
In C++11, an alternative is to use the operator+= with std::to_string to append to a string, which can also be chained like the operator<< of the output stream. You can wrap both += and to_string in a nice operator<< for string if you like:
template <class Number>
std::string& operator<<(std::string& s, Number a) {
return s += std::to_string(a);
}
std::string& operator<<(std::string& s, const char* a) {
return s += a;
}
std::string& operator<<(std::string& s, const std::string &a) {
return s += a;
}
Your example, updated using this method: http://ideone.com/4zbVtD
Probably lost in the depths of time now but formatted output was always associated with streams in C (since they didn't have "real" strings) and this may have been carried over into C++ (which was, after all, C with classes). In C, the way to format to a string is to use sprintf, a variation on fprintf, the output-to-stream function.
Obviously conjecture on my part but someone probably thought similarly to yourself that these formatting things in the streams would be brilliant to have on strings as well, so they subclassed the stream classes to produce one that used a string as it's "output".
That seems the elegant solution to getting it working as quickly as possible. Otherwise, you would have had formatting code duplicated in streams and strings.

To print something without using cout, printf or puts()

I had learned that
inline ostream & _Cdecl ostream::operator<< (const signed char * _s) {
outstr(_s, (const signed char *)0);
return *this;
}
is how the insertion operator (<<) is declared(overloaded) in the iostream.h header file. Can I possibly use the same function to print a string value on screen?
I tried
#include<iostream.h>
int main() {
outstr("Hello world!", (const signed char *)0);
return 0;
}
it ended up in error. I would like to use something like this in order to see if there is some possible way to answer this query of printing something on screen without using printf, cout or puts().
Update: I would welcome if you have any suggestions other than
#include<stdlib.h>
void main() {
system("echo /"Hello world!/"");
}
NB: I have no restrictions if you can provide the C equivalent code that can print without a printf(), cout or puts()
Yes you could call the function directly, however your reasoning to do so is flawed. The time you save by eliminating the subroutine call to the operator is negligible when compared to the time taken to perform the actual function; this would be like closing the windows of your car while the convertible roof is down in order to reduce the rain.
If you want portability across all standards compliant C++ implementations, you can print a string to standard output in the following ways
const char * str = "Hello World\n";
fprintf(stdout, str);
fputs(str, stdout);
for (int i=0; str[i]!=0; ++i)
putchar(str[i]);
for (int i=0; str[i]!=0; ++i)
putc(str[i], stdout);
for (int i=0; str[i]!=0; ++i)
fputc(str[i], stdout);
fwrite(str, sizeof(*str), strlen(str), stdout);
Additionally, you can use std::cerr and std::clog. They write to stderr instead of stdout, but from the user's perspective, that's often the same place:
std::cerr << str;
std::clog << str;
From an efficiency perspective, I doubt any of these are going to help you. For that purpose, you might want to look at something a bit more platform specific. For POSIX systems, see the answer given by Dave S. For Windows, see this link.
What you shouldn't do, is open up your header files and imitate what they use. At least, not at the middle levels, where they are using different various obscure functions within their own implementation. Those functions might not exist upon the next release. However, if you go to the deepest levels, you will find OS specific calls like the ones in the link I provided above. Those should be safe to use as long as you stay on the same OS, or even between OS versions.
On a UNIX type system, you can do the following.
#include <unistd.h>
#include <stdio.h>
int main()
{
const char x[] = "Hello World!";
write(STDOUT_FILENO, x, strlen(x)); // Feel free to check the return value.
return 0;
}
I'm curious what your motivation for doing this would be. Outside of signal handlers, I'm reluctant to go to the lower level calls like this. The performance of the I/O will be the primary driver of time, not the intermediate function calls which are usually fairly heavily optimized / inlined.
The time required to make a function call is much, much smaller than the amount of time it takes to print your string. The amount of time you might save with your approach can (and usually should) be ignored.
You can directly use system calls.
http://docs.cs.up.ac.za/programming/asm/derick_tut/syscalls.html
This page, for example, explains linux system calls. You can start from the link I copied, and use many methods using assembly, or to say it in the other way, do something without calling the function of it.
But I'm guessing that was a trick question and if I had a company, I would never hire a person that uses system calls instead of functions.
This is an example of using sys_write(4) with standart output(1). You can inline assembly codes into your C/C++ code.
http://docs.cs.up.ac.za/programming/asm/derick_tut/#helloworld
The extraction operator is overloaded in the ostream class. So you cannot actually use it without having an object of that class with it.
It is implemented in the following manner:
cout<<"Hii";
is equivalent to:
cout.operator<<("Hii")

printf with std::string?

My understanding is that string is a member of the std namespace, so why does the following occur?
#include <iostream>
int main()
{
using namespace std;
string myString = "Press ENTER to quit program!";
cout << "Come up and C++ me some time." << endl;
printf("Follow this command: %s", myString);
cin.get();
return 0;
}
Each time the program runs, myString prints a seemingly random string of 3 characters, such as in the output above.
C++23 Update
We now finally have std::print as a way to use std::format for output directly:
#include <print>
#include <string>
int main() {
// ...
std::print("Follow this command: {}", myString);
// ...
}
This combines the best of both approaches.
Original Answer
It's compiling because printf isn't type safe, since it uses variable arguments in the C sense1. printf has no option for std::string, only a C-style string. Using something else in place of what it expects definitely won't give you the results you want. It's actually undefined behaviour, so anything at all could happen.
The easiest way to fix this, since you're using C++, is printing it normally with std::cout, since std::string supports that through operator overloading:
std::cout << "Follow this command: " << myString;
If, for some reason, you need to extract the C-style string, you can use the c_str() method of std::string to get a const char * that is null-terminated. Using your example:
#include <iostream>
#include <string>
#include <stdio.h>
int main()
{
using namespace std;
string myString = "Press ENTER to quit program!";
cout << "Come up and C++ me some time." << endl;
printf("Follow this command: %s", myString.c_str()); //note the use of c_str
cin.get();
return 0;
}
If you want a function that is like printf, but type safe, look into variadic templates (C++11, supported on all major compilers as of MSVC12). You can find an example of one here. There's nothing I know of implemented like that in the standard library, but there might be in Boost, specifically boost::format.
[1]: This means that you can pass any number of arguments, but the function relies on you to tell it the number and types of those arguments. In the case of printf, that means a string with encoded type information like %d meaning int. If you lie about the type or number, the function has no standard way of knowing, although some compilers have the ability to check and give warnings when you lie.
Please don't use printf("%s", your_string.c_str());
Use cout << your_string; instead. Short, simple and typesafe. In fact, when you're writing C++, you generally want to avoid printf entirely -- it's a leftover from C that's rarely needed or useful in C++.
As to why you should use cout instead of printf, the reasons are numerous. Here's a sampling of a few of the most obvious:
As the question shows, printf isn't type-safe. If the type you pass differs from that given in the conversion specifier, printf will try to use whatever it finds on the stack as if it were the specified type, giving undefined behavior. Some compilers can warn about this under some circumstances, but some compilers can't/won't at all, and none can under all circumstances.
printf isn't extensible. You can only pass primitive types to it. The set of conversion specifiers it understands is hard-coded in its implementation, and there's no way for you to add more/others. Most well-written C++ should use these types primarily to implement types oriented toward the problem being solved.
It makes decent formatting much more difficult. For an obvious example, when you're printing numbers for people to read, you typically want to insert thousands separators every few digits. The exact number of digits and the characters used as separators varies, but cout has that covered as well. For example:
std::locale loc("");
std::cout.imbue(loc);
std::cout << 123456.78;
The nameless locale (the "") picks a locale based on the user's configuration. Therefore, on my machine (configured for US English) this prints out as 123,456.78. For somebody who has their computer configured for (say) Germany, it would print out something like 123.456,78. For somebody with it configured for India, it would print out as 1,23,456.78 (and of course there are many others). With printf I get exactly one result: 123456.78. It is consistent, but it's consistently wrong for everybody everywhere. Essentially the only way to work around it is to do the formatting separately, then pass the result as a string to printf, because printf itself simply will not do the job correctly.
Although they're quite compact, printf format strings can be quite unreadable. Even among C programmers who use printf virtually every day, I'd guess at least 99% would need to look things up to be sure what the # in %#x means, and how that differs from what the # in %#f means (and yes, they mean entirely different things).
use myString.c_str() if you want a c-like string (const char*) to use with printf
thanks
Use std::printf and c_str()
example:
std::printf("Follow this command: %s", myString.c_str());
You can use snprinft to determine the number of characters needed and allocate a buffer of the right size.
int length = std::snprintf(nullptr, 0, "There can only be %i\n", 1 );
char* str = new char[length+1]; // one more character for null terminator
std::snprintf( str, length + 1, "There can only be %i\n", 1 );
std::string cppstr( str );
delete[] str;
This is a minor adaption of an example on cppreference.com
printf accepts a variable number of arguments. Those can only have Plain Old Data (POD) types. Code that passes anything other than POD to printf only compiles because the compiler assumes you got your format right. %s means that the respective argument is supposed to be a pointer to a char. In your case it is an std::string not const char*. printf does not know it because the argument type goes lost and is supposed to be restored from the format parameter. When turning that std::string argument into const char* the resulting pointer will point to some irrelevant region of memory instead of your desired C string. For that reason your code prints out gibberish.
While printf is an excellent choice for printing out formatted text, (especially if you intend to have padding), it can be dangerous if you haven't enabled compiler warnings. Always enable warnings because then mistakes like this are easily avoidable. There is no reason to use the clumsy std::cout mechanism if the printf family can do the same task in a much faster and prettier way. Just make sure you have enabled all warnings (-Wall -Wextra) and you will be good. In case you use your own custom printf implementation you should declare it with the __attribute__ mechanism that enables the compiler to check the format string against the parameters provided.
The main reason is probably that a C++ string is a struct that includes a current-length value, not just the address of a sequence of chars terminated by a 0 byte. Printf and its relatives expect to find such a sequence, not a struct, and therefore get confused by C++ strings.
Speaking for myself, I believe that printf has a place that can't easily be filled by C++ syntactic features, just as table structures in html have a place that can't easily be filled by divs. As Dykstra wrote later about the goto, he didn't intend to start a religion and was really only arguing against using it as a kludge to make up for poorly-designed code.
It would be quite nice if the GNU project would add the printf family to their g++ extensions.
Printf is actually pretty good to use if size matters. Meaning if you are running a program where memory is an issue, then printf is actually a very good and under rater solution. Cout essentially shifts bits over to make room for the string, while printf just takes in some sort of parameters and prints it to the screen. If you were to compile a simple hello world program, printf would be able to compile it in less than 60, 000 bits as opposed to cout, it would take over 1 million bits to compile.
For your situation, id suggest using cout simply because it is much more convenient to use. Although, I would argue that printf is something good to know.
Here’s a generic way of doing it.
#include <string>
#include <stdio.h>
auto print_helper(auto const & t){
return t;
}
auto print_helper(std::string const & s){
return s.c_str();
}
std::string four(){
return "four";
}
template<class ... Args>
void print(char const * fmt, Args&& ...args){
printf(fmt, print_helper(args) ...);
}
int main(){
std::string one {"one"};
char const * three = "three";
print("%c %d %s %s, %s five", 'c', 3+4, one + " two", three, four());
}