I was trying to affirm my knowledge about how streams work with doubles and various manipulators, and stumbled upon G++ doing something strange:
int main() {
double v = 10.0/3.;
//std::cout << v << '\n';
std::cout << std::setw(5) << std::setprecision(2) << v << '\n';
std::cout << std::setw(5) << std::setprecision(2) << std::fixed << v << '\n';
}
Output:
3.3 //why is this left aligned?
3.33 //why is this right aligned? which is right?
See it live here
Then I uncommented that first cout and got different results!
3.33333 //which alignment is this?
3.3 //now this is right aligned?!
3.33 //that implies right-aligned is correct
Subsequent tests show that the first double I stream out is left aligned, and all subsequent doubles are right aligned:
double v = 10.0/3.;
std::cout << std::setw(10) << v << '\n';
std::cout << std::setw(5) << std::setprecision(2) << std::fixed << v << '\n';
std::cout << std::setw(5) << std::setprecision(2) << v << '\n';
std::cout << std::setw(5) << std::setprecision(2) << v << '\n';
Output:
3.33333 //well, this is left aligned
3.33
3.3
3.3 //all subsequent tests are right aligned
Clang++ on Coliru is doing the same thing, I presume because they're using the same library.
I know the answer to "is this a G++ bug" is no 99.9% of the time, so can someone potentially explain the behavior I'm seeing?
After running this on other IDEs I would conclude that this is a problem not with g++/clang++, nor is any standard behavior, but a problem with the Coliru editor itself. I ran this on several different sites including Rextester and Ideone and I got the correct output for all of them. This seems to be a problem only belonging to the Coliru editor.
Streams are, by default, right-justified. When setting the width using std::setw(), the stream will insert padding characters as specified by out.fill() into the beginning (or end) of the output stream. I've noticed that Coliru doesn't add any padding characters on the first output operation when the stream is using the default fill character (a simple space). But when I change the fill character to anything other than a space, it works just fine.
Related
How can I format my output in C++? In other words, what is the C++ equivalent to the use of printf like this:
printf("%05d", zipCode);
I know I could just use printf in C++, but I would prefer the output operator <<.
Would you just use the following?
std::cout << "ZIP code: " << sprintf("%05d", zipCode) << std::endl;
This will do the trick, at least for non-negative numbers(a) such as the ZIP codes(b) mentioned in your question.
#include <iostream>
#include <iomanip>
using namespace std;
cout << setw(5) << setfill('0') << zipCode << endl;
// or use this if you don't like 'using namespace std;'
std::cout << std::setw(5) << std::setfill('0') << zipCode << std::endl;
The most common IO manipulators that control padding are:
std::setw(width) sets the width of the field.
std::setfill(fillchar) sets the fill character.
std::setiosflags(align) sets the alignment, where align is ios::left or ios::right.
And just on your preference for using <<, I'd strongly suggest you look into the fmt library (see https://github.com/fmtlib/fmt). This has been a great addition to our toolkit for formatting stuff and is much nicer than massively length stream pipelines, allowing you to do things like:
cout << fmt::format("{:05d}", zipCode);
And it's currently being targeted by LEWG toward C++20 as well, meaning it will hopefully be a base part of the language at that point (or almost certainly later if it doesn't quite sneak in).
(a) If you do need to handle negative numbers, you can use std::internal as follows:
cout << internal << setw(5) << setfill('0') << zipCode << endl;
This places the fill character between the sign and the magnitude.
(b) This ("all ZIP codes are non-negative") is an assumption on my part but a reasonably safe one, I'd warrant :-)
Use the setw and setfill calls:
std::cout << std::setw(5) << std::setfill('0') << zipCode << std::endl;
In C++20 you'll be able to do:
std::cout << std::format("{:05}", zipCode);
In the meantime you can use the {fmt} library, std::format is based on.
Disclaimer: I'm the author of {fmt} and C++20 std::format.
cout << setw(4) << setfill('0') << n << endl;
from:
http://www.fredosaurus.com/notes-cpp/io/omanipulators.html
or,
char t[32];
sprintf_s(t, "%05d", 1);
will output 00001 as the OP already wanted to do
Simple answer but it works!
ostream &operator<<(ostream &os, const Clock &c){
// format the output - if single digit, then needs to be padded with a 0
int hours = c.getHour();
// if hour is 1 digit, then pad with a 0, otherwise just print the hour
(hours < 10) ? os << '0' << hours : os << hours;
return os; // return the stream
}
I'm using a ternary operator but it can be translated into an if/else statement as follows
if(c.hour < 10){
os << '0' << hours;
}
else{
os << hours;
}
Well i was creating a program in c++ .Then i noticed that there were some error due to which my program was not functioning correctly
compiler used:- Dev c++
Error was this:-
Suppose n1=274877906944 which is 2^38. so log(n1) should be 38 which is correct. now let n2=274877906942 which is smaller than n1.
So that means if we calculate log(n2) then it must give me less than log(n1).But log(n2) gave me the same result as log(n1) which is 38.
Which is wrong!!!
Someone please explain this stuff..
You see the result that you do because of rounding. If you increase your precision, it'll be okay (note that the log2 function in <cmath> will cast your integers to double):
std::cout << std::fixed;
std::cout << std::setprecision(16);
std::cout << static_cast<double>(274877906944) << std::endl;
std::cout << static_cast<double>(274877906942) << std::endl;
std::cout << static_cast<double>(274877906948) << std::endl;
std::cout << log2(274877906944) << std::endl;
std::cout << log2(274877906942) << std::endl;
std::cout<< log2(274877906948) << std::endl;
Produces:
274877906944.0000000000000000
274877906942.0000000000000000
274877906948.0000000000000000
38.0000000000000000
37.9999999999895053
38.0000000000209965
Demo
I am using Qt, and I have an unsigned char *bytePointer and want to print out a number-value of the current byte. Below is my code, which is meant to give the int-value and the hex-value of the continuous bytes that I receive from a machine attached to the computer:
int byteHex=0;
byteHex = (int)*bytePointer;
qDebug << "\n int: " //this is the main issue here.
<< *bytePointer;
std::cout << " (hex: "
<< std::hex
<< byteHex
<< ")\n";
}
This gives perfect results, and I get actual numbers, however this code is going into an API and I don't want to use Qt-only functions, such as qDebug. So when I try this:
int byteHex=0;
byteHex = (int)*bytePointer;
std::cout << "\n int: " //I changed qDebug to std::cout
<< *bytePointer;
std::cout << " (hex: "
<< std::hex
<< byteHex
<< ")\n";
}
The output does give the hex-values perfectly, however the int-values return symbols (like ☺, └, §, to list a few).
My question is: How do I get std::cout to give the same output as qDebug?
EDIT: for some reason the symbols only occur with a certain Qt setting. I have no idea why it happened but it's fixed now.
As others pointed out in comment, you change the outputting to hex, but you do not actually set it back here:
std::cout << " (hex: "
<< std::hex
<< byteHex
<< ")\n";
You will need to apply this afterwards:
std::cout << std::dec;
Standard output streams will output any character type as a character, not a numeric value. To output the numeric value, convert to a non-character integer type:
std::cout << int(*bytePointer);
In C a space can be included in a printf formatting flag which results in positive numbers being prefixed with a space. This is a useful feature for aligning signed values. I can't figure out how to do the same in C++. In C:
double d = 1.2;
printf("%f\n",d);
printf("%+f\n",d);
printf("% f\n",d);
produces:
1.2
+1.2
1.2
Using ostream, I can do the first two, but how do I do the third?
int d = 1.2;
std::cout << d << std::endl;
std::cout << std::showpos << d << std::endl;
// ??????????????
EDIT: There seems to be some confusion about whether or not I just want to prefix all of my values with a space. I only want to prefix positive values with a space, similar to a) like the printf space flag does and b) similar to what showpos does, except a space rather than a '+'. For example:
printf("%f\n", 1.2);
printf("%f\n", -1.2);
printf("% f\n", 1.2);
printf("% f\n", -1.2);
1.2
-1.2
1.2
-1.2
Note that the third value is prefixed with a space while the fourth (negative) value is not.
You can use setfill and setw, like this:
cout << setw(4) << setfill(' ') << 1.2 << endl;
cout << setw(4) << setfill(' ') << -1.2 << endl;
This produces the following output:
1.2
-1.2
Don't forget to include <iomanip> in order for this to compile (link to ideone).
I don't have my standard with me and I'm doing this stuff too rarely to be confident about it: There are two ingredients to achieving this with IOStreams:
Use std:: showpos to have an indicator of positive values be shown. By default this will use +, of course.
I think the + is obtained using std::use_facet<std::ctype<char> >(s.get_loc()).widen('+'). To turn this into a space you could just use a std::locale with a std::ctype<char> facet installed responding with a space to the request to widen +.
That is, something like this:
struct my_ctype: std::ctype<char> {
char do_widen(char c) const {
return c == '+'? ' ': this->std::ctype<char>::do_widen(c);
}
};
int main() {
std::locale loc(std::locale(), new my_ctype);
std::cout.imbue(loc);
std::cout << std::showpos << 12.34 << '\n';
}
(the code isn't tested and probably riddled with errors).
How about
std::cout << (d >= 0 ? " ":"") << d << std::endl;
std::cout << " " << my_value;
If you need space only for positive:
if (my_value >=0 ) cout << " "; cout << my_value;
Is there a way to align text when printing using std::cout? I'm using tabs, but when the words are too big they won't be aligned anymore.
Sales Report for September 15, 2010
Artist Title Price Genre Disc Sale Tax Cash
Merle Blue 12.99 Country 4% 12.47 1.01 13.48
Richard Music 8.49 Classical 8% 7.81 0.66 8.47
Paula Shut 8.49 Classical 8% 7.81 0.72 8.49
The ISO C++ standard way to do it is to #include <iomanip> and use io manipulators like std::setw. However, that said, those io manipulators are a real pain to use even for text, and are just about unusable for formatting numbers (I assume you want your dollar amounts to line up on the decimal, have the correct number of significant digits, etc.). Even for just plain text labels, the code will look something like this for the first part of your first line:
// using standard iomanip facilities
cout << setw(20) << "Artist"
<< setw(20) << "Title"
<< setw(8) << "Price";
// ... not going to try to write the numeric formatting...
If you are able to use the Boost libraries, run (don't walk) and use the Boost.Format library instead. It is fully compatible with the standard iostreams, and it gives you all the goodness for easy formatting with printf/Posix formatting string, but without losing any of the power and convenience of iostreams themselves. For example, the first parts of your first two lines would look something like:
// using Boost.Format
cout << format("%-20s %-20s %-8s\n") % "Artist" % "Title" % "Price";
cout << format("%-20s %-20s %8.2f\n") % "Merle" % "Blue" % 12.99;
Another way to make column aligned is as follows:
using namespace std;
cout.width(20); cout << left << "Artist";
cout.width(20); cout << left << "Title";
cout.width(10); cout << left << "Price";
...
cout.width(20); cout << left << artist;
cout.width(20); cout << left << title;
cout.width(10); cout << left << price;
We should estimate maximum length of values for each column. In this case, values of "Artist" column should not exceed 20 characters and so on.
IO manipulators are what you need. setw, in particular. Here's an example from the reference page:
// setw example
#include <iostream>
#include <iomanip>
using namespace std;
int main () {
cout << setw (10);
cout << 77 << endl;
return 0;
}
Justifying the field to the left and right is done with the left and right manipulators.
Also take a look at setfill. Here's a more complete tutorial on formatting C++ output with io manipulators.
See also: Which C I/O library should be used in C++ code?
struct Item
{
std::string artist;
std::string c;
integer price; // in cents (as floating point is not acurate)
std::string Genre;
integer disc;
integer sale;
integer tax;
};
std::cout << "Sales Report for September 15, 2010\n"
<< "Artist Title Price Genre Disc Sale Tax Cash\n";
FOREACH(Item loop,data)
{
fprintf(stdout,"%8s%8s%8.2f%7s%1s%8.2f%8.2f\n",
, loop.artist
, loop.title
, loop.price / 100.0
, loop.Genre
, loop.disc , "%"
, loop.sale / 100.0
, loop.tax / 100.0);
// or
std::cout << std::setw(8) << loop.artist
<< std::setw(8) << loop.title
<< std::setw(8) << fixed << setprecision(2) << loop.price / 100.0
<< std::setw(8) << loop.Genre
<< std::setw(7) << loop.disc << std::setw(1) << "%"
<< std::setw(8) << fixed << setprecision(2) << loop.sale / 100.0
<< std::setw(8) << fixed << setprecision(2) << loop.tax / 100.0
<< "\n";
// or
std::cout << boost::format("%8s%8s%8.2f%7s%1s%8.2f%8.2f\n")
% loop.artist
% loop.title
% loop.price / 100.0
% loop.Genre
% loop.disc % "%"
% loop.sale / 100.0
% loop.tax / 100.0;
}
At the time you emit the very first line,
Artist Title Price Genre Disc Sale Tax Cash
to achieve "alignment", you have to know "in advance" how wide each column will need to be (otherwise, alignment is impossible). Once you do know the needed width for each column (there are several possible ways to achieve that depending on where your data's coming from), then the setw function mentioned in the other answer will help, or (more brutally;-) you could emit carefully computed number of extra spaces (clunky, to be sure), etc. I don't recommend tabs anyway as you have no real control on how the final output device will render those, in general.
Back to the core issue, if you have each column's value in a vector<T> of some sort, for example, you can do a first formatting pass to determine the maximum width of the column, for example (be sure to take into account the width of the header for the column, too, of course).
If your rows are coming "one by one", and alignment is crucial, you'll have to cache or buffer the rows as they come in (in memory if they fit, otherwise on a disk file that you'll later "rewind" and re-read from the start), taking care to keep updated the vector of "maximum widths of each column" as the rows do come. You can't output anything (not even the headers!), if keeping alignment is crucial, until you've seen the very last row (unless you somehow magically have previous knowledge of the columns' widths, of course;-).
C++20 std::format options <, ^ and >
According to https://en.cppreference.com/w/cpp/utility/format/formatter#Standard_format_specification the following should hold:
#include <format>
// left: "42 "
std::cout << std::format("{:<6}", 42);
// right: " 42"
std::cout << std::format("{:>6}", 42);
// center: " 42 "
std::cout << std::format("{:^6}", 42);
The existing fmt library implements it for before it gets official support: https://github.com/fmtlib/fmt Install on Ubuntu 22.04:
sudo apt install libfmt-dev
Modify source to replace:
<format> with <fmt/core.h>
std::format to fmt::format
main.cpp
#include <iostream>
#include <fmt/core.h>
int main() {
std::cout << "123456.\n";
// left
std::cout << fmt::format("{:<6}.\n", 42);
// right
std::cout << fmt::format("{:>6}.\n", 42);
// center
std::cout << fmt::format("{:^6}.\n", 42);
}
and compile and run with:
g++ -std=c++11 -o main.out main.cpp -lfmt
./main.out
Output:
123456.
42 .
42.
42 .
More information at: std::string formatting like sprintf
The easiest way, without the use of "complex" functions, is to format by hand:
an example:
std::cout << std::setprecision(10)
<< "integral of sin(x) from x = 0 to x = pi" << '\n'
<< "approximate: " << integrate(a, b, nbins) << '\n'
<< "exact: " << 2.0 << '\n';