I know I can limit the size of a particular string argument, but IDK how to do it for the entire output.
In other words can this program be written with only one call to format?
#include <string>
#include <iostream>
#include <fmt/format.h>
int LOG_MAX_MESSAGE_LENGTH = 11;
void f(const char* p) {
std::cout << fmt::format("{:.{}}", fmt::format("ABI is {}", p),
LOG_MAX_MESSAGE_LENGTH)
<< std::endl;
}
int main() {
f("hellooooo");
}
There's also format_to_n, which takes an output iterator and a size (and then the usual format string and arguments).
format_to_n(
std::ostream_iterator(std::cout), LOG_MAX_MESSAGE_LENGTH,
"ABI is {}", p);
Related
I wanted a simple function that takes a collection of strings and appends them and returns one string which is each string item appended together. I thought std::accumulate was the way to go but interested to hear if this code could be improved.
Is there a simpler append type function that can be used here instead of the lambda?
Is this overly complicated and best achieved by some other code?
#include <string>
#include <vector>
#include <iostream>
#include <numeric>
std::string concatenate(std::vector<std::string> strings)
{
return std::accumulate(strings.begin(), strings.end(), std::string(""), [](std::string s1, std::string s2) { return s1 + s2; });
}
int main() {
std::vector<std::string> vec2{ "aaa","bbb","ccc","ddd","eee","fff" };
std::cout << concatenate(vec2) << std::endl;
}
Yes, you can omit the lambda entirely (or use std::plus<>{}).
Also the "" can be removed from std::string(""), or the whole third argument can be removed if you switch to std::reduce:
std::reduce(strings.begin(), strings.end());
Also concatenate should take the vector by a const reference, or even better a std::span<const std::string> (by value).
Also libfmt can be used for this:
fmt::format("{}", fmt::join(strings, ""));
It can be simplified with C++17 fold expression, there is no an intermediate or a temporary collection needed.
#include <string>
#include <iostream>
template<typename ...S>
std::string concatenate(S&&... strings) {
using namespace std::string_literals;
return (""s + ... + strings);
}
int main() {
std::cout << concatenate("aaa","bbb","ccc","ddd","eee","fff") << std::endl;
}
I made a array of words and made a function to return a random word from the array. But it shows this error -
hangman.cpp: In function 'std::__cxx11::string get_random_word(std::__cxx11::string*)':
hangman.cpp:17:33: warning: 'sizeof' on array function parameter 'words' will return size of 'std::__cxx11::string* {aka std::__cxx11::basic_string<char>*}' [-Wsizeof-array-argument]
size_t length = sizeof(words) / sizeof(words[0]);
^
hangman.cpp:15:47: note: declared here
std::string get_random_word(std::string words[])
^
Here is the code -
#include <iostream>
#include <string>
#include <ctime>
std::string get_random_word(std::string words[]);
int main()
{
srand(time(0));
std::string words[] = {"cpp", "python", "java"};
std::cout << get_random_word(words);
return 0;
}
std::string get_random_word(std::string words[])
{
size_t length = sizeof(words) / sizeof(words[0]);
return words[rand() % length];
}
The sizeof operator may not be doing exactly what you think. According to cppreference: (sizeof) Yields the size in bytes of the object representation of type. This may include any internal members needed for the class, and not just how many characters are used in the string for example. std::string has the size() and length() functions for this, which are the same, and instead of using an array you can use a vector that also provides a size() function.
#include <iostream>
#include <string>
#include <ctime>
#include <vector>
std::string get_random_word(std::vector<std::string>& words)
{
return words[rand() % words.size()];
}
int main()
{
srand(time(0));
std::vector<std::string> words = {"cpp", "python", "java"};
std::cout << get_random_word(words);
return 0;
}
I am writing a code to insert an integer at an index of the string, but after providing the integer to add as string, insert function is not giving the correct output.
It is giving the error that :
no matching member function to call for insert string
This is my code:
#include <iostream>
using namespace std;
int main()
{
string s = "45564528";
int x = 8;
s.insert(s.begin()+5,to_string(x));
cout<<s<<endl;
return 0;
}
The expected output is 455648528.
Looking at the documentation for std::string::insert() shows that it takes a char or an iterator range, not a std::string, which std::to_string() naturally returns. At least, this is the case for the overloads that take an iterator for the first argument.
#include <iostream>
#include <string> // CHANGED: Include what you use
// using namespace std; // CHANGED: Bad practice
int main()
{
std::string s = "45564528";
int x = 8;
// CHANGED: Create string from the int, and use the iterator range overload
// to account for multi-digit numbers
auto tmp = std::to_string(x);
s.insert(s.begin()+5, tmp.begin(), tmp.end());
std::cout << s << '\n'; // CHANGED: std::endl is rarely actually needed
return 0;
}
There is an overload that lets you insert another std::string, but the first argument must be an index and not an iterator. So this would work as well:
#include <iostream>
#include <string>
int main()
{
std::string s = "45564528";
int x = 8;
s.insert(5, std::to_string(x));
std::cout << s << '\n';
return 0;
}
I'm trying check if a string representation equals given integer. I'm meant to use stringstream for this in a function. I also have an operator= for this as well.
I'm a little confused on how to execute these together and if I'm missing something. This is the last bit to an assignment I have, this is just a small snippet of my whole program. I can't find many guides on this, and I sense they all direct me to atoi or atod, which I'm not allowed to use.
#ifndef INTEGER
#define INTEGER
using std::string;
class Integer
{
private:
int intOne;
string strOne;
public:
Integer() {
intOne = 0;
}
Integer(int y) {
intOne = y;
}
Integer(string x) {
strOne = x;
}
void equals(string a);
Integer &operator=(const string*);
string toString();
};
#endif
In this header I'm not sure what argument I'm to use for the = operator.
#include <iostream>
#include <sstream>
#include <string>
#include "Integer.h"
using namespace std;
Integer &Integer::operator=(const string*)
{
this->equals(strOne);
return *this;
}
void Integer::equals(string a)
{
strOne = a;
toString(strOne);
}
string Integer::toString()
{
stringstream ss;
ss << intOne;
return ss.str();
}
#include <iostream>
#include <cstdlib>
#include <conio.h>
#include <string>
#include <ostream>
using namespace std;
#include "Menu.h"
#include "Integer.h"
#include "Double.h"
int main()
{
Integer i1;
i1.equals("33");
cout << i1;
}
Sorry if its a bad question I'm not too familiar with this type of assignment and will take any help I can get. Thanks.
you can use std::to_strig() that lets you convert from int to a string that represents the same number.
So if i understand correctly, you want to overload operator =, and that is a bad idea, since operator= is used for assignment not for comparison.
The correct operator signature is:
ReturnType operator==(const TypeOne first, const TypeSecond second) [const] // if outside of class
ReturnType operator==(const TypeSecond second) [const] // if inside class
Since you can't compare string to integer (they are different types), you need to write your comparisment function, since you don't have one i will write one for you:
bool is_int_equal_string(std::string str, int i)
{
std::string tmp;
tmp << i;
return tmp.str() == i;
}
Last but not least, you need to merge both of those, into one convenient operator:
// inside your Integer class
bool operator==(std::string value) const
{
std::stringstream tmp;
tmp << intOne;
return tmp.str() == ref;
}
Now you can use this operator, just like any other:
Integer foo = 31;
if (foo == "31")
cout << "Is equal" << endl;
else
cout << "Is NOT equal" << endl;
I hope this helps.
If you are allowed to use std::to_string then it would be the best.
Otherwise, you could create a function to handle the equality between the string and the integer with the use of std::stringstream:
Example:
bool Integer::equal(const string& str)
{
stringstream ss(str);
int str_to_int = 0;
ss >> str_to_int;
if (intOne == str_to_int)
return true;
else
return false;
}
Combine this with an if statement:
int main()
{
Integer i{100};
if (i.equal("100"))
cout << "true" << endl;
else
cout << "false" << endl;
}
Like this function in C:
size_t fwrite ( const void * ptr, size_t size, size_t count, FILE * stream );
I've looked in C++ file stream and found this one:
ostream& write ( const char* s , streamsize n );
this one only accepts char* instead of void*
but does it really matter if I use a C-style fwrite function in c++?
Streams are probably what you're looking for unless I misunderstand your question. There are many flavors that handle different jobs, like outputting to a file:
#include <cstdlib>
#include <fstream>
#include <string>
using namespace std;
int main()
{
ofstream f("c:\\out.txt");
const char foo[] = "foo";
string bar = "bar";
int answer = 42;
f << foo << bar<< answer;
return 0;
}
...building strings like you would with printf:
#include <cstdlib>
#include <sstream>
#include <string>
#include <iostream>
using namespace std;
int main()
{
stringstream ss;
const char foo[] = "foo";
string bar = "bar";
int answer = 42;
ss << foo << bar<< answer;
string my_out = ss.str();
return 0;
}
...and they can even handle your own types, if you tell them how:
#include <cstdlib>
#include <string>
#include <iostream>
using namespace std;
class MyGizmo
{
public:
string bar_;
int answer_;
MyGizmo() : bar_("my_bar"), answer_(43) {};
};
ostream& operator<<(ostream& os, const MyGizmo& g)
{
os << g.bar_ << " = " << g.answer_;
return os;
}
int main()
{
MyGizmo gizmo;
cout << gizmo;
return 0;
}
You can use either one. Using char * instead of void * doesn't make much real difference -- both fwrite and ostream::write are typically used for a variety of data types (with with C++, you need to add an explicit cast to char *, where in C the cast will happen implicitly, assuming you've included a proper prototype for fwrite).
In C++ you will want to use std::ofstream objects to write to a file. They can accept any type of data using the << operator, in much the same way that std::cout works for writing to the console. Of course, just like std::cout, if you want to print a custom type, you will need to define an operator<< overload for it.
An example:
std::ofstream outfile("myfile.txt");
int i = 5;
double d = 3.1415926535898;
std::string s = "Hello, World!";
outfile << i << std::endl;
outfile << d << std::endl;
outfile << s << std::endl;
To use std::ofstream, you need to #include <fstream>.
The outfile object will automatically close the file when it destructs, or you can call its close() method.
Contrary to already given answers, there is an important difference between fwrite() and ostream::write().
fwrite() writes binary data unmodified (well, on those poor non-Unix platforms there is endline translation, unless the file is opened in binary mode).
ostream::write() uses locale to transform every character, this is why it accepts char* rather than void*. Normally, it uses the default "C" locale, which does not do any transformation.
Just keep in mind that basic_ostream is a formatter on top of basic_streambuf, not a binary sink.