c++ equivalent of this code below (sprintf)

Question

What is the C++ equivalent of the code below?

sprintf(name,"jari%d.jpg",rank+42+size);

help me how to change it to c++. can anyone help me?

`std::stringstream` is a rough equivalent, but not an exact one since it lacks the danger of buffer overflows. If you want to simulate that in C++, you can just do `std::string("")[100] = 'a';`. — , Jan 21 '14 at 23:54
For something like this with just one argument, `std::to_string` works well also. — chris, Jan 21 '14 at 23:55
@chris, `to_string` is a fine choice if you're working in C++11 or later (which I usually am), but it's not available in C++03. Just a caveat. — Adam H. Peterson, Jan 22 '14 at 00:10
this is my code char name[100]; sprintf(name,"E:/tigaouts/Debug/jari2.jpg",rank+42+size); how to make it in c++ — Monic92, Jan 22 '14 at 00:24

score 4 · Accepted Answer · answered Jan 22 '14 at 00:02

4

I would suggest an ostringstream:

string name;
{
    ostringstream oss;
    oss << "jari" << (rank+42+size) << ".jpg";
    name = oss.str();
}

Details: To use this solution, you'll want #include <sstream> and pull ostringstream into scope with using std::ostringstream; (or just use std::ostringstream directly qualified).

answered Jan 22 '14 at 00:02

Adam H. Peterson

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how to change it if we use std::cout ? – Monic92 Jan 22 '14 at 00:05
2

@user3021930, I don't understand the question. But if you want to print the name out once you've constructed it, you can just write `cout << name << "\n";`. Alternatively, you could just stream it into `cout` rather than `oss` in the first place. – Adam H. Peterson Jan 22 '14 at 00:06

SirDarius · Answer 2 · 2014-01-22T08:58:27.907

The equivalent would be:

#include <sstream>

std::ostringstream s;
s << "jari" << rank + 42 + size << ".jpg";
std::string name = s.str();

Not exactly equivalent in terms of data types, since the final result is a std::string instead of a char*, but the most idiomatic.

Outputting the formated string directly is even simpler:

std::cout << "jari" << rank + 42 + size << ".jpg";

Alternatively, there is also the Boost Format library that offers similar functionality:

#include <boost/format.hpp>

boost::format formatter("jari%1%.jpg");
formatter % rank+42+size;
std::string name = formatter.str();

or to directly output the string:

std::cout << boost::format("jari%1%.jpg") % rank+42+size;

score 0 · Answer 3 · answered Jan 22 '14 at 00:10

0

or you can use boost format library

cout << boost::format("jari%1%.jpg") % (rank + 42 + size)

answered Jan 22 '14 at 00:10

pm100

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c++ equivalent of this code below (sprintf)

3 Answers3