Printing integers with a fixed number of zeros in C++

Question

The question is simple. I have messed up with std::fixed , std::setprecicion , std::setw , but none of them solves my problem.

If I have a integer variable equal to 1, I want it to be printed as: 1000

Thank you.

Answer 1

I think you mean you want it to print as 0001, not as 1000.

Look at printf / sprintf with a format string. For example:

#include <iostream>

int main(int argc, char **argv) {
    int i = 1;

    printf("%04d\n", i);
}

Answer 2

The question is not very clear on what you are trying to achieve.

If you want to print the value of your variable with 3 zeroes appended to it, you should just print the variable and then 3 zeroes.

cout << x << "000" << endl;

If you need to do any computation with it, you could always just multiply it by 1000 .

int y = x * 1000;
cout << y << endl;

Keep in mind that this may cause an overflow, though...

Answer 3

You can use std::left together with std::setw and std::setfill to add a number of specific fill characters on the right. Example with std::cout :

#include <iostream>
#include <iomanip>
#include <ios>

int main() {
    std::cout << std::setw(4) << std::setfill('0') << std::left << 1;
    return 0;
}

I believe it should do the job.