Question

#include <vector>
#include <iostream>
#include <stdio.h>
using namespace std;
int main(int argc, const char *argv[])
{
    vector<bool> a;
    a.push_back(false);
    int t=a[0];
    printf("%d %d\n",a[0],t);
    return 0;
}

This code give output “5511088 1”. I thought it would be “0 0”.

Anyone know why is it?

Answer 1

The %d format specifier is for arguments the size of integers, therefore the printf function is expecting two arguments both the size of an int. However, you’re providing it with one argument that isn’t an int, but rather a special object returned by vector<bool> that is convertible to bool.

This is basically causing the printf function to treat random bytes from the stack as part of the values, while in fact they aren’t.

The solution is to cast the first argument to an int:

printf("%d %d\n", static_cast<int>(a[0]), t);

An even better solution would be to prefer streams over printf if at all possible, because unlike printf they are type-safe which makes it impossible for this kind of situation to happen:

cout << a[0] << " " << t << endl;

And if you’re looking for a type-safe alternative for printf-like formatting, consider using the Boost Format library.