使用cygwin在64位Windows上C ++中的int指针大小

Question

I tried to print the size of int pointer on a 64 bit windows 7 machine. I am using cygwin. I expected it to give output 8, but the actual output is 4.

#include <iostream>
using namespace std;

int main()
{
    cout<< sizeof(int*)<<endl;
    return 0;
}

Possibly related question What the pointer size in 64 bits computer in C++?

But it is about compiling as 64 bit project in Visual Studio. I couldn't find any such option in cygwin. How is this explained?

Answer 1

The "bitness" of the machine or the OS does not matter. It makes no difference whatsoever.

The only thing that matters is what kind of code you asked your compiler to generate. In your experiment you either asked GCC to generate 32-bit code or it defaulted to 32-bit code by itself. This is why your pointers are 32-bit wide.

You have to explicitly ask GCC to generate 64-bit code, if you want to see 64-bit pointers. Specify -m64 in compiler's command line.