为什么size_type在64位平台上是32位整数？

Question

The std::vector::resize member function is like this:

void resize (size_type n);

Per my understanding, size_type should be a 64-bit long type on a 64-bit platform. But compiling the following program:

#include <iostream>
#include <climits>
#include <vector>

using namespace std;

vector<char> v;

int main() {
    // your code goes here
    v.resize(INT_MAX +1);
    for (auto i = 0; i < v.size(); i++ ) {
        cout << i << endl;
    }
    return 0;
}

The following warning is generated:

g++ -std=c++11 hello.cpp
hello.cpp: In function ‘int main()’:
hello.cpp:11:19: warning: integer overflow in expression [-Woverflow]
  v.resize(INT_MAX +1);

So the size_type is still 32-bit int even though we're working on a 64-bit platform?

Answer 1

The vector's size_type is probably a typedef for allocator::size_t which is (probably) a typedef for std::size_t which is an unsigned type. The generated warning has nothing to do with the vector's resize() signature. You are overflowing the max integer limit and your INT_MAX + 1 expression invokes undefined behaviour . Also the for loop deduces the type of i to be int which will also issue a warning when comparing signed and unsigned values. If you really want to you can cast to size_type and add 1 :

v.resize(static_cast<std::vector<char>::size_type>(INT_MAX) + 1);

and append the u literal to initial value inside the for loop:

for (auto i = 0u; i < v.size(); i++)

You can get the underlying type name with:

std::cout << typeid(std::vector<char>::size_type).name();