使用 operator[] 插入时 std::map 大小不同（vc++ vs g++）

Question

这个代码

#include <iostream>
#include <map>

int main()
{
    std::map<int, std::size_t> m;
    m[0] = m.size();
    std::cout << m[0] << std::endl;
}

将打印 0 和 vc++ and 1 with g++。

此代码有效吗？
如果是，哪个编译器是正确的？
直觉上我预计 1。 vc++ 如何以 0 结尾？

Answer 1

Assignment operators - cppreference.com

When the left operand has reference type, the assignment operator modifies the referred-to object.

If the left and the right operands identify overlapping objects, the behavior is undefined (unless the overlap is exact and the type is the same)

Answer 2

自 C++17 以来，order of evaluation 得到保证，m.size() 排在 m[0] 之前；结果保证为 0.

In every simple assignment expression E1=E2 and every compound assignment expression E1@=E2, every value computation and side-effect of E2 is sequenced before every value computation and side effect of E1

在 C++17 之前，行为是 unspecified。

顺便说一句，您可以使用 Gcc C++17 mode and Gcc C++14 mode.

观察不同的行为

使用 operator[] 插入时 std::map 大小不同（vc++ vs g++）

Different std::map size when inserting with operator[] (vc++ vs g++)

c++

gcc

stdmap

visual-c++

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