运算符<<中的执行顺序

I thought that the call std::cout<< b->fooA() << b->fooB() << std::endl was equivalent to call

  std::cout.operator<<( b->fooA() ).operator<< ( b->fooB() )

但我可以看出事实并非如此.你能帮助我更好地理解它是如何工作的以及与全局 operator<< 的关系吗?这是此序列中的最后一次调用吗?

but I can see that this is not the case. Can you help me understanding better how this does it work and the relationship with the global operator<<? Is this last ever called in this sequence?

#include <iostream>

struct cbase{
    int fooA(){
        std::cout<<"A";
        return 1;
    }
    int fooB(){
        std::cout <<"B";
        return 2;
    }
};

void printcbase(cbase* b ){
    std::cout << b->fooA() << b->fooB() << std::endl;
}

int main(){
    cbase b;
    printcbase( &b );
}

推荐答案

编译器可以对函数 printcbase() 求值:

The compiler can evaluate the function printcbase() as this:

void printcbase(cbase* b ){
    int a = b->FooA();    // line 1
    int b = b->FooB();    // line 2
    std::cout << a;       // line 3
    std::cout << b;       // line 4
    stc::cout << std::endl;
}

或标记为 1 - 4 行的许多排列中的一些.您只能保证第 1 行在第 3 行之前完成，第 2 行在第 4 行之前完成(当然还有第 3 行在第 4 行之前).标准没有多说，确实可以预期不同的 C++ 编译器会产生不同的结果.

or some of many permutatins of lines marked as 1 - 4. You are only guaranteed that that the line 1 is done before the line 3, and line 2 before the line 4 (and of course line 3 before line 4). Standard does not say more and indeed you can expect different results with different C++ compilers.

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