运算符<<中的执行顺序 [英] Order of execution in operator <<

问题描述

我很难理解下面代码中的调用顺序. 我期待看到下面的输出

I have difficulties in understanding the sequence of calls in the code below. I was expecting to see the output below

    A1B2

虽然我看到的输出是

    BA12

我认为通话std::cout<< b->fooA() << b->fooB() << std::endl等同于通话

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

但是我可以看到情况并非如此.您能帮助我更好地了解它是如何工作的以及与全局operator<<的关系吗?这是此顺序中最后一次被调用吗?

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

编译器可以按以下方式评估函数printcbase():

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的行的许多置换中的一些.仅保证行1在行3之前完成，行2在行4之前完成(当然行3在行4之前完成). Standard并没有多说什么，实际上您可以期望使用不同的C ++编译器得到不同的结果.

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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