在操作符中的执行顺序< [英] order of execution in operator<<

问题描述

我在下面的代码中理解调用顺序有困难。
我期待看到下面的输出

  A1B2 
  pre> 
 
 
虽然我可以看到我得到的输出是
  BA12 
  
我以为呼叫 std :: cout< b  - > fooA（）<< b  - > fooB（）<< std :: endl 等同于调用
  std :: cout.operator<< - > fooA（））.operator<< （b-> fooB（））
  
但我可以看到情况并非如此。你能帮助我更好地理解它是如何工作的，以及与全局运算符<< 的关系？ 
 
 
 
注意
 
 
 
 AFAG 
  #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）; 
} 
  
 
 
解决方案
编译器可以 printcbase（）如下：
  void printcbase（cbase * b） {
 int a = b-> FooA（）; // line 1 
 int b = b-> FooB（）; // line 2 
 std :: cout<<一个; // line 3 
 std :: cout<< b; // line 4 
 stc :: cout<< std :: endl; 
} 
  
或许多标记为1  -  4的行中的一些保证线1在线3之前完成，线2在线4之前（当然线3之前在线4之前）。标准不说更多，确实可以预期不同的C ++编译器的不同结果。
 
I have difficulties in understanding the sequence of calls in the code below.
I was expecting to see the output below
    A1B2
While I can see that the output I get is
    BA12
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() )
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?

Regards

AFAG
    #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 );
    }

解决方案
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;
}
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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