C++条件运算符的运算问题，求指教

Question

1. 问题：为什么输出的y是5，而不是6？

2. 相关代码

   #include <iostream>
   
   int main() {
       int someValue = 1, x = 1, y = 5;
       std::cout << (someValue ? ++x, ++y : --x, --y) << std::endl;
       
       return 0;
   }

3. 相关贴图（测试环境：Xcode 7.3.1 OSX EI Capitan）
   

4. 个人思考

   • 思考1：因为条件运算符?:是右结合，所以先执行--y,y变成4，再执行++y，y变成5。疑问：条件运算符的执行顺序是：先求解表达式1，若为非0(真)则求解表达式2，此时表达式2的值就作为整个条件表达式的值。那为什么还会出现这样的问题

   • 思考2：这句语句中的第二第三个运算对象指向同一个对象x，y；而且两个都修改了对象的值，这种行为是未定义的。疑问：但是？：运算符不是已经确定了运算顺序码？为什么还会出现这样的问题？

Answer 1

Foreword

Every time I meet people who ask questions about C++, they are all relatively basic questions like this. I always say this:
Disassembly is a skill that C++ programmers must master! ! !
Disassembly is a skill that C++ programmers must master! ! !
Disassembly is a skill that C++ programmers must master! ! !
Otherwise, you will be like a blind man when debugging a program, you can only rely on guessing.

Analysis

In fact, based on experience, I think it is far less complicated, and it is not an undefined behavior, but a problem of operator scope.
The correct execution order of this expression is as follows:

if(1){
    ++x;//这句没有用，在release模式下会被编译器优化掉
    ++y;//y+1
}else{
    --x;
}
--y;//y-1
cout<<y<<endl;

is obvious at a glance because the : operator has higher precedence than ,. The conditions before : are true and can only be executed sequentially, while the subsequent , exceeds the scope of the : operator. Therefore, whether the condition is true or not, --y will definitely be executed.
So the correct way to write it is to add 括号.

auto y=someValue ? (++x, ++y) : (--x, --y);
//y=6

Disassembly

00F0625E  mov         dword ptr [someValue],1  
00F06265  mov         dword ptr [x],1  
00F0626C  mov         dword ptr [y],5  
00F06273  cmp         dword ptr [someValue],0  
00F06277  je          main+56h (0F06296h)  ;判断条件是否成立
00F06279  mov         eax,dword ptr [x]  
00F0627C  add         eax,1  
00F0627F  mov         dword ptr [x],eax  
00F06282  mov         ecx,dword ptr [y]  
00F06285  add         ecx,1  ;y的值+1变成了6
00F06288  mov         dword ptr [y],ecx  
00F0628B  mov         edx,dword ptr [y]  
00F0628E  mov         dword ptr [ebp-0E8h],edx
00F062A8  mov         edx,dword ptr [y]  
00F062AB  sub         edx,1  ;再减一变成5
00F062AE  mov         dword ptr [y],edx ;传值返回   

Answer 2

Because the ?: operator has a higher priority than the , operator, the expression (someValue ? ++x, ++y : --x, --y) is equivalent to ((someValue ? (++x, ++y) : --x), --y) so the conditional operation ++y will be followed by --y, so y outputs 5

Answer 3

First of all, a correction, any order of operations is defined. Most of the undefined behavior is for the evaluation order rather than the operation order.

There are three evaluation orders defined in

c: A ? B : C, A, B, A && B (of course there is also ||). Your code above can be accused of being unreadable, but it cannot be said to be undefined.

The result is pretty clear, it’s 5.