增量递减运算符的奥秘？ [英] Increment decrement operator mystery?

问题描述

可能重复：结果
  <一href=\"http://stackoverflow.com/questions/949433/could-anyone-explain-these-undefined-behaviors-i-i-i-i-i-etc\">Could谁能解释这些不确定的行为，（I = I + + + + I，I = I ++等＆hellip;）结果
  未定义行为和顺序点

不管我多么努力理解他们的行为，我永远不会成功。今天我遇到了另一个例子。

No matter how much I try to understand their behavior, I never succeed. I encountered another example today.

void fun(int x, int y, int z){
    cout<<x<<endl<<y<<endl<<z;   
}

int main(){
    int x=10;
    fun(++x, x++, ++x);
}

这我$ P $的输出pdicted是11，11，13，但是，其13，11，11，任何人都可以解释如何？更重要的是，有一个傻瓜证明的方式来理解这些运营商。就像，重写以其他方式除权pression。如何编译器实际上codeS它们将也有帮助。

The output of this I predicted to be 11, 11, 13. But, its 13, 11, 11. Can anyone explain how? More importantly, is there a fool proof way to understand these operators. Like, rewriting the expression in some other way. How compiler actually codes them will also help.

推荐答案

有没有什么都懂，至少在C ++的行为是，因为它的未定义。什么事情都可能发生。这里有两个问题：

There is nothing to understand about the behaviour in C++ at least, since it is undefined. Anything could happen. There are two issues here:

1. 函数参数的评估顺序为未指定，这意味着排序可以是任何东西，不能依靠。


2. 您正在修改和读取变量 X 多次没有任何干预序列点。这导致的未定义行为

1. The order of evaluation of function arguments is unspecified, meaning the ordering could be anything and cannot be relied on.
2. You are modifying and reading variable x multiple times without any intervening sequence points. This leads to undefined behaviour.

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