在单个语句中的多个复合赋值：是未定义的行为还是不行？ [英] Multiple compound assignments in a single statement: is it Undefined Behavior or not?

问题描述

我找不到一个明确的答案：以下代码是否有未定义的行为？

  int x = 2 ; 
 x + = x + = x + = 2.5; 
  

解决方案

行为未定义。让我们看看稍微简单的表达式：

  x + =（x + = 1）
  

在C ++ 11中，左侧 x 的值计算表达式（x + = 1）的值计算。这意味着 x 的值计算与 x （由于这是因为 = x <= x <= x <= x <= x <= x <= code> + = 运算符不相对于彼此排序（因为标准没有另外指定）。和1.9p15状态：

如果对标量对象的副作用相对于对同一标量对象的另一个副作用

在C ++ 03中，行为是未定义的，因为 x 被修改两次而没有中间序列点。

I can't find a definitive answer for this: does the following code have undefined behavior?

int x = 2;
x+=x+=x+=2.5;

解决方案

The behavior is undefined. Let's look at the slightly simpler expression:

x += (x+=1)

In C++11, the value computation of the left x is unsequenced relative to the value computation of the expression (x+=1). This means that value computation of x is unsequenced relative to the assignment to x (due to x+=1), and therefore the behavior is undefined.

The reason for this is that the value computation of the two sides of the += operator are unsequenced relative to each other (as the standard doesn't specify otherwise). And 1.9p15 states:

If a side effect on a scalar object is unsequenced relative to either another side effect on the same scalar object or a value computation using the value of the same scalar object, the behavior is undefined.

In C++03 the behavior is undefined because x is modified twice without an intervening sequence point.

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