基于短路逻辑操作条件执行 [英] Conditional execution based on short-circuit logical operation

问题描述

由于逻辑运算符的计算＆放大器;＆安培; 和 || 被定义为短路，我假设code以下两件是等价的：

As the evaluation of logical operators && and || are defined as "short circuit", I am assuming the following two pieces of code are equivalent:

p = c || do_something();

和

if (c) {
   p = true;
}
else {
   p = do_something();
}

给 P 和 C 是布尔，和 do_something（）是返回一个函数布尔和可能有副作用。根据C标准，人们可以依靠假设的片段是相同的？特别是具有第一片段，难道是承诺，如果 C 为真，功能将不被执行，它没有副作用会发生？

given p and c are bool, and do_something() is a function returning bool and possibly having side effects. According to the C standard, can one rely on the assumption the snippets are equivalent? In particular, having the first snippet, is it promised that if c is true, the function won't be executed, and no side effects of it will take place?

推荐答案

一些搜索后，我会回答我的问题我自己参考的标准：
该 C99标准，部分 6.5 0.14逻辑或运营商的说明：

After some search I will answer my question myself referencing the standard: The C99 standard, section 6.5.14 Logical OR operator is stating:

不同的是按位|操作者，在||运营商保证
  左到右的评价;有一个序列点之后的
  第一个操作数的评价。如果第一个操作数比较不等
  为0，第二个操作数不评估

Unlike the bitwise | operator, the || operator guarantees left-to-right evaluation; there is a sequence point after the evaluation of the first operand. If the first operand compares unequal to 0, the second operand is not evaluated.

和类似的部分约＆放大器;＆安培; 。
因此，答案是肯定的，code可以安全地认为是等效的。

And a similar section about &&. So the answer is yes, the code can be safely considered equivalent.

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