c - 为什么 "if (i++ && (i == 1))"是假的，而我是一个值为 1 的整数？

Question

{
    int i = 1;
    if (i++ && (i == 1))
        printf("Yes\n");
    else
        printf("No\n");
}

根据我的理解，在 if 条件下，首先表达式 (i==1) 将被评估，它应该返回 1 , 然后它与 1 逻辑与，这是 i 的值，所以表达式应该返回 1 && 1 == 1，但是else 部分被执行。

有人可以解释一下为什么执行 else 部分吗？

Answer 1

没有。在 C 中，有一个 sequence point &&的LHS评估之间运算符和 RHS 的评估，增量必须在评估 RHS 之前发生并完成。所以，i++ (相当于 i++ != 0 )被执行并完成增量(并且表达式的计算结果为真)，因此在计算 RHS 时，i == 2因此整体表达式是错误的，你得到“否”打印。如果 && 的 LHS运算符评估为 false (0)，则不会评估 RHS，因为 && 的“短路”属性运营商。

只有少数运算符具有在 LHS 和 RHS 的评估之间具有序列点的属性:&& , || , 和 , (作为运算符，而不是作为参数列表中的分隔符)— 还有 ? :它也不是二元运算符，但在条件求值之后和 ? 之后的表达式之前有一个序列点或 : 之后的表达式被评估(其中一个或另一个，但不是两者，总是被评估)。

&&和 ||运算符是唯一具有“短路”属性的运算符。 && 的 RHS仅在 LHS 评估为真时才评估； || 的 RHS仅在 LHS 评估为 false 时评估。

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序列点说明

Iwillnotexist Idonotexist正确asserted :

The C11 standard hasn't done away with sequence points, only the C++11 standard did.

C++11(ISO/IEC 14882:2011)说:

1.9 Program execution

¶13 Sequenced before is an asymmetric, transitive, pair-wise relation between evaluations executed by a single thread (1.10), which induces a partial order among those evaluations. Given any two evaluations A and B, if A is sequenced before B, then the execution of A shall precede the execution of B. If A is not sequenced before B and B is not sequenced before A, then A and B are unsequenced. [Note: The execution of unsequenced evaluations can overlap. —end note] Evaluations A and B are indeterminately sequenced when either A is sequenced before B or B is sequenced before A, but it is unspecified which. [Note: Indeterminately sequenced evaluations cannot overlap, but either could be executed first. —end note]

术语“序列点”根本没有出现在 C++11 中(唯一接近匹配的是“序列指针”)。

C11(ISO/IEC 9899:2011)说:

5.1.2.3 Program execution

¶3 Sequenced before is an asymmetric, transitive, pair-wise relation between evaluations executed by a single thread, which induces a partial order among those evaluations. Given any two evaluations A and B, if A is sequenced before B, then the execution of A shall precede the execution of B. (Conversely, if A is sequenced before B, then B is sequenced after A.) If A is not sequenced before or after B, then A and B are unsequenced. Evaluations A and B are indeterminately sequenced when A is sequenced either before or after B, but it is unspecified which.¹³⁾ The presence of a sequence point between the evaluation of expressions A and B implies that every value computation and side effect associated with A is sequenced before every value computation and side effect associated with B. (A summary of the sequence points is given in annex C.)

¹³⁾ The executions of unsequenced evaluations can interleave. Indeterminately sequenced evaluations cannot interleave, but can be executed in any order.

因此，C11 确实保留了序列点，但使用与 C++11 基本相同的术语添加了“之前排序”和相关术语。