英文:
Is the output of this C code compiler dependent?
问题
变量 a 的值被更新了两次，这违反了C标准。因此，我认为这将取决于编译器。我在许多在线C编译器中运行了这段代码，它们都给出了相同的输出。
英文:
#include <stdio.h>
int main(void)
{
int a = 0, b = 1;
a = (a = 5) && (b = 0);
printf("%d", a);
printf("%d", b);
}
The value of variable a is getting updated twice, which violates the C standards. So I think it will be compiler dependent. My I ran this code in lot of online C compilers and all are giving same output.
答案1
得分: 8
这段代码的行为是由&&运算符引入的序列点所定义的。在C标准的第6.5.13p4节中明确规定了逻辑AND运算符&&的行为：
与按位二进制
&运算符不同，&&运算符保证从左到右进行评估；**如果评估第二个操作数，则在评估第一个和第二个操作数之间存在一个序列点。**如果第一个操作数等于0，则不评估第二个操作数。
对于这段代码的分析：
a = (a = 5) && (b = 0);
首先评估&&运算符的左侧，即(a = 5)，它将a设置为5。这评估为true，因此在右侧，即(b = 0)被评估之前存在一个序列点。这将b设置为0，并评估为false，因此&&运算符的结果为0。然后将这个值赋给a。
英文:
The behavior of this code is well defined due to the sequence point introduced by the && operator. This is spelled out in section 6.5.13p4 of the C standard regarding the logical AND operator &&:
> Unlike the bitwise binary & operator, the && operator guarantees
> left-to-right evaluation; if the second operand is evaluated, there is
> a sequence point between the evaluations of the first and second
> operands. If the first operand compares equal to 0, the second operand
> is not evaluated.
Going through the code:
a = (a = 5) && (b = 0);
The left side of the && operator, i.e. (a = 5) is evaluated first, which sets a to 5. This evaluates to true, so there is then a sequence point before the right side, i.e. (b = 0) is evaluated. This sets b to 0 and evaluates to false, so the && operator results in the value 0. This value is then assigned to a.
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