C语言运算符 [英] C Language Operators
问题描述
#include <stdio.h>
int main()
{
int a=-1?2:5 + 8?4:5;
printf("%d\n",a);
return 0;
}
以上程序的输出为2.为什么?请解释
The output of above program is 2. But why ? Please explain
推荐答案
编写易于理解的代码. ( Atleast,尝试... )
Write human-readable and understandable code. (Atleast, try to...)
int a=-1?2:5 + 8?4:5;
与
int a = (-1) ? 2 : ( 5 + ( 8 ? 4 : 5) );
参考:操作员优先
现在,让我们将其与三元运算符条件进行比较,如C11
第6.5.15章
Now, let's compare that with the ternary operator condition, as mentioned in C11
, chapter §6.5.15,
第一个操作数被求值;它的评估与 第二个或第三个操作数的评估(以所评估的为准).第二个操作数 仅当第一个比较值不等于0时才进行评估;仅在以下情况下才对第三个操作数求值 第一个比较等于0;结果是第二个或第三个操作数的值 (以评估为准)
The first operand is evaluated; there is a sequence point between its evaluation and the evaluation of the second or third operand (whichever is evaluated). The second operand is evaluated only if the first compares unequal to 0; the third operand is evaluated only if the first compares equal to 0; the result is the value of the second or third operand (whichever is evaluated),
所以,就您而言,
- 第一个操作数不等于零
- 因此,它计算第二个操作数,并返回结果(该操作数的值)并将其存储到赋值运算符的LHS变量中.
这篇关于C语言运算符的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!