对包含前缀和后缀运算符的 printf() 感到困惑 [英] confused about printf() that contains prefix and postfix operators

问题描述

如果 int var=20 那么如何

If int var=20 then how

printf("%d %d %d", var--, ++var, --var); 

执行发生在 C 编程语言中.

execution happens in C programming language.

推荐答案

这是未定义的行为，因为 var 被多次修改，中间没有序列点.例如，序列点可以是 ;.然而，参数列表中的逗号并没有引入序列点，操作数的计算顺序也是未定义的(你可以说，代码是双重未定义的......).

It is undefined behaviour because var is modified several times without a sequence point in between. A sequence point would be, for example, a ;. The commas in parameter lists, do, however, not introduce sequence points, also the order in which the operands are evaluated is undefined (you could say, the code is doubly undefined ...).

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