比较操作符的数学般的链接 - 同在，与QUOT;若（1 LT; A＆LT; = 5）QUOT; [英] Math-like chaining of the comparison operator - as in, "if (1<A<=5)"

问题描述

int j=42;
if(1<j<=5) {
    printf("yes");
} else {
    printf("no");
}

输出：

yes

为什么它的输出是吗？结果
是不是条件只有一半是真的吗？

Why does it output yes?
Isn't the condition only half true?

推荐答案

C不理解数学的语法，所以

C does not understand math-like syntax, so

if(1<j<=5)

如您所愿，想不跨preTED;它应该是

is not interpreted as you expect and want; it should be

if (1 < j && j <= 5)

或相似的。

在其他的答案解释，前pression被评为

As explained in other answers, the expression is evaluated as

 ((1 < j) <= 5)

 =>  ("true" <= 5)

 =>  "true"

其中，真（布尔值）隐式转换为1，explaneid例如这里，具有标准也引用了，这解释了为什么真必须少比5（虽然用C可能没有完全正确的谈论，从布尔隐式转换为int）

where "true" (boolean value) is implicitly converted to 1, as explaneid e.g. here, with references to standards too, and this explain why "true" has to be "less than" 5 (though in C might not be totally correct to speak about "implicit conversion from bool to int")

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