比较运算符的类似数学的链接 - 例如，“if（（< j< = 1））" [英] Math-like chaining of the comparison operator - as in, "if ( (5<j<=1) )"

问题描述

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

输出：

yes

为什么输出是？

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不懂数学式语法，所以

C does not understand math-like syntax, so

if(1<j<=5)

未被解释为您期望和想要的;它应该是

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

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

或类似。

如其他答案中所述，表达式已被评估as

As explained in other answers, the expression is evaluated as

 ((1 < j) <= 5)

 =>  ("true" <= 5)

 =>  "true"

其中true（布尔值）被隐式转换为1，例如，例如此处，也参考了标准，这解释了为什么真实必须少比5（虽然在C中可能不完全正确地说从bool到int的隐式转换）

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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