本文介绍了比较运算符的类似数学的链接 - 例如,“if((< j< = 1))"的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
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,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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