为什么true==151被认为是不安全的操作，而true==true ，它实际上持有相同的表达，不是。 此外，为什么true==151永远不会是真的，不像true&&151 ？
cout<<(true==151); //0 and it gives compiler warning C4806 cout<<(true==true); //1 no warning if(true==151)cout<<"1"; //"0" and warning again if(true&&151)cout<<"1"; //1 no warning if(151)cout<<"1"; //1警告C4806本身
'=='：不安全的操作：升级为类型'int'的类型'bool'的值不能等于给定的常量
Why is true==151 regarded as unsafe operation while true==true, which is virtually holding the same expression, isn't. Moreover, why can true==151 never be true, unlike true&&151?
cout<<(true==151); //0 and it gives compiler warning C4806 cout<<(true==true); //1 no warning if(true==151)cout<<"1"; //"0" and warning again if(true&&151)cout<<"1"; //1 no warning if(151)cout<<"1"; //1The warning C4806 itself
'==': unsafe operation: no value of type 'bool' promoted to type 'int' can equal the given constant
最满意答案
首先让我们回答第二个问题： true==151总是为false，因为true定义为值1 。
一般情况下，如果你将一个数值解释为一个布尔值，那么0将是错误的，其他的将是真的; 但字面上的true必须具有特定的值，并且该值为1。
那么为什么要警告关于操作数是布尔型和整型的== ？ 那么，首先，因为151==true ，如果你注意它，看起来应该是真的（因为151是一个“真实”值），但是 - 如上所述 - 事实并非如此。 这可能是错误的根源，所以值得警告。 更一般地说，你直接比较两种不同的类型，其中隐式转换可能不明显/可能具有非显而易见的后果。
First let's answer your second question: true==151 is always false because true is defined to have the value 1.
In general if you interpret a numeric value as a boolean, 0 will be false and anything else will be true; but the literal true has to have a specific value, and that value is 1.
So why warn about an == where the operands are a bool and an integer? Well, first of all because 151==true, if you eyeball it, looks like it ought to be true (because 151 is a "true" value) but - as noted above - it's not. It's a likely source of error, so that merits a warning. More generally you're directly comparing two different types where the implicit conversion may not be obvious / may have non-obvious consequences.
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