按位'&'链比较中的运算符 [英] Bitwise '&' operator in chained comparison

问题描述

var = 86
print((var < 90) & (var >= 80))

打印 True 。

但是为什么要这样做所有这些打印 False ？

But why do all these print False?

print(var < 90 & var >= 80)
print(var < 90 & (var >= 80))
print((var < 90) & var >= 80)
print(var < 90 & True)

推荐答案

您应该使用和运算符代替布尔运算。

You should be using the and operator instead for boolean operations.

由于python支持链接关系运算符（即，您可以使用 0< var< 100 代替 0< var和var< 100 ），并在关系运算符之前将二元运算（即加，减，按位运算等）处理为

Since python supports chaining relational operators (i.e. you can use 0 < var < 100 instead of 0 < var and var < 100) and processes binary operations (i.e. addition, subtraction, bitwise operations, etc.) before relational operators due to operator precedence, all of the failing cases you posted actually mean something else.

1. var<优先级，您发布的所有失败案例实际上都具有其他含义。 90& var> = 80 等效于（var< 90& var）和（90& var> = 80）


2. var< 90& （var> = 80）等效于 var< 90&真，要看第四种情况。


3. （var< 90）& var> = 80 也类似于第4种情况（此解析为 True& var> = 80 然后将解析为 0> = 80 ）。


4. var< 90& True 等效于 var< （90&True）。 & 运算符的实现被设计为，如果其中一个操作数不是整数，则返回 0 是动态类型化的另一个最大陷阱），这就是为什么所有这些类似的语句都解析为 var< 0 这是错误的。

1. var < 90 & var >= 80 is equivalent to (var < 90 & var) and (90 & var >= 80)
2. var < 90 & (var>=80) is equivalent to var < 90 & True for which look at the 4th case.
3. (var<90) & var>=80 is also similar to the 4th case (this resolves to True & var >= 80 which will then resolve to 0 >= 80).
4. var < 90 & True is equivalent to var < (90 & True). The implementation of the & operator is designed to return 0 if either one of the operands is not an integer (which is another one of the biggest pitfalls of dynamic typing), which is why all such similar statements resolve to var < 0 which is false.

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