运算符优先级和关联性 [英] Operator Precedence and associativity

问题描述

当一个表达式具有两个具有相同优先级的运算符时，该表达式将根据其关联性进行求值.我想知道以下工作原理:

When an expression has two operators with the same precedence, the expression is evaluated according to its associativity. I want to know how the following works:

i=b + b + ++b

i这是4 因此++b并没有更改前2个b值，而是先执行，因为执行是从左到右.

i here is 4 So ++b didn't change the first 2 b values, but it executed first, because the execution is from left to right.

但是，在这里:

int b=1;
i= b+ ++b + ++b ;

i是6

根据关联性，我们应该执行第三个b，因此它应该是: 1+ (++1) + ( ++1 should be done first).这样就变成了: 1 ++++ 1 + 2 = 5 但是，这是不对的，那么它如何工作?

According to associativity, we should execute the 3rd b so it should be: 1+ (++1) + ( ++1 should be done first). so it becomes: 1 + ++1 + 2 =5 However, this is not right, so how does this work?

推荐答案

您正在将优先级与执行顺序混淆.

示例:

a[b] += b += c * d + e * f * g

优先级规则规定*在+之前在+=之前.关联性规则(优先级规则的一部分)规定*是左关联的，而+=是右关联的.

Precedence rules state that * comes before + comes before +=. Associativity rules (which are part of precedence rules) state that * is left-associative and += is right-associative.

优先/关联规则基本上定义了隐式括号的应用，将上述表达式转换为:

Precedence/associativity rules basically define the application of implicit parenthesis, converting the above expression into:

a[b] += ( b += ( (c * d) + ((e * f) * g) ) )

但是，此表达式仍然是从左到右求值的.

However, this expression is still evaluated left-to-right.

这意味着表达式a[b]中b的索引值将使用执行b += ...之前的b值.

This means that the index value of b in the expression a[b] will use the value of b from before the b += ... is executed.

对于更复杂的示例，混合使用++和+=运算符，请参见问题 Incrementor logic ，以及有关其工作原理的详细答案.

For a more complicated example, mixing ++ and += operators, see the question Incrementor logic, and the detailed answer of how it works.

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