可交换数学运算的顺序 [英] Order of commutative mathematical operations

问题描述

我遇到了一个奇怪的问题(在阅读粗略的代码时自己问).让我们看一下表达式:

I've into curious question (asking it myself while reading a crude piece of code). Let's look at expression:

double a =  c*d*e*2/3*f;

其中c，d，e，f是类型为double的初始化变量.标准是否保证将其视为c*d*e*2(双重结果)，然后除以3，再乘以f(或某些类似行为).显然，不希望将2/3计算为0.

where c, d, e, f are initialized variables of type double. Does standard guarantee that it would be treated as c*d*e*2 (double result) then divided by 3 and multiplied by f (or some similar behavior). Obviously, 2/3 being calculated to 0 is undesirable.

标准的哪个段落对此进行了定义?

Which paragraph of standard defines that?

推荐答案

基于 [简介]-注释7 (非规范性内容):

可以根据通常的数学规则对运算符进行重新组合 仅在运算符真正具有关联性或可交换性的情况下.

Operators can be regrouped according to the usual mathematical rules only where the operators really are associative or commutative.

MDAS的数学规则是从左到右(考虑运算符的关联性和优先级).因此，其评估如下:

Mathematical rule for MDAS is from left to right (considering the associativity and precedence of operators). So it is evaluated as follows:

(((((c * d) * e) * 2) / 3) * f)

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