是否有与放大器定义的计算顺序; =和| =？ [英] Is there a defined evaluation order for &= and |=?

问题描述

如果你有一个C函数返回一个整数，你可以写这样的语句：

 敏＆安培; = MYFUNC（）;
 

...这里我们使用按位与赋值运算符。

现在的问题是：是MYFUNC（）保证会执行，即使敏等于零

Likwise，如果我们使用按位OR赋值运算符？（| =），将MYFUNC（）总是被执行，即使是敏设置为所有的人

换句话说：是在C允许按位运算符懒惰的评价

解决方案

 敏＆安培; = MYFUNC（）;
 

相当于：

 敏=敏和放大器; MYFUNC（）;
 

语言指出＆放大器;经营者不得短路。然而，
一个优化器可以产生code不调用函数如敏
是零，这是肯定的功能没有副作用。我怀疑任何编译器acrtually做到这一点，因为运行时测试可能使它成为pessimisation。

If you have a C function which returns an integer, you could write a statement like this:

MyInt &= MyFunc();

...where we're using the bitwise-AND assignment operator.

The question is: is MyFunc() guaranteed to be executed, even if MyInt equals zero?

Likwise, if we used the bitwise-OR assignment operator (|=), would MyFunc() always be executed, even if MyInt were set to all ones?

Put another way: is lazy evaluation permitted in C for bitwise operators?

解决方案

MyInt &= MyFunc();

is equivalent to:

MyInt = MyInt & MyFunc();

The language states that the & operator is not short-circuited. However, an optimiser could generate code not to call to the function if MyInt was zero and it was sure that the function had no side effects. I doubt any compilers acrtually do this, as the runtime test probably makes it a pessimisation.

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