为什么有些c ++编译器让你拿一个字面量的地址？ [英] Why do some c++ compilers let you take the address of a literal?

问题描述

一个C ++编译器，我不会命名，让你取一个文字的地址，int * p =&42;

A C++ compiler that I will not name lets you take the address of a literal, int *p = &42;

清楚地，42是r值，大多数编译器拒绝这样做。

Clearly 42 is an r-value and most compilers refuse to do so.

为什么编译器允许这个？

Why would a compiler allow this? What could you do with this other than shoot yourself in the foot?

推荐答案

如果你需要一个指向整数的指针，值42？ ：）

What if you needed a pointer to an integer with the value of 42? :)

C ++引用与自动引用的指针非常相似。可以创建一个对文字的常量引用，如下：

C++ references are much like automatically dereferenced pointers. One can create a constant reference to a literal, like this:

const int &x = 42;

它有效地要求编译器使用值为42的整数的地址初始化指针，因为你可以随后这样做：

It effectively requires the compiler to initialize a pointer with the address of an integer with the value 42, as you might subsequently do this:

const int *y = &x;

结合这一点，编译器需要有逻辑来区分没有地址采取，一个有，所以它知道将其存储在内存中。第一个不需要具有存储器位置，因为它可以是完全临时的并且存储在寄存器中，或者可以通过优化来消除。取值的地址可能会引入别名，编译器无法跟踪和禁止优化。因此，应用 & 运算符可能会将值（无论是什么）强制到内存中。

Combine that with the fact that compilers need to have logic to distinguish between a value which has not had its address taken, and one which has, so it knows to store it in memory. The first need not have a memory location, as it can be entirely temporary and stored in a register, or it may be eliminated by optimization. Taking the address of the value potentially introduces an alias the compiler can't track and inhibits optimization. So, applying the & operator may force the value, whatever it is, into memory.

所以，你可能发现一个错误，结合这两个效果。

So, it's possible you found a bug that combined these two effects.

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