可以GCC / g ++的告诉我，当它忽略了我的注册？ [英] Can gcc/g++ tell me when it ignores my register?

问题描述

在编译C / C ++ codeS使用gcc / g ++的，如果它忽略了我的寄存器，可以将它告诉我吗？
例如，在此code

When compiling C/C++ codes using gcc/g++, if it ignores my register, can it tell me? For example, in this code

int main()
{
    register int j;
    int k;
    for(k = 0; k < 1000; k++)
        for(j = 0; j < 32000; j++)
            ;
    return 0;
}

Ĵ将被用作寄存器，但是在这个code

j will be used as register, but in this code

int main()
{
    register int j;
    int k;
    for(k = 0; k < 1000; k++)
        for(j = 0; j < 32000; j++)
            ;
    int * a = &j;
    return 0;
}

Ĵ将是一个正常的变量。
它可以告诉我，我使用的寄存器变量是否真的存储在CPU寄存器？

j will be a normal variable. Can it tell me whether a variable I used register is really stored in a CPU register?

推荐答案

您可以相当假定GCC忽略注册关键字也许除了在 - O0 。但是，它不应该有所作为这种或那种方式，如果你是在这样的深度，你应该已经阅读组装code。

You can fairly assume that GCC ignores the register keyword except perhaps at -O0. However, it shouldn't make a difference one way or another, and if you are in such depth, you should already be reading the assembly code.

下面是关于这一主题的信息线索：的http：// GCC .gnu.org /毫升/ GCC / 2010-05 / msg00098.html 。早在过去，注册的确有助于编译器将变量分配到寄存器，但今天寄存器分配可以自动优化来实现，而不提示。关键字并继续为用C两个目的：

Here is an informative thread on this topic: http://gcc.gnu.org/ml/gcc/2010-05/msg00098.html . Back in the old days, register indeed helped compilers to allocate a variable into registers, but today register allocation can be accomplished optimally, automatically, without hints. The keyword does continue to serve two purposes in C:

1. 在C中，prevents你采取变量的地址。因为寄存器没有地址，这个限制可以帮助一个简单的C编译器。 （简单的C ++编译器不存在。）


2. A 注册对象不能声明限制。因为限制涉及的地址，他们的交集是没有意义的。 （C ++还没有限制，反正，这个规则是有点微不足道。）

1. In C, it prevents you from taking the address of a variable. Since registers don't have addresses, this restriction can help a simple C compiler. (Simple C++ compilers don't exist.)
2. A register object cannot be declared restrict. Because restrict pertains to addresses, their intersection is pointless. (C++ does not yet have restrict, and anyway, this rule is a bit trivial.)

有关C ++中，关键字一直以来C ++ 11 pcated德$ P $和的从原定的2017年标准修订提出了去除。

For C++, the keyword has been deprecated since C++11 and proposed for removal from the standard revision scheduled for 2017.

一些编译器使用了注册的参数声明确定的函数调用约定，与ABI允许混合stack-和基于寄存器的参数。这似乎是不合格的，它往往与像扩展语法寄存器（A1）来发生了，我不知道任何这样的编译器是否仍在使用中。

Some compilers have used register on parameter declarations to determine the calling convention of functions, with the ABI allowing mixed stack- and register-based parameters. This seems to be nonconforming, it tends to occur with extended syntax like register("A1"), and I don't know whether any such compiler is still in use.

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