Question

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;
}

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;
}

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

Answer 1

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.

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. 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.)

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

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.