cases for gcc to ignore register declaration?

Question

K&R says

compilers are free to ignore the advice (register declaration).

In what cases, gcc would ignore if I define register int x = 4; ?

Answer 1

This is totally implementation-dependent.

In general, you should trust the compiler to put variables to register and not define them yourself.

C99 6.7.1 Storage-class specifiers

A declaration of an identifier for an object with storage-class specifier register suggests that access to the object be as fast as possible. The extent to which such suggestions are effective is implementation-defined.

Plus, C++11 has deprecated the use of the register keyword as a storage-class-specifier, maybe sometime in the future C will do the same.

Answer 2

By declaring a variable as 'register' , we are only requesting (not forcing) the compiler to store variable a in CPU register. Compiler will decide where to store the variable.

If you have declared more variables as register , then few of them might get into the CPU register as the CPU registers are limited. That is again implementation dependent.

Answer 3

If we refer to the C99 draft standard we see that this is implementation defined, section 6.7.1 Storage-class specifiers paragraph 4 says:

A declaration of an identifier for an object with storage-class specifier register suggests that access to the object be as fast as possible. The extent to which such suggestions are effective is implementation-defined .

This thread from gcc mailing list seems to indicate if you follow the thread a bit that register has been ignored for a while except in the case that you do not use optimization( using -O0 ).

Note that using register does have another impact in that it prevents the use of address operator( & ).