What does ASM_ARGS_##nr mean?

Question

When I read the softirq of Linux, I found the following code:

    asm volatile ("swi  0x0 @ syscall " #name   \
             : "=r" (_a1)               \
             : "r" (_nr) ASM_ARGS_##nr          \
             : "memory");               \
       _a1; })

I knew “r” (_nr) is the input list, but what does ASM_ARGS_##nr mean?

Answer 1

The operator ## is being used for the concatenation of identifiers within a preprocessor macro, as can be seen by the line continuation character \, ASM_ARGS_ or nr or both of them are arguments of that macro.

The whole macro will probably look similar to this fragment (that I found in another question):

#define ASM_ARGS_1      ASM_ARGS_0, "r" (_a1)
#define ASM_ARGS_2      ASM_ARGS_1, "r" (_a2)
#define ASM_ARGS_3      ASM_ARGS_2, "r" (_a3)
#define LOADREGS_5(a1, a2, a3, a4, a5)          \
register int _v1 asm ("v1") = (int) (a5);     \
LOADREGS_4 (a1, a2, a3, a4)

#define LOADREGS_6(a1, a2, a3, a4, a5, a6)      \

register int _v2 asm ("v2") = (int) (a6);     \

LOADREGS_5 (a1, a2, a3, a4, a5)

#define MYLIBC_SYSCALL(name, nargs, args...)               \
({                                                        \

    unsigned int retval;                              \

    {                                                 \

    register int _a1 asm ("r0"), _nargs asm ("r7");   \
    LOADREGS_##nargs(args)                            \
    _nargs = __NR_##name;                             \
    asm volatile (                                    \
            "swi    0x0"                              \
            :"=r"(_a1)                                \
            :"r"(_nargs) ASM_ARGS_##nargs             \
            : "memory" );                             \

But this fragment will help to understand your question about ##. Here ASM_ARGS_##nargs results in the combination of the argument nargs (which is an integer literal between 0 and 3) as to get an expression like ASM_ARGS_0, ASM_ARGS_1 and so on, which itself is a macro. In fact, ASM_ARGS_0 will probably be replaced by nothing, and ASM_ARGS_3 will be replaced by
, "r" (_a1), "r" (_a2), "r" (_a3).