about linux kernel __asm__ grammar

Question

When I analyzed kernel code from include/asm/system.h of linux 0.11 kernel,

I have had some question.

There are some code like

#define _set_gate(gate_addr,type,dpl,addr) \
__asm__ ("movw %%dx,%%ax\n\t" \
    "movw %0,%%dx\n\t" \
    "movl %%eax,%1\n\t" \
    "movl %%edx,%2" \
    : \
    : "i" ((short) (0x8000+(dpl<<13)+(type<<8))), \
    "o" (*((char *) (gate_addr))), \
    "o" (*(4+(char *) (gate_addr))), \
    "d" ((char *) (addr)),"a" (0x00080000))

#define set_intr_gate(n,addr) \
    _set_gate(&idt[n],14,0,addr)

#define set_trap_gate(n,addr) \
    _set_gate(&idt[n],15,0,addr)

#define set_system_gate(n,addr) \
    _set_gate(&idt[n],15,3,addr)

It needs to set idt. some code that set idt use the macro like

void trap_init(void)
{
    int i;

    set_trap_gate(0,&divide_error);
    set_trap_gate(1,&debug);
    set_trap_gate(2,&nmi);
    set_system_gate(3,&int3);   /* int3-5 can be called from all */
    set_system_gate(4,&overflow);
    set_system_gate(5,&bounds);
    set_trap_gate(6,&invalid_op);
    set_trap_gate(7,&device_not_available);

I have question about c-grammar at this point:"o" (*((char ) (gate_addr))). and "o" ( (4+(char *) (gate_addr)))

Does this code make the output one byte???

for example If &idt[0] is 0x00006620, Does "o" (*((char *) (gate_addr))) code make output like 0x20 because of char type??

but, It seems that the code makes the output like 0x00006620.

I don't know about this asm grammar. why does this asm code work like this??? what is the rule and grammar?

Answer 1

That isn't "C grammar" per se, that is part of GCC's extended asm syntax . The "o" is a constraint that limits the access method that the compiler will attempt to use when that variable is referenced in the assembly.