Task Switching - Linux 0.01

[thepowersgang]

I have been trying to get task switching implemented in my OS for quite a while now, despite the immense amount of tutorials I have read.

To try to solve my problem I started to look through the Linux 0.01 source code to see how it was done. I came across a macro called switch_to() that did the actual task switching and noticed that it seemed to copy an undefined value to %dx and then jump to another undefined value.

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#define switch_to(n) {\
struct {long a,b;} __tmp; \
__asm__("cmpl %%ecx,_current\n\t" \
	"je 1f\n\t" \
	"xchgl %%ecx,_current\n\t" \
	"movw %%dx,%1\n\t" \
	"ljmp %0\n\t" \
	"cmpl %%ecx,%2\n\t" \
	"jne 1f\n\t" \
	"clts\n" \
	"1:" \
	::"m" (*&__tmp.a),"m" (*&__tmp.b), \
	"m" (last_task_used_math),"d" _TSS(n),"c" ((long) task[n])); \
}

The macro defines __tmp and then proceeds to use the values undefined, is there some other place in the code that these magically get a useful value inserted into them?

[cyr1x]

No,
in the Gnu(AT&T) syntax the format is like this

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mnemonic	source, destination

that means that the value from dx is copied into the var.

[thepowersgang]

Actually I noticed that part about 5 mins later, but I still don't understand why it would copy %dx to %1 (__tmp.b) and then long jump to %0 (__tmp.a)

[cyr1x]

I'm not sure on this, but I think it is in the reverse order(e.g %0 = task, %1 = TSS, ...)

[thepowersgang]

Then why not use %ecx

the 'm' code (i think) means that the data is placed in a temporary memory address instead of a register. the 'd' and 'c' before the tss and task mean to put the value passed into %edx and %ecx respectively.