CODEWITHSUNDEEP
cgccassemblyx86
VELVETDETH
VELVETDETH

Reputation: 314

How to understand atomic test_and_set in assembly level?

Hi I'm new to the gcc's built-in atomic functions. And I'm using the test-and-set one. You may find the reference here

Here's the question:

I've done this code:

#define bool int
#define true 1
#define false 0

int main() {
    bool lock = true;
    bool val = __sync_lock_test_and_set(&lock, true);

    return 0;
}

What I intend to do is to check the assembly instruction of __sync_lock_test_and_set. I've used:

gcc -S [filename].c

And the result is:

        .file   "test_and_set.c"
        .file   "test_and_set.c"
        .text
        .globl  main
        .type   main, @function
main:
.LFB0:
        .cfi_startproc
        pushl   %ebp
        .cfi_def_cfa_offset 8
        .cfi_offset 5, -8
        movl    %esp, %ebp
        .cfi_def_cfa_register 5
        subl    $16, %esp
        movl    $1, -8(%ebp)
        movl    $1, %eax
        xchgl   -8(%ebp), %eax
        movl    %eax, -4(%ebp)
        movl    $0, %eax
        leave
        .cfi_restore 5
        .cfi_def_cfa 4, 4
        ret
        .cfi_endproc
.LFE0:
        .size   main, .-main
        .ident  "GCC: (GNU) 4.8.1"

However, I can't find where the test_and_set instruction is...

As you can see, I'm using gcc-4.8.1, and the environment is MAC OSX 10.10(I'm sure that this gcc is not what Apple provides. I've compiled this by myself)

Thanks!

Upvotes: 5

Views: 2676

Answers (1)

Jester
Jester

Reputation: 58782

    movl    $1, -8(%ebp)    # lock = true
    movl    $1, %eax        # true argument
    xchgl   -8(%ebp), %eax  # the test-and-set

It is an atomic exchange, which returns the previous value (that's the test part) and writes 1 into the variable (the set part). This is used to implement mutexes. After the operation the lock will be held by somebody - either the original owner or your code that has just acquired it. That's why it's safe to write a value of 1. The original value is returned, so you can distinguish between those two events. If the original value was 0 then you got the lock and can proceed, otherwise you need to wait because somebody else has it.

Upvotes: 7

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