``continue`` breaks label placement

Question

This works just fine:

#include <stdio.h>

int main(){
    volatile int abort_counter = 0;
    volatile int i = 0;
    while (i < 100000000) {
        __asm__ ("xbegin ABORT");
        i++;
        __asm__ ("xend");
        __asm__ ("ABORT:");
        ++abort_counter;
    }

    printf("%d\n", i);
    printf("nof abort-retries: %d\n",abort_counter-i);
    return 0;
}

However, what I originally wrote was

#include <stdio.h>

int main(){
    volatile int abort_counter = 0;
    volatile int i = 0;
    while (i < 100000000) {
        __asm__ ("xbegin ABORT");
        i++;
        __asm__ ("xend");
        continue;
        __asm__ ("ABORT:");
        ++abort_counter;
    }

    printf("%d\n", i);
    printf("nof abort-retries: %d\n",abort_counter);
    return 0;
}

but this led to

/tmp/cchmn6a6.o: In function `main':
rtm_simple.c:(.text+0x1a): undefined reference to `ABORT'
collect2: error: ld returned 1 exit status

Why?

(Compiled with gcc rtm_simple.c -o rtm_simple.)

Answer 1

You might be able to trick it:

        continue;
        reachable:
        __asm__ ("ABORT:");
        ++abort_counter;
    }

    printf("%d\n", i);
    printf("nof abort-retries: %d\n",abort_counter);
    if (abort_counter < 0) goto reachable;
    return 0;
}

The goto with the label tells gcc that the code is reachable, and abort_counter being volatile should prevent gcc from being able to optimize the goto away.