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Home/ Questions/Q 8987103
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Editorial Team
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Editorial Team
Asked: June 15, 20262026-06-15T21:42:35+00:00 2026-06-15T21:42:35+00:00

Consider the following code: const char *s = a b c d !; const

  • 0

Consider the following code:

const char *s = "a   b    c  d !";
const char *p = s;

top:for(; *p; p++) {
    switch(*p) {
    case 0x20: 
    case '\n': 
        goto top;
    default: 
        putchar(*p);
    }
}

Can someone explain why it enters an infinite loop instead of stopping when *p is NULL? I had in mind the following: when *p is 0x20 or \n go to the beginning of the loop again, since it tests the condition and evaluates the expression p++. So, I don’t see a reason for it to loop infinitely, or I really don’t get how the goto statement and labels work in the C programming language.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-15T21:42:36+00:00Added an answer on June 15, 2026 at 9:42 pm

    When you goto top, p++ is not executed, because the for loop is started again from the beginning. Then you goto top again. And then again. And then again. And forever.

    If you want your increment to work, use continue instead of goto. Or, better yet, do something even clearer:

    for(p = s; *p != '\0'; p++) {
        switch(*p) {
        case 0x20:
        case '\n':
            // Do nothing.
            break;
    
        default: 
            putchar(*p);
        }
    }
    

    Oh, and by the way, avoid goto statements like the plague. Unless you’re generating C code in an automated way and it’s not supposed to be human readable, goto is virtually never a good idea.

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