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Home/ Questions/Q 8133793
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Editorial Team
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Editorial Team
Asked: June 6, 20262026-06-06T09:45:01+00:00 2026-06-06T09:45:01+00:00

Please look at the following code: char* test ( ) { char word[20]; printf

  • 0

Please look at the following code:

char* test ( )
{
    char word[20];
    printf ("Type a word: ");
    scanf ("%s", word);
    return word;
}

void main()
{
    printf("%s",test());
}

When the function returns, the variable word is destroyed and it prints some garbage value. But when I replace

char word[20];

by char *word;

it prints the correct value. According to me, the pointer variable should have been destroyed similar to the character array and the output should be some garbage value. Can anyone please explain the ambiguity?

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  1. Editorial Team
    Editorial Team
    2026-06-06T09:45:03+00:00Added an answer on June 6, 2026 at 9:45 am

    Undefined behavior is just that – undefined. Sometimes it will appear to work, but that is just coincidence. In this case, it’s possible that the uninitialized pointer just happens to point to valid writeable memory, and that memory is not used for anything else, so it successfully wrote and read the value. This is obviously not something you should count on.

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