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Editorial Team
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Asked: 2026-06-04T05:18:24+00:00

int main() { char *p; p = (char* ) malloc(sizeof(char) * 0); printf(Hello Enter

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int main()
{
    char *p;
    p = (char* ) malloc(sizeof(char) * 0);
    printf("Hello Enter the data without spaces :\n");
    scanf("%s",p);
    printf("The entered string is %s\n",p);
    //puts(p);
}

On compiling the above code and running it , the program is able to read the string even though we assigned a 0 byte memory to the pointer p .

What actually happens in the statement p = (char* ) malloc(0) ?

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1 Answer

  1. Editorial Team

    It is implementation defined what malloc() will return but it is undefined behavior to use that pointer. And Undefined behavior means that anything can happen literally from program working without glitch to a crash, all safe bets are off.

    C99 Standard:

    7.22.3 Memory management functions
    Para 1:

    If the size of the space requested is zero, the behavior is implementation-defined: either a null pointer is returned, or the behavior is as if the size were some nonzero value, except that the returned pointer shall not be used to access an object.

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