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Editorial Team
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Asked: 2026-05-13T16:53:06+00:00

C-strings are null-terminated which means that in a char-array, the char at index strlen()

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C-strings are null-terminated which means that in a char-array, the char at index strlen() is a byte with all bits set to 0. I’ve seen code where, instead of '\0', the integer 0 is used. But since sizeof(int) > sizeof(char), this might actually write beyond the allocated space for the array – am I wrong? Or does the compiler implicitely cast a an int to char in such a situation?

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1 Answer

  1. Editorial Team

    Yes you are wrong – the zero value will be truncated. After all, if you write:

    char c = 0;

    you would not expect the compiler to write beyond the bounds of the variable c, I hope. If you write something like this:

    char c = 12345;

    then the compiler should warn you of the truncation. GCC produces:

    c.c:2: warning: overflow in implicit constant conversion

    but still nothing will be written beyond the bounds of the variable.

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