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Asked: 2026-05-19T00:13:39+00:00

Let’s say I write char c[99] = {‘Stack Overflow’}; in C or C++. It

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Let’s say I write char c[99] = {'Stack Overflow'}; in C or C++. It compiles fine but is this valid? By valid I meant not invoking any kind of undefined or unspecified behavior.

Again if I write char c[99] = 'Stack Overflow'; gcc complains about multicharacter constant which is obvious but in the above when I am enclosing within curly brackets compiler is happy! Why is it so?

I also notice that puts(c); after the first statement will output ‘w’ precisely the last character of a general string in-place of Stack Overflow. Why so?

Could somebody explain these behaviors separately?

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1 Answer

  1. Editorial Team

    They both are only a single literal, so c[0] gets set to the literal and c[1] … c[98] get filled with zero (NUL character).

    I think what value actually gets stuffed into c[0] is implementation dependent, but it should at least compile on any compliant compiler.

    EDIT: Verified against the standard, in C++0x at least:

    A multicharacter literal has type int and implementation-defined value.

    And in C99 (using the draft, cause it’s free):

    The value of an integer character constant containing more than one character (e.g.,
    'ab'), or containing a character or escape sequence that does not map to a single-byte
    execution character, is implementation-defined.

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