On Sunday, November 21, 2021, Harald van Dijk <har...@gigawatt.nl> wrote: > On 21/11/2021 07:12, Kang-Che Sung wrote: >> >> On Sunday, November 21, 2021, David Laight <david.lai...@aculab.com <mailto:david.lai...@aculab.com>> wrote: >> > >> > You shouldn't need the strdup(). >> > I think you can even do: >> > if (!(opt & 1)) >> > prompt = ""; >> > because (IIRC and for historic reasons) quoted strings are char[] not const char[]. >> >> I don't know why you are messing up with the "constness" of the strings. C standard says string literal is of type const char[], and the const keyword didn't exist before C89. >> Note the compiler is free to merge string literals with identical content so they share the same buffer in the .rodata section (that's why they are const). > > It does not say that, it says the opposite: > > The multibyte character sequence is then used to initialize an array > of static storage duration and length just sufficient to contain the > sequence. For character string literals, the array elements have type > char, and are initialized with the individual bytes of the multibyte > character sequence. > > Note the "char" as opposed to "const char".
It refers to the initialization like this: char str[] = "abc"; str[0] = 'd'; // Vaild not this: char arr[5]; arr = "abc"; arr[0] = 'd'; // Undefined behavior nor this: char *ptr; ptr = "abc"; *ptr = 'd'; // Undefined behavior And messing with the "constness" is always a bad idea. The compiler actually treats the string literals of the latter two examples to have the type 'const char[]', so modifying the contents yields undefined behavior, matching what the standard says.
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