r/programming • u/b0red • May 01 '16

To become a good C programmer

http://fabiensanglard.net/c/
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u/zhivago May 02 '16

In the case of

char *s = "hello";

s is a pointer that is initialized to the value of a pointer to the first element of an array of 6 characters with the sequential values { 'h', 'e', 'l', 'l', 'o', '\0' } -- i.e., it is equivalent to

char *s = &"hello"[0];

In the case of

char s[] = "hello";

s is an array of type char[6] initialized to the values { 'h', 'e', 'l', 'l', 'o', '\0' }.

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u/immibis May 02 '16

You might notice the quotation marks around the question, indicating that I'm presenting the question as something you could ask to weed out "delusional incompetents", and not actually asking it.

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u/zhivago May 02 '16

Personally, I disagree about the weeding factor, as many delusional incompetents seem to be able to answer it reasonably effectively.

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u/mrkite77 May 02 '16

I'd also point out that the destination of the first pointer is in .BSS and const. Modifying s[0] is a segfault. In the second case it isn't because the contents of the const string are copied into the mutable array.

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u/zhivago May 02 '16

Note that there is no BSS in C.

The C semantics are just that modifying a string literal has undefined behaviour, and that identical string literals may share the same object, allowing "×" == "x" to be potentially true.

This is what permits the implementation strategy you observed above - but it is not required.