I know this is, but what about this?Code:#include <stdio.h> int main(void) { char v[] = "blablabla"; v[0] = 'b'; return 0; }
I think it is ok but I just want to make sure. Thank you.Code:#include <stdio.h> #include <string.h> #include <stdlib.h> int main(void) { char *v = malloc(strlen("blablabla") + 1); if (v) strcpy(v, "blablabla"); v[0] = 'b'; free(v); return 0; }
Neither is UB. You are thinking of the UB of writing to a string literal.
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
if char *v = "blablabla";
this one is undefined behaviour.
Code:The C99 Draft (N869, 18 January, 1999) J.2 Undefined behavior #1 The behavior is undefined in the following circumstances: ... An attempt is made to modify a string literal of either form (6.4.5).
EDIT=edited.thanks xeddiex and eerok.
Last edited by cbastard; 01-15-2006 at 02:04 PM.
Long time no C. I need to learn the language again.
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Complex problems have simple, easy to understand wrong answers.
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Code:char *s1 = "string literal"; char s2[] = "not a string literal";
immutable: char *s1Originally Posted by eerok
They're both string literals but the difference is that in the first one, s1 is being assigned a literal that is located in the data segment (always?) where all your programs execution instructions are stored along with other life time program duration stufff and you cannot and shouldn't try to change anything there.
mutable: char s2[]
The latter one has two stages; first the memory for s2 is allocated on the stack and then the string literal (which is located in the data segment) on the right is copied character for character to each block in memory in s2. And the values in s2 are now different from the string literal because they're both different copies, so, it's safe to write to.
