Why changing string characters causes segmentation fault(core dumped):
char *str = "string";
str[0] = 'S'; //segmentation fault(core dumped)
-
1Please use search first stackoverflow.com/questions/1704407/…Abhi– Abhi2014-09-11 18:09:35 +00:00Commented Sep 11, 2014 at 18:09
Add a comment
|
2 Answers
The solution is simple, declare your string in the following way instead
char str[] = "string";
The reason why you should do this is because of the Undefined behavior. Creating a string with pointers will make your string locate at the read only memory part, so you cannot modify it, whereas another way will also make a copy of your string on the stack. Also check What is the difference between char s[] and char *s in C?
2 Comments
abelenky
You cannot say both "undefined behavior" and "will make" (a definitive statement). The point of "undefined behavior" is that anything can happen, and nothing is required to happen. It is likely that your string will be located in read-only memory, but not required.
betteroutthanin
@abelenky Thanks for pointing out, I'm not a native english speaker, and the undefined behavior says that the one like the question is one of the undefined behavior.
char *str = "string"; points to a read-only part of memory and because of that the string can't be changed.
You need to declare an array instead of a pointer which points to an array if you want to change the array like this
char str[] = "string";
