Question

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Why does simple C code receive segmentation fault?

Hey Everyone, I’m sure this is a very basic question, but apparently I’m not quite understanding something here.

I’ve been playing around with C a lot over winter break and just came across something that I thought would work but is giving me a segmentation fault.

if I declare a string as:
char name[5] = “Mike”;
I can manipulate the string: *(name+1) = ‘a’; This works fine, name becomes “Make”.

If I declare as:
char *name = “Mike”;
and then try the same thing: *(name+1) = ‘a’; I get a segmentation fault. Why can’t I do that?

If I malloc the space for the string: char *name = (char*)malloc(5*sizeof(char)); and then copy the string to name: strcpy(name,”Mike”); I can manipulate it like above just fine. *(name+1) = ‘a’; works.

What is the difference between char *name = “Mike”‘, and char *name = (char*)malloc(5*sizeof(char)); strcpy(name,”Mike”);? Aren’t they both just pointing to memory containing the string?

Sorry for the noobish question!

Answer 1

char name[5] = "Mike" declares a local array and copies the string “Mike” into it. char* name = "Mike" assigns a pointer to “Mike” without copying. In both cases, “Mike” is a constant string held in a read-only page, and so in the second case you are trying to modify the original constant.