C code crashes from memmove

Question

My code does not crash when I write:

char s[44] = "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa";
memmove(s, "asdf", 5);

But it does when I write:

char* s = "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa";
memmove(s, "asdf", 5);

Does anyone know why?

Refer this http://stackoverflow.com/questions/1704407/what-is-the-difference-between-char-s-and-char-s-in-c — Anantha Krishnan, Mar 16 '12 at 10:32
Trying to modify a string literal is Undefined Behaviour. In your case, apparently, the UB makes your program crash; in another computer it might change the contents of the string literal; in yet another computer it might make lemon juice ooze out of the USB port. — pmg, Mar 16 '12 at 11:13

score 7 · Answer 1 · answered Mar 16 '12 at 10:14

7

first one allocates space and puts the a's in

second one is a pointer to constant memory, you aren't allowed to change it.

answered Mar 16 '12 at 10:14

Keith Nicholas

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score 1 · Answer 2 · answered Mar 16 '12 at 10:15

1

In the first case, 44 bytes are allocated on stack and the string "aa..a" is copied to this space. But in the second space, the string "aa..a" is a constant value and stored in the read only data segment. So a page fault will occur when you try to write a read only memory address.

answered Mar 16 '12 at 10:15

Jinghao Shi

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@OliCharlesworth:Yes, more specifically, SIGSEGV sigal in UNIX – Jinghao Shi Mar 16 '12 at 10:38

score 1 · Answer 3 · answered Mar 16 '12 at 11:08

char* s = "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa";

the string constant "aaaa" whatever is stored in a the memory which is readonly. For example in elf executables they will be stored in the .rodata section, which is nor writable. Therefore when you attempt to write at such a location it results in an errorhe

On the other hand char s[] will have the string stored in the local stack area, which you can modify.

C code crashes from memmove

3 Answers3