Question

Possible Duplicate:
What is the difference between char a[] = “string” and char *p = “string”?

What is the difference between using memcpy on stack memory vs on heap memory?
The following code works on Tru64 but segfaults on LINUX

char * string2 = "            ";
(void)memcpy((char *)(string2),(char *)("ALT=---,--"),(size_t)(10));

The second version works on LINUX

char * string2 = malloc(sizeof(char)*12);
(void)memcpy((char *)(string2),(char *)("ALT=---,--"),(size_t)(10));

Can someone explain the segfault on LINUX?

Answer 1

The first example has an Undefined Behavior. And so it might work correctly or not or show any random behavior.

Explanation:
The first example, declares a pointer string2to a string literal. String literals are stored in Implementation defined read only memory locations. A user program is not allowed to modify this memory. Any attempt to do so results in Undefined Behavior.

Reference:

C99 standard 6.4.5/5 "String Literals – Semantics":

In translation phase 7, a byte or code of value zero is appended to each multibyte character sequence that results from a string literal or literals. The multibyte character sequence is then used to initialize an array of static storage duration and length just sufficient to contain the sequence. For character string literals, the array elements have type char, and are initialized with the individual bytes of the multibyte character sequence; for wide string literals, the array elements have type wchar_t, and are initialized with the sequence of wide characters…

It is unspecified whether these arrays are distinct provided their elements have the appropriate values. If the program attempts to modify such an array, the behavior is undefined.