Question

Possible Duplicate:
What is the difference between char s[] and char *s in C?

int main()
{
    char *t1 = "Hi hello";
    char t2[] = " strcat";

    printf("%s", strcat(t1, t2));
}

When I run this program it gives me segmentation fault

int main()
{
    char t1[] = "Hi hello";
    char *t2 = " strcat";

    printf("%s", strcat(t1, t2));
}

Whereas when I run the next program it runs properly and shows up the concatenated string.
Why is that so ?

Thanks in advance 🙂

SUMMARY OF THIS QUESTION [SOLVED]

This question is very much closed. Just wanted to add summary. The points which I understood are:
For variables declared in this manner

char *t1 = "hi hello";  

Just make sure to add type qualifier const. Since by default it is read-only memory. At any cost we cannot modify the data. For example

t1[0] = "L"; 

is disallowed. Variables declared in this manner are not under our control and will remain forever during the life time of program. We cannot even free that memory.

char t1[10] = "hi hello";

This way we have a better control of the memory. We are permitted to modify the array. Whenever the scope is gone, the allocated memory gets deallocated.

char t1[] = "hi hello"; char t2[8] = "hi hello";

The two declarations t1 and t2 are very much same with 8 character locations allocated sequentially.

Hope I made sense. If not please edit. 🙂

Answer 1

t1 is a pointer to a read-only sequence of characters. strcat wants to write into the array pointed to by the first argument, and this is undefined behaviour.

(The second version is also undefined behaviour because you are overrunning the array t1.)