Question

Possible Duplicate:
Why do I get a segmentation fault when writing to a string?

If I have a pointer and I know the indexes of both the chars, how would I swap the chars(I didn’t actually allocate an array)

i.e.

char *str = "hello" and I know and I wanted to swap the ‘h’ and ‘o’ which is index 0 and 4, and then return the pointer as well.

I’m used to doing the convention of

temp = array[i];
array[i] = array[j];
array[j] = temp;

Answer 1

The following declaration:

char *str = "hello";

Should be, to warn you about unsafe behaviors:

const char *str = "hello";

Because attempt to modify a string litteral is an undefined behavior according to C standard, you should rather use an array.

char str[] = "hello";

Then, swap the array’s elements is easy and safe:

char tmp = str[0];
str[0] = str[4];
str[4] = tmp;