Swap chars only given pointer

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;

This is not clear what you are asking. – MByD Oct 13 '12 at 15:55 — MByD, Oct 13 '12 at 15:55

score 3 · Answer 1 · answered Oct 13 '12 at 15:56

3

Well, you can't with pointer to a string literal. Literals are immutable. Stick with the way you've always done it.

answered Oct 13 '12 at 15:56

Lews Therin

So, If I was given a pointer to a string literal I would have to copy it into a char array? – Brandon Ling Oct 13 '12 at 15:58
Yeah, and then you can swap it. Did you try it? You should have gotten a segfault. – Lews Therin Oct 13 '12 at 15:59
exactly I got a seg fault, that's why I was confused! – Brandon Ling Oct 13 '12 at 16:02
Yep, anyway some of the links in the comment box of your question should help ;) Trying to change the contents of a read only memory location is not good! :P – Lews Therin Oct 13 '12 at 16:05

score 1 · Answer 2 · answered Oct 13 '12 at 15:55

1

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

answered Oct 13 '12 at 15:55

Nico

But I never said char[] s = {'h','e' ....}; How could I do it that way? – Brandon Ling Oct 13 '12 at 15:56
char* foo = "bar"; is equal to char foo[] = {'b','a','r','\0' }; – Nico Oct 13 '12 at 15:58
2

It's not equal. Different type, different behavior. – Alexey Frunze Oct 13 '12 at 16:03
Sorry I must be new to c then, could you explain the difference between the two char pointers for me? – Nico Oct 13 '12 at 23:56

score 1 · Accepted Answer · answered Oct 13 '12 at 17:11

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;

score 0 · Answer 4 · answered Oct 14 '12 at 22:06

0

Some operating systems / compilers allow you to do modify string literals, mainly AIX 5.1.

If modify the string literal in your code, it makes it a pain to have your code cross platform as you have to rewrite your code to compile for Linux.

answered Oct 14 '12 at 22:06

Doug

4 Answers4