Confusion with Pointers Increment

Question

char *str = "string" ;
char *end = str;
while(*(end+1) != '\0') {
    end++;
}

while(str < end) {
    char temp = *str;
    *str = *end;
    *end = temp;
    str++;
    end--;
}`

EDIT: Are both these*str = *end, *str++ = *end invalid?

The above code gives error at that line.

Aren't str and end pointing to a read only part in memory whether it be post increment?

The first line should be `char str[] = "string";`. That way the string is in writable memory, and you can change it. Technically, your code has undefined behavior, which means that it may or may not work, depending on whether it does or not. — user3386109, Oct 02 '15 at 22:31

score 1 · Accepted Answer · edited May 23 '17 at 12:14

1

char *str = "string" ;

Is a string literal placed in a read-only data section.

In this line:

*str = *end;

is trying to modify the string, you would probably get a segfault..

Same for *str++ = *end, both are invalid.

To make this code work, change

char *str = "string";

to

char str[] = "string";

But note that you can not use *str++ because array names are constant (not modifiable lvalue)

edited May 23 '17 at 12:14

Community

1
1

answered Oct 02 '15 at 22:15

David Ranieri

39,972
7
52
94

My bad. I meant *str++ = *end; – rldl Oct 02 '15 at 22:38
1

`*str++ = *end;` is also modifying the string before incrementing the pointer. – David Ranieri Oct 02 '15 at 22:41
Right, so it is invalid. How can I make this code work while using char pointers? – rldl Oct 02 '15 at 22:46
1

Change `char *str = "string" ;` to `char str[] = "string";` – David Ranieri Oct 02 '15 at 22:47
Makes sense. Thank you. – rldl Oct 02 '15 at 22:48

Confusion with Pointers Increment

1 Answers1