copying pointer characters not working as expected

Question

In the following code, what am I doing wrong? I run the code in eclipse and using MinGW C compiler. When I run it eclipse stops responding. When I debug the code, it breaks on the line

*start = *end;

I verified the value of *start and *end in debug mode and none is null.

void func1(char *str)
{
    char *end, *start;
    end = start = str;
    char tmp;

    if (str)
    {
        while (*end)
            ++end;
        --end;

        while (start < end) 
        {
            tmp = *start;
            *start = *end;
            *end = tmp;

            start++;
            end--;
        }
    }
}

Any tips/ideas?

Could be a lot of things, maybe you are using a string literal? Providing a [SSCCE](http://www.sscce.org/) would allow us to help you better. — Shafik Yaghmour, Aug 07 '13 at 19:10
It's almost certainly due to passing in a read-only string literal, but you also have another problem: if the string is the empty string `""`, then the decrement of `--end` puts `end` before the start of the string, and that's technically Undefined Behavior, though that's highly unlikely to cause any practical problems. — Adam Rosenfield, Aug 07 '13 at 19:15
I'm passing "Hello World". When the code reaches that line, *start has 'H' and *end has 'd' in debug window. and "Hello World" is also defined as char pointer in the main function that calls func1. — Big Ali, Aug 07 '13 at 19:43

score 3 · Accepted Answer · answered Aug 07 '13 at 19:46

So according to your feedback your are passing a string literal, "Hello World" to func1, modifying a string literal is undefined behavior, you could alternatively use something like this and it will work:

char arr1[] = "hello world" ;
func1(arr1) ;

Although as Adam and Kerrek pointed out you need to add more error checking to you code but this should fix your immediate problem.

copying pointer characters not working as expected

1 Answers1