Swap problem

Hi
I had been working on some code and I had to reverse a string. I tried to use the swap function as follows:

Code:

---

void swap(char *i, char *j)
{
  char temp = *i;
  *i = *j;
  *j = c;
}

---

No matter what swapping technique I use, the compiler won't let me do it. I checked the addresses being passed and they were correct. This is what I get.

Code:

---

    55 [main] newrev 3716 _cygtls::handle_exceptions: Error while dumping state
 (probably corrupted stack)
Segmentation fault (core dumped)

---

Obviously I dont need to assign memory to the arguments because that would make them point to a new address in memory. What am I missing?

Thanks

How are you calling it?

Code:

---

int main( void )
{
    char a = 'a', b = 'b';

    swap( &a, &b );
    printf( "a is %c, and b is %c\n", a, b );

    return 0;
}

---

Oh, also you have a 'temp', but you try to use a 'c' variable. I'll assume that was a typo.


Quzah.

Code:

---

void swap(char &a, char &b) {
    char temp = a;
    a = b;
    b = temp;
}

---

Up to your tutorial reading skills to figure out how/why this works.

i m calling them from main.

Code:

---

int main()
{
 char *i,*j;
 char *str = "something";
 
 //At the point of calling swap below i = str and j is pointing to the last char of str
 //I've checked for this using addresses of str and j.
 
 swap(i,j);
 
 return 0;
}

---

I know how that works.

Quote:

---

Originally Posted by Wraithan

Code:

---

void swap(char &a, char &b) {
    char temp = a;
    a = b;
    b = temp;
}

---

Up to your tutorial reading skills to figure out how/why this works.

---

Then read up on it again to see why it doesn't work. Here's a hint, this is the C board.


Quzah.

Quote:

---

Originally Posted by devilhaunts

i m calling them from main.

Code:

---

int main()
{
 char *i,*j;
 char *str = "something";
 
 //At the point of calling swap below i = str and j is pointing to the last char of str
 //I've checked for this using addresses of str and j.
 
 swap(i,j);
 
 return 0;
}

---

---

If inside a function, you make changes to an argument, and you want those changes to keep once your function exits, you have to have a pointer to that type of object.

Thus, 'char *' will let you make changes to a char. If you need to make a 'char *' point to something else, then your arguments need to be 'char **'.


Quzah.

That certainly makes more sense.

Quote:

---

Originally Posted by quzah

Then read up on it again to see why it doesn't work. Here's a hint, this is the C board.


Quzah.

---

I know, that's how C++ does it. I hope the comment was not directed at me.

There is an easy solution to this problem. The pointer str points to a string literal, something that you cannot modify only look at and access. What you are doing is comparable to changing the value of the number 4 to 0, or something.

Anyway, change your pointer to an array, and your problem should be solved. That also works.

@quzah & citizen

Both of your ideas work. Thanks.

Nvm me... I should learn better to hit new posts and reply.