Thread: swap char pointers

I have the following code.According to this the values of pointers p[0] and p[1] remains unchanged since the swap is made to local variables in swap function.Now my doubt is how can I swap the pointers p[0] and p[1] inside the function swap??

Code:

#include<stdio.h>
int main()
{
char*p[2]={"hello","good morning"};
  swap(p[0],p[1]);
  printf("%s %s",p[0],p[1]);
return0;
}
void swap(char*a,char*b)
{
char*t;
  t=a;
  a=b;
  b=t;
}

Originally Posted by Malcolm McLean

swap needs to make two char **s.

As a more general solution, write the function

Code:

/* swap two memory buffers */
void memswap(void *a, void *b, size_t len)

Then you can swap any two variables, structures or buffers in a single call like swap(&x, &y, sizeof(x));

This wont work in case of char pointers since you cannot change contents of read only block...

The first line of Malcolm's post is what you should be looking at.

Code:

int a = 0;

printf("a is %d\n",a);

updateVariable(a);

printf("a is %d\n",a);

// ---------------

void updateVariable(int a)
{
    a = 4;
}

(1) Do you understand why this snippet will print "a is 0" twice (not "a is 4" the second time around)?

(2) Do you know how to modify "updateVariable()" so that it will correctly update the value of a?