Thread: Please help me! Pointer!

I run this code and the program give me this result. But i don't know how the program can have this result. Please explain for me.

Code:

#include <stdio.h> 

void swap1(int *i1,int *i2) { 
   int tmp; 

   tmp=*i1; 
   *i1=*i2; 
   *i2=tmp; 
} 

void swap2(int i1,int i2) { 
   int tmp; 

   tmp=i1; 
   i1=i2; 
   i2=tmp; 
} 

main(){ 
   int i1,i2; 

   i1=100; 
   i2=200; 
   printf("i1=%d i2=%d\n",i1,i2); 

   swap1(&i1,&i2); 
   printf("i1=%d i2=%d\n",i1,i2); 

   swap2(i1,i2); 
   printf("i1=%d i2=%d\n",i1,i2); 
}

The result is:
i1=100 i2=200
i1=200 i2=100
i1=200 i2=100

When i run swap1 i think that i1=200 and i2=100. So when i run swap2 they must be i1=100 and i2=200???

In swap1 you are passing the arguements by reference and hence the swap occurs, where in swap2 you are passing the arguements by value : that means i1 and i2 are created locally in scope of swap2 , the values you pass are copied there and swaped but not written back to i1 and i2 of main..

Thanks. And about this one, i don't know why they define tmp in the swap but not use it? So why there is tmp here?

Code:

#include <stdio.h> 
 
void swap(int *i1,int *i2) { 
  int tmp; 
 
  *i1=*i1^*i2; 
  *i2=*i1^*i2; 
  *i1=*i1^*i2; 
} 
 
main(){ 
  int i1,i2; 
 
  i1=100; 
  i2=200; 
 
  printf("i1=%d i2=%d¥n",i1,i2); 
 
  swap(&i1,&i2); 
  printf("i1=%d i2=%d¥n",i1,i2); 
}

What is the purpose of this program?
It definitely does not perform a swap...whatever it does.. remove tmp, you'll still get the same result ...

cool........ I didn't realize, that performs a swap.. , and I am surprised why I haven't come across XOR-swap till now...

Thanks for your reply. They are very useful. Thank you very much.