why can't I pass in variables to implement a swap function? - c

Question

I just started learning C and I was confused about why do I have to use pointers to implement a swap function?
Version where swap does not work:
#include <stdio.h>
void swap(int i, int j) {
int t = i;
i = j;
j = t;
}
int main() {
int a = 23, b = 47;
printf("Before. a: %d, b: %d\n", a, b);
swap(a,b);
printf("After. a: %d, b: %d\n", a, b);
return 0;
}
Version where swap works:
#include <stdio.h>
void swap(int *i, int *j) {
int t = *i;
*i = *j;
*j = t;
}
void main() {
int a = 23, b = 47;
printf("Before. a: %d, b: %d\n", a, b);
swap(&a, &b);
printf("After . a: %d, b: %d\n", a, b);
}

Answer 1

By default function arguments are passed by value in C and C++. This basically means a copy of the passed argument is received in the function. This copy is distinct from the passed variable and any change on this copy is not reflected back to the passed argument.
So your first snippet actually swaps copies of the passed arguments and not the arguments themselves.
For the swapping to be reflected back to the arguments being passed, You need to pass by reference, In C this is acheived by using pointers. Where the address of the arguments is passed to function and the function actually operates on the memory address passed to it, i.e: the original variable who's value is being passed.

Answer 2

swap(a,b);
void swap(int i, int j) {
int t = i;
i = j;
j = t;
}
The above code fails because, the variables i and j are local to the function swap , which means the variables have their scope limited to the function swap , and their life is lost once the function terminates.
One workaround is to use pass by address as shown in the code below...
swap(&a,&b);
void swap(int* i, int* j) {
int t = *i;
i* = *j;
j* = t;
}
Now the variables i and j, are pointers to original variables a and b , which means they are acting on behalf of a and b, and hence even if the pointers have limited scope within the function, they are able to indirectly modify variables a and b.

Answer 3

Why have you presented two programs back to back?Those are two separate programs.The first one ,despite all its fallacies about void main() and stuff, intends to(but doesn't,as it deals with local variables) swap values by using a function that is passed copies of the integer variables a and b.This is pass-by-value.In fact, C only has pass-by-value, despite what your book by Yeshwant Kanetkar says.Anyways, to make the first program work,use the printf() to display the swapped values inside swap() function itself,as being local variables the change in values won't show in main().
void swap(int i, int j) {
int t = i;
i = j;
j = t;
printf("After. a:%d,b:%d",i,j);
}
The second program, which he may claim uses pass-by-reference, is essentially pass-by-value where copies of the addresses of the variables are passed,not the actual addresses themselves.And again, despite it's non-portable eccentricities like void main(), it does swap the values of a and b.
void swap(int *i, int *j) {
int t = *i;
*i = *j;
*j = t;
}
Output Before. a:23 b:47
After. a:47 b:23
I ran it on CodeBlocks.But then your author Mr.Void Main must have suggested you to use 16-bit Turbo C eh?

Answer 4

You can not do swap this way:
void swap(int i, int j) {
int t = i;
i = j;
j = t;
}
because i and j are passed by value. That means that it is the same as local variable tmp. Thus, modifying of the i and j does not actually modify variables that you passed, as local copies of the i and j do not exist outside the function swap().
Because of it, if you need to modify i and j from main(), you have to pass to swap() addresses of i and j of the main().
That's why the following version is actually works.
void swap(int *i, int *j) {
int t = *i;
*i = *j;
*j = t;
}
EDIT: In order to understand these things better, I would like to recommend you Stanford Lectures Programming Paradigms, that are available at YouTube.