Swap Issue in C [Beginner] [duplicate] - c

Question

This question already has answers here:
Why does this code not swap the numbers? [duplicate]
(2 answers)
Closed 6 years ago.
I am asked to make a swap between 2 integers.
#include <stdio.h>
#include <stdlib.h>
void swaplol(int a, int b)
{
int tmp;
tmp = a;
a = b;
b = tmp;
printf("After the swap : a= %d, b=%d\n",a,b);
}
I dont see where the problems lies. Everything seems good synthax wise...
int main(void)
{
int a,b;
a = 666;
b = 998;
printf("Before the swap a = %d, b = %d\n",a,b);
swaplol(a,b);
return 0;
}
Thanks

Answer 1

You need to pass the address of the integer to the swap function, so that when the value is swapped, you get swapped a and b values back in main. You therefore have to change the swaplol function to accept pointers instead of ints.
#include <stdio.h>
#include <stdlib.h>
void swaplol(int *a, int *b)
{
int tmp;
tmp = *a;
*a = *b;
*b = tmp;
printf("After the swap : a= %d, b=%d\n",*a,*b);
}
int main(void) {
// your code goes here
int a,b;
a = 666;
b = 998;
printf("Before the swap a = %d, b = %d\n",a,b);
swaplol(&a,&b);
printf("After the swap : a= %d, b=%d\n",a,b);
return 0;
}

Answer 2

The swap function should swap values of the original variables declared in main.
To do this the function has to accept the variables by reference.
And you should remove the output statement from the function and place it in main to demonstrate that the original variables were swapped.
For example
#include <stdio.h>
void swaplol( int *a, int *b )
{
int tmp = *a;
*a = *b;
*b = tmp;
}
int main( void )
{
int a, b;
a = 666;
b = 998;
printf( "Before the swap a = %d, b = %d\n", a, b );
swaplol( &a, &b );
printf( "After the swap : a = %d, b = %d\n", a, b );
return 0;
}
There is no need to include header <stdlib.h>

Answer 3

Use pointers:
void swaplol( int * a, int * b )
{
int tmp;
tmp = *a;
*a = *b;
*b = tmp;
}
int main(void)
{
int a, b;
a = 666;
b = 998;
printf( "Before the swap a = %d, b = %d\n", a ,b );
swaplol( &a, &b);
printf( "After the swap : a= %d, b=%d\n", a, b );
return 0;
}

Answer 4

Just add pointers.. i'm providing a code snippet without using a temp variable.
int main()
{
....
....
swaplol(&a,&b);
}
int swaplol(int *a,int *b)
{
*a=*a+*b;
*b=*a-*b;
*a=*a-*b;
printf("a=%d,b=%d",a,b);
}

Answer 5

In C there is The call by value and call by address with help of the Pointers. There you can "simulate" a call by reference which is present in other Languages.
Any way by default, C uses call(pass) by value to pass arguments. This means that the code within a function cannot alter the arguments used to call the function.
Lets take the following example:
#include <stdio.h>
void foo( int x );
int main( void ){
int x = 5;
foo(x);
printf("X = %d\n", x);
}
void foo( int x ){
x = 10;
}
The output will be 5.
If you are using a good compiler with settings turned ON you will see this kind of warning:
program.c:14:15: error: parameter ‘x’ set but not used [-Werror=unused-but-set-parameter]
void foo( int x ){
^
Now what did just happen?
By definition, call(pass) by value means you are making a copy in memory of the actual parameter's value that is passed in, a copy of the contents of the actual parameter.
If you really need to modify a value you need to pass the address of that variable and use a pointer to point to it:
#include <stdio.h>
void foo( int *x );
int main( void ){
int x = 5;
foo(&x);
printf("X = %d\n", x);
}
void foo( int *x ){
*x = 10;
}
Now the Output will be 10.