Call by value in C

Question

I'm new to programming and I am currently working on C.

I learned that C does not have call by reference. The programs that we write to pass the address of actual parameters to the formal parameters is also call by Value in C.

Correct me if I'm wrong.. However, I ran this program :

//Swapping of two numbers using functions. #include

void swap(int *,int *);

void main()
{
    int x,y;
    printf ("Enter the values of x and y : ");
    scanf("%d %d",&x,&y);

    swap(x,y);

    printf("The value of x = %d and y = %d",x,y);
}

void swap(int *a,int *b)
{
    int temp;

    temp=*b;
    *b=*a;
    *a=temp;
}

It compiles just fine.. however, I'm getting a Segmentation Fault in the Output.

It asks me the enter the value of X and Y and then gives, Segmentation fault..

Please help!!

Answer 1

you are sending an int to a function that expects int*, thus when you are dereferencing - temp=*b; you are trying to access memory you don't own -> segfault. Call swap like this: swap(&x,&y);

Answer 2

So close

swap(&x,&y);

You were not passing references (pointers)