CODEWITHSUNDEEP
cpointerssegmentation-faultswap
Gaurav
Gaurav

Reputation: 397

Segmentation fault in C. Trying to swap two values using pointer

I am getting a segmentation fault while trying to attempt swapping values in two variables. My code is :

void swap(int *a,int *b){
    int *temp;
    *temp=*a;
    *a=*b;
    *b=*temp;
}
int main(){
    int i=1,j=0;
    printf("Before %d,%d\n",i,j);
    swap(&i,&j);
    printf("After %d,%d\n",i,j);
    return 0;
}

I am getting the following error:

Before 1,0
After 0,1
Segmentation fault (core dumped)

What looks mysterious to me is the error is being produced after the values have been swapped successfully. What is the bug? Do I need to typecast the pointers anywhere?

Upvotes: 3

Views: 1980

Answers (3)

Ilya
Ilya

Reputation: 4689

Your pointer int *temp; points on nothing.

So, when your program does *temp=*a;, it puts value of a into random memory block.

Try this fix:

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

Updated:

Additional question:

Suppose I want to use temp as a pointer variable and not as a regular variable, is there any way to get my program executed or I need to leave my stubbornness?

Answer: you can try this:

void swap(int *a,int *b){
    int *temp;
    temp = malloc(sizeof(int));
    if (temp == NULL)
      return;
    *temp=*a;
    *a=*b;
    *b=*temp;
    free(temp);
}

Upvotes: 11

Ajay
Ajay

Reputation: 4971

Use this swap function :-

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

Upvotes: 1

Basile Starynkevitch
Basile Starynkevitch

Reputation: 1

Your swap function is wrong. It should be

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

Upvotes: 3

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