Clever Geek Handbook

Segmentation fault in C. Attempt to exchange two values ​​using a pointer

I am getting a segmentation error when I try to swap values ​​in two variables. My code :

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 puzzling to me is the error occurring after successfully replacing the data. What is the error? Do I need to point pointers somewhere?

+3


source to share


3 answers


The pointer int *temp;

doesn't point to anything.

So, when your program *temp=*a;

, it puts a value a

in an arbitrary block of memory.

Try this fix:

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

Update



Additional question:

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

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);
}
+11


source


Your function swap

is wrong. It should be



void swap(int *a,int *b){
  int temp = *a;
  *a= *b;
  *b= temp;
}
+3


source


Use this swap feature: -

void swap(int *a,int *b){
    int temp;
    temp=*a;
    *a=*b;
    *b=temp;
   }
+1


source






More articles:


All Articles