usage of double pointers as arguments

Question

Please find the code snippet as shown below:

#include <stdio.h> 

int My_func(int **); 

int main() 
{ 
     int a =5;
     int *p = &a;
     My_Func(&p);
     printf("The val of *p is %d\n,*p);
}

void My_Func(int **p)
{
     int val = 100;
     int *Ptr = &val;
     *p = Ptr;
}

How does by using a double pointer as a argument in my_Func function and making change of value reflects the same in the main function but if we use a single pointer in My_Func does not change the value in main?Please do explain me with examples if possible

Advanced thanks
Maddy

Answer 1

int **p is a pointer to a pointer-to-int. My_Func(int **p) works by changing the value of integer that the pointer-to-int points to i.e. int a.

Without changing the implementation, the function will not work with a pointer-to-int parameter int *p as there is a second level of indirection. In addition, you're setting the value to a local variable that is created on the stack. When the function is completed the memory used for the variable will be reclaimed, therefore making the value of a invalid.

void My_Func(int **p)
{
     int val = 100; // Local variable.
     int *Ptr = &val; // This isn't needed.
     *p = Ptr;
} // val dissapears.

Remove the second level of indirection and copy val by value instead of pointing to it:

#include <stdio.h>

void My_Func(int *p) 
{
    int val = 100;
    *p = val;
}

int main(void) 
{
    int a = 5;
    My_Func(&a);
    printf("The val of a is %d\n", a);
    return 0;
}