C Programming - Functions - Discussion
#include<stdio.h>
void fun(int*, int*);
int main()
{
int i=5, j=2;
fun(&i, &j);
printf("%d, %d", i, j);
return 0;
}
void fun(int *i, int *j)
{
*i = *i**i;
*j = *j**j;
}
Step 1: int i=5, j=2; Here variable i and j are declared as an integer type and initialized to 5 and 2 respectively.
Step 2: fun(&i, &j); Here the function fun() is called with two parameters &i and &j (The & denotes call by reference. So the address of the variable i and j are passed. )
Step 3: void fun(int *i, int *j) This function is called by reference, so we have to use * before the parameters.
Step 4: *i = *i**i; Here *i denotes the value of the variable i. We are multiplying 5*5 and storing the result 25 in same variable i.
Step 5: *j = *j**j; Here *j denotes the value of the variable j. We are multiplying 2*2 and storing the result 4 in same variable j.
Step 6: Then the function void fun(int *i, int *j) return back the control back to main() function.
Step 7: printf("%d, %d", i, j); It prints the value of variable i and j.
Hence the output is 25, 4.
int main()
{
float a=13.5;
float *b,*c;
b=&a;
c=b;
printf("\n%u %u %u", &a, b,c);
printf("\n%f %f %f %f %f %f", a,*(&a),*&a, *b,*c);
return 0;
}
In this problem, what would be printed because of *c i.e last argument of second printf statement.
Can anybody help me out in this?
What is the purpose of "void fun (int*, int*);" ?
so that when we calling the function i and j values are entered into formal arguments *i and *j
so,
*i**i=25
*j**j=4