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?
The output will be like this:
1229823404 1229823404 1229823404 // address of a.
13.500000 13.500000 13.500000 13.500000 13.500000 // value of a.
& - Takes the address of the variable.
* - Takes the data which is stored in the given address.
How printf in main will print 25, 4?
It's a recursive function, so it will call himself again n again .. in your question.
Till condition get satisfied i==10,
by i++ ;
Value of I will get increment in stack downwards. Once i become 10,
If condition will get sastisfty. So return statement comes in picture and loop will get terminated. exit.
As per the stacks, Value will print down to up, In reverse order i.e. 10 9 8 7 6 5 4 3 2....
Hope you get it.