C programming

Question

void main (){
int a =3;
fun(a);
}

void fun(int n){
if(n>0){
fun(--n);
printf("%d ",n);
}
}

Why the output : 012

Answer 1

FIRST READ THIS:
Recursive functions. What an interesting part of programming. Look:
in void main you call the function with n = 3
Now you call fun:
void fun(3) {
if (n > 0 (true)) {
fun(--n); // n now is 2
prinf(“%d”, n); // this line isn’t executed yet because you called fun in the previous line
}
}

//NOW n = 2
void fun(2) {
if (n > 0 (true)) {
fun(--n); // n now is 1
prinf(“%d”, n); // this line isn’t executed yet because you called fun in the previous line
}
}

void fun(1) {
if (n > 0 (true)) {
fun(--n); // n now is 0, but when n reached if block it was 1. Calls fun(0)
prinf(“%d”, n); // this line isn’t executed yet because you called fun in the previous line
}
}

void fun(0) {
if (n > 0 (false)) {
// this isn’t executed
}
}
Wait a moment I’m continuing this in the next message

Answer 2

LAST READ THIS:
Now that void fun(0) didn’t continued the recursion, the printf line of each fun() call is executed.

void fun(1) {
if (n > 0 (true)) {
fun(--n); // n now is 0. This line is already executed.
prinf(“%d”, n); // it prints the value of n, 0. It doesn’t print a new line. Now the output is 0.
}
}

void fun(2) {
if (n > 0 (true)) {
fun(--n); // n now is 1. This line is already executed.
prinf(“%d”, n); // it prints the value of n, 1. It doesn’t print a new line. Now output is 01.
}
}

void fun(3) {
if (n > 0 (true)) {
fun(--n); // n now is 2. This line is already executed.
prinf(“%d”, n); // it prints the value of n, 2. It doesn’t print a new line. So the output is 012.
}
}
// END OF RECURSION
Output of the code:
012

Answer 3

Thank you , for answer

Answer 4

The code use the stack memory , by branches instructions