C Programming - Functions - Discussion
#include<stdio.h>
int reverse(int);
int main()
{
int no=5;
reverse(no);
return 0;
}
int reverse(int no)
{
if(no == 0)
return 0;
else
printf("%d,", no);
reverse (no--);
}
Step 1: int no=5; The variable no is declared as integer type and initialized to 5.
Step 2: reverse(no); becomes reverse(5); It calls the function reverse() with '5' as parameter.
The function reverse accept an integer number 5 and it returns '0'(zero) if(5 == 0) if the given number is '0'(zero) or else printf("%d,", no); it prints that number 5 and calls the function reverse(5);.
The function runs infinetely because the there is a post-decrement operator is used. It will not decrease the value of 'n' before calling the reverse() function. So, it calls reverse(5) infinitely.
Note: If we use pre-decrement operator like reverse(--n), then the output will be 5, 4, 3, 2, 1. Because before calling the function, it decrements the value of 'n'.
if(x>0)myfunc(--x);
printf("%d",x);
}
int main()
{
myfunc(5);
return 0;
}
output: 001234
Anyone, please explain this.
D is correct answer.
As no is 5. So it is transferred to function, where it does not equal to 0. Then control goes to else part and prints 5. After that post operator used. Hence, it simply transfers same value i.e. 5 and again same process continue. It continuously produces 5.
Because the characteristic of post decrement is that it gets implement after all operation executed so it fist call recusively then it decrement the no but now again reverse function gets 5 as a fresh no and tgis process continues.
How it becomes infinite?