Recursive Functions

Show the exact screen-output of each of the following programs

Code:

#include<stdio.h>
void rev( ) //function
{
   char ch;
   scanf("%c", &ch); //reads what you entered in main
   if (ch < 'z')    //general case   
      rev ();
   printf ("%c", ch); //base case
} //end of function

void main()
{
   printf("Enter any word and then press enter\n");
   rev (); //function call
} //end of main

Hint: Assume the user’s  input was: ysaezr

Answer:

Code:

Enter any word and then press enter
ysaezr
zeasy

i dont konw how to get the above answer

Modify the statement
printf ("%c", ch); //base case
to
printf ("%c, address = %d\n", ch, &ch); //base case

In the above e.g
if (ch < 'z') keeps calling rev() till it finds 'z'. Every call to rev() creates a new set of variable ch, which is getting printed after all the calls to rev() are completed.
The statement
printf ("%c, address = %d\n", ch, &ch); //base case
shows the address of ch associated with each rev() call.

finally it prints all the var in the reverse order as the user entered, because the var are stored after each recursion in buffer.
so 'y' stored first goes to the last position.

1. main returns int, not void
2. This looks like C not C++

Originally Posted by jack999

Show the exact screen-output of each of the following programs

Answer:

Code:

Enter any word and then press enter
ysaezr
zeasy

Is there something weird with this program ? Why doesn't it stop when it gets 'z' as the base case. Here we see the program let user enter the character 'r' after it received 'z'. What is wrong with this ?

> What is wrong with this ?
It's doing the reverse thing in a strange way, yes, but it was probably written to demonstrate recursion on some test from school. You fail.

Originally Posted by citizen

It's doing the reverse thing in a strange way, yes, but it was probably written to demonstrate recursion on some test from school. You fail.

So would you please explain why doesn't it stop calling itself after received 'z' ? when user entered 'z' as input, the program would meet the base case's condition, how come it lets us enter the next char after 'z' ?

> So would you please explain why doesn't it stop calling itself after received 'z' ?

Try running it. It does stop calling itself after the user enters z. You don't see the results until you hit [enter] though, so scanf just leaves a bunch of jargon in the input buffer. scanf is horrible at its job as you know.

Originally Posted by citizen

> So would you please explain why doesn't it stop calling itself after received 'z' ?

Try running it. It does stop calling itself after the user enters z. You don't see the results until you hit [enter] though, so scanf just leaves a bunch of jargon in the input buffer. scanf is horrible at its job as you know.

Actually, I didn't test this program, just wondered when I was looking at the way it works as the sample represented above. I am not similar with C code (scanf, printf,...), that was why I haven't run it.

>scanf is horrible at its job as you know.
scanf is wonderful at its job. Most people just don't understand its job.

>i dont konw how to get the above answer
Trust it or trace it. There's a certain measure of trust that comes with recursion, mostly because all but the simplest of functions are difficult to trace in a reasonable amount of time. Here's the trace for your function:

Code:

Read 'y', less than 'z', call rev()
  Read 's', less than 'z' call rev()
    Read 'a', less than 'z' call rev()
      Read 'e', less than 'z', call rev()
        Read 'z', not less
        Print 'z'
      Print 'e'
    Print 'a'
  Print 's'
Print 'y'

Keep in mind that when one call to rev returns, it plops execution right back where it left off. So upon the return of rev, the next statement is the printf.