How to write a recursive void-function that has one parameter which is a positive integer When called the function writes its argument to the screen backward That is if the argument is 1234 it outputs?

Answer 1

I am a beginner at C++ but my professor gave us a similar one to this which was to display 4321. So i just modified it to fit your satisfaction.

Code:

void f(int n)

{

cout<<n;

if(n<=4) f(n+1);

}

calling statement: f(1) Output:1234

Trace :

f(1)

/ \

1 f(2)

/ \

2 f(3)

/ \

3 f(4)

/

4