Can someone explain me how this C language code?

Question

#include <stdio.h>
void printPattern ( int n );
int main(){
     int n =3;
     printPattern(n);
     return 0;
}
// for n = 3
// *
// ***
// *****
// 1-1
// 2-3
// 3-5
void printPattern ( int n ){

if (n ==1 ){
    printf("*\n");
    return;
}

printPattern(n-1);
for (int i = 0;i<(2*n-1);i++){
    printf("*");
}
printf("\n");
}

please explain to me this C language code I copied from a youtube tutorial it prints like this

*
***
*****

but how ?? I can't understand this recursion and please don't share another method to do this because I am trying to learn recursion in C but these types of projects shake my head!

Answer 1

Pattern of N - 1

*
***
...
***...***

Pattern of N

*                  // these
***                // form
...                // the
***...***          // pattern of N - 1
****...****

Pattern of N

<Pattern of N - 1>
****...****

Answer 2

The function printPattern is recursively defined. It has a base case for n equals 1, meaning that the recursion stops here. It has a recursive case for other values of n in which it first calls itself with n - 1 and then prints the asterisk character 2*n-1 times.

If we "unroll" the function for n equal to 3, we get:

1. printPattern(3) -> call printPattern(2), print 5 asterisks afterwards.
2. printPattern(2) -> call printPattern(1), print 3 asterisks afterwards.
3. printPattern(1) -> base case, print 1 asterisk.
4. The program now continues in the call at Step 2 and thus prints 3 asterisks.
5. The program now continues in the call at Step 1 and thus prints 5 asterisks.

Answer 3

For n == 3, the execution looks like this:

printPattern( 3 ):
  if ( 3 == 1 ) // false, call printPattern( 3 - 1 )
  printPattern( 2 ):
    if ( 2 == 1 ) // false, call printPattern( 2 - 1 )
    printPattern( 1 ):
      if ( 1 == 1 )
        print "*\n"  //  first row
        return
    // back in printPattern( 2 )
    for( i = 0; i < 3; i++ ) // second row
      print "*"
    print "\n"
    return
  // back in printPattern( 3 )
  for ( i = 0; i < 5; i++ )  // third row
    print "*"  
  print "\n"
  return