Question

In: Computer Science

#include <stdio.h> int sum(int n); //prototype int main() { int number, result; printf("Enter a positive integer:...

#include <stdio.h>

int sum(int n); //prototype

int main() {

int number, result;

printf("Enter a positive integer: ");

scanf("%d", &number);

result = sum(number);

printf("sum = %d", result);

return 0;

}

int sum(int n) {

if (n != 0)

return n + sum(n-1);

else

return n;

}

What does the code above do?

Solutions

Expert Solution

Implementation of given program;

Screenshots

Output:

The given code calculates the sum of all natural numbers from 1 to given number where number is given by the user. The given code uses recursive function to calculate the sum of all natural numbers.

i.e., For a given value of number, the program calculates sum as

sum=(number*(number+1))/2

For example when number = 10

then sum = (10*11)/2

= 110/2

= 55

Iterations in program for number = 10

sum(10)

= 10+sum(9) [ sum(9) is a function call again ]

= 10+9+sum(8) [ sum(8) is a function call again ]

= 10+9+8+sun(7) [ sum(7) is a function call again ]

= 10+9+8+7+sum(6) [ sum(6) is a function call again ]

= 10+9+8+7+6+sum(5) [ sum(5) is a function call again ]

= 10+9+8+7+6+5+sum(4) [ sum(4) is a function call again ]

= 10+9+8+7+6+5+4+sum(3) [ sum(3) is a function call again ]

= 10+9+8+7+6+5+4+3+sum(2) [ sum(2) is a function call again ]

= 10+9+8+7+6+5+4+3+2+sum(1) [ sum(1) is a function call again ]

= 10+9+8+7+6+5+4+3+2+1+sum(0) [ sum((0) returns 0 ]

= 10+9+8+7+6+5+4+3+2+1+0

= 55

venereology answered 1 week ago

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