why is my variable storing value from previous recursion runs? - c

Question

the first image is of the Program with regular recursion format, I am returning values for both cases but self call to the function is made only when the argument is a natural number. The result is as expected
in the second program, however, I am not returning any value for natural numbers, the only return case is when the argument is 0.
Ideally, (taking 5 as the input) recursion will happen till num value equals 0. First with
sum = 5 + add(5-1);
sum = 4 + add(4-1);
sum = 3 + add(3-1);
sum = 2 + add(2-1);
sum = 1 + add(1-1);
At this stage, the function will return 0 as flow moves to else block,and value of sum for that loop is 0.
Now, with 0 being returned we rise one step up and 0 will take place of add(1-1) in the line
sum = num + add(num-1);
it should look like
sum = 1 + 0;
the function add(num) should terminate now, after assigning 1 to variable sum.
Two main questions.
No value should rise up for sum = 2 + add(2-1), as in the previous step no value is returned. for add(1), after getting value from add(0), summation of 1 and returned value is stored in variable sum and then exits out of function body.
Every time recursion occurs new variable sum is declared and initiated to 0, so, I don't see why the computer is storing the values of the sum from previous calls to the function, when instead it should start a fresh new variable sum.
to me, after completion of code, the value of variable sum in add() should be 1.
Please someone, if you can make sense of this code. Let me know. Much appreciated
P.s: you can try this on any online C compiler as well. Here is the code:
#include <stdlib.h>
#include <stdio.h>
int add(int num);
int main(){
int number,sum;
printf("enter a number of your choice: ");
scanf("%d",&number);
sum = add(number);
printf("SUM = %d",sum);
}
int add(int num){
int sum = 0;
if(num != 0)
sum = num + add(num-1);
else
return sum;
}

Answer 1

sum = add(number);
Using the return value of a function that flowed off the end without explicitly returning a value is undefined behavior. UB means anything is allowed to happen, including giving you the (false) illusion that a variable is "storing value from previous recursion runs". Quoting from C11 6.9.1/12:
If the } that terminates a function is reached, and the value of the function call is used by the caller, the behavior is undefined.

Answer 2

A function declared to return an int must always return a value (i.e. an int value). Your code doesn't.
int add(int num){
int sum = 0;
if(num != 0)
// If the execution gets in here
sum = num + add(num-1);
else
return sum;
// it will continue here (when the recursion ends).
// And here it is WITHOUT a returned value - that's bad.
}
You probably want:
int add(int num){
int sum = 0;
if(num != 0)
sum = num + add(num-1);
return sum;
}
or
int add(int num){
if(num != 0)
return num + add(num-1);
else
return sum;
}
or (more clear and secure):
int add(int num){
if(num <= 0)
{
return 0;
}
return num + add(num-1);
}