Source Code - 1 month ago 4x

C Question

Where will the value of

`n`

When

`n=1`

`1`

`res`

`1`

`#include<stdio.h>`

int factorial(int n)

{

if(n<=1)

return 1;

else

return n*factorial(n-1);

}

int main()

{

int res =factorial(3);

printf("%d",res) ;

}

Answer

The run-time stack will have four functions in total: main(), factorial(3), factorial(2), and factorial(1). Each of these gets its own variable space. There are *three* instances of the variable **n**, one for each call to **factorial**.

When factorial(1) returns 1, that value isn't returned to main(), because main() is not what called it: factorial(2) called it. *That* instance performs the operation 2*1 and returns 2. The caller of that instance is factorial(3), which receives the 2, performs the operation 3*2, and returns a value of 6 to *its* caller, main().

Does that help clear it up? If not, I have two suggestions:

(1) Search this topic on StackOverflow and the internet in general; there are plenty of detailed walk-through examples of **factorial**.

(2) Insert **print** statements to trace the flow of data and execution. Within factorial, separate the computation and the return, so you can put a **print** in between:

```
catch = factorial(n-1)
result = n * catch
print n, catch, result
return result
```

Source (Stackoverflow)

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