How does recursion work in this code and not terminate after the if condition is not satisfied?

Try thinking of this in terms of code substituation.

At each point that someFunction() is called swap in the actual code of the function.

include <iostream> using namespace std; void someFunction(int[], int, int); int main() { int a[]={1,2,3,4,5,6,7,8,9,10}; int value=10; someFunction(a,0,value); return 0; } void someFunction(int b[], int x, int y){ if(x<y){ someFunction(b,x+1,y); cout<<b[x]<<endl; } } 

Lets look at the one line in main that matters:

someFunction(a,0,10); 

Step 1: Replace this by the function body (replacing the variables with the actual paramters)

if(0<10){ someFunction(a,1,10); cout<<a[0]<<endl; } 

Step 2: That if statement is always true so lets remove it:

someFunction(a,1,10); cout<<a[0]<<endl; 

Repeat Steps 1

if(1<10){ someFunction(a,2,10); cout<<a[1]<<endl; } cout<<a[0]<<endl; 

Repeat Steps 2

someFunction(a,2,10); cout<<a[1]<<endl; cout<<a[0]<<endl; 

We still have a call to someFunction() so lets repeat again:

if(2<10){ someFunction(a,3,10); cout<<a[2]<<endl; } cout<<a[1]<<endl; cout<<a[0]<<endl; 

And:

someFunction(a,3,10); cout<<a[2]<<endl; cout<<a[1]<<endl; cout<<a[0]<<endl; 

OK. You see the pattern. This repeats until you get to fail condition.

someFunction(a,10,10); cout<<a[9]<<endl; cout<<a[8]<<endl; cout<<a[7]<<endl; cout<<a[6]<<endl; cout<<a[5]<<endl; cout<<a[4]<<endl; cout<<a[3]<<endl; cout<<a[2]<<endl; cout<<a[1]<<endl; cout<<a[0]<<endl; 

Now if we replace that last call

if(10<10){ someFunction(a,11,10); cout<<a[10]<<endl; } cout<<a[9]<<endl; cout<<a[8]<<endl; cout<<a[7]<<endl; cout<<a[6]<<endl; cout<<a[5]<<endl; cout<<a[4]<<endl; cout<<a[3]<<endl; cout<<a[2]<<endl; cout<<a[1]<<endl; cout<<a[0]<<endl; 

This time we see that the condition is false. So we can remove it and the code.

cout<<a[9]<<endl; cout<<a[8]<<endl; cout<<a[7]<<endl; cout<<a[6]<<endl; cout<<a[5]<<endl; cout<<a[4]<<endl; cout<<a[3]<<endl; cout<<a[2]<<endl; cout<<a[1]<<endl; cout<<a[0]<<endl; 

That's it.