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I have difficulties to understand this code. I can fully understand why foo() is printing that values but I just can't get my head around why bar() is printing those in reverse. Can anyone please anyhow explain this so that I can feel it intuitively or at least give me a direction where to go for to reach the absolution.
#include<iostream>
using namespace std;
void bar(int a){
cout<<"bar: "<<a<<endl;
}
void foo(int a){
if(a>0){
a -=1;
cout<<"foo: "<<a<<endl;
foo(a);
bar(a);
}else{
cout<<"Foo exited"<<endl;
}
}
int main(){
foo(10);
}
[Output]:
foo: 9
foo: 8
foo: 7
foo: 6
foo: 5
foo: 4
foo: 3
foo: 2
foo: 1
foo: 0
Foo exited
bar: 0
bar: 1
bar: 2
bar: 3
bar: 4
bar: 5
bar: 6
bar: 7
bar: 8
bar: 9
560
Here's the intuition: the return value of any method call depends on the return value of another method call. Therefore, all the method calls must be stored in memory until a base case is reached. When the base case is reached, the values start becoming well-defined instead of recursively defined.
Recursion is the process of repeating items in a self-similar way. In programming languages, if a program allows you to call a function inside the same function, then it is called a recursive call of the function. The C programming language supports recursion, i.e., a function to call itself.
Recursion is the process of defining a problem (or the solution to a problem) in terms of (a simpler version of) itself. For example, we can define the operation "find your way home" as: If you are at home, stop moving. Take one step toward home.
Under the hood, your program makes infinite recursive calls, and thus a stack gets created for each call. The stack memory is limited. Hence the base case is the key for a recursive function. The recursive call is difficult to get right.
Recursion is best understood if you don’t try to „run the whole callstack in your head“. Think in abstractions:
As such the output of a "single level" would be (e.g for foo(10)):
Foo 9
output of foo(9)
Bar 9
Resolving one more level by filling in the partial output of foo(9)
Foo 9
Foo 8
output of foo(8)
Bar 8
Bar 9
This pattern continues until we reach the end of recursion.
The code might look like it is sequential foo();bar(); (which it is) but foo() first descents which leads to bar() being called just before ascending the callstack.
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