从C ++ / C函数返回的问题

Question

i have this problem with returning from function. it's a search function, it should return 1 if the function find the value that i give but No, it's returning the 0 instead of 1. (sorry for my bad english.)

int rech(int tab[],int n,int i,int r){
    if(i<n){
        if(tab[i]==r){
            return 1;
        }
        i++;
        rech(tab,n,i,r);
    }

    return 0;
}

int main(int argc, char** argv) {
    int tab[5]={1,2,3,4,5};
    printf("%d",rech(tab,5,0,2));
    return 0;
}

Answer 1

When you use return you will return to the caller, which in a recursive function will mean the same function most of the times.

Change

rech(tab,n,i,r);

to

return rech(tab,n,i,r);

Answer 2

Let's walk through it together.

When rech gets called the first time:

First time we call rech: rech(tab = {1,2,3,4,5}, n = 5, i = 0, r = 2) is i < n or 0 < 5 ? This is true. Now we check if tab[i] == r or 1 == 2 . This is false. We increase i and start over again

Second time we call rech: rech(tab = {1,2,3,4,5}, n = 5, i = 1, r = 2) is tab[i] < r or 1 < 5 ? This is true. Now we check if tab[i] == r or 2 == 2 . This is true. We return 1 back to the first time rech got called.

Back to the first time we call rech: rech(tab, n, i, r) returned 1. So, in coding terms this would look like 1; which does nothing.

Now we are done with the if statement and return 0 to main.

main now prints out a 0 and the program ends.

This is the process of debugging the code.

To have the program return the 1 back to main instead you'll want to have it return the rech function because otherwise, the recursive call does nothing.