sum of digits at odd and even places of a string - c++

Question

The output for the variable sum_e is negative which isn't what I expect it to be.
I have simply added the values at even and odd places and stored them in two variables. I checked for solutions and found ones with digit extraction from a number. None of them had a string input.
#include<iostream>
using namespace std;
int main(){
string s;
cin>>s;
int sum_e=0,sum_o=0;
int l=s.length();
for(int i=0;i<=l;i=i+2){
sum_o+=(s[i]-'0');
}
for(int j=1;j<=l;j=j+2){
sum_e+=(s[j]-'0');
}
cout<<sum_o<<endl<<sum_e;
return 0;
}
I subtracted '0' from the string index to convert it into int. One of the variables shows the right output and the other shows a negative one.

Answer 1

Your for loops run one time longer than the length of the array, so when i = l, s[i] will get an undefined/garbage value from memory. Use i < l and j < l rather than i <= l and j <= l, since the index in C++ begins at zero.
#include <iostream>
using namespace std;
int main(){
string s;
cin >> s;
int sum_e = 0, sum_o = 0;
int l = s.length();
for(int i = 0; i < l; i = i + 2){
sum_o += (s[i] - '0');
}
for(int j = 1; j < l; j = j + 2){
sum_e += (s[j] - '0');
}
cout << sum_o << endl << sum_e;
return 0;
}
To improve your code, use one for loop instead of two.
for(int i = 0; i < l; i++){
// Check if even (i%2 returns the remainder of i/2, so here i%2==1 means even)
if(i%2 == 1){
sum_e += (s[i] - '0');
}else{
sum_o += (s[i] - '0');
}
}

Answer 2

Array indexing in C++ starts from 0. You store the length of string as l, so elements of your string lies from s[0] to s[l-1]. At s[l] some garbage value is present which gets added to one of your variables, hence producing undesired results.
#include<iostream>
using namespace std;
int main(){
string s;
cin>>s;
int sum_e=0,sum_o=0;
int l=s.length();
for(int i=0;i<l;i=i+2){ // use <
sum_o+=(s[i]-'0');
}
for(int j=1;j<l;j=j+2){ // use <
sum_e+=(s[j]-'0');
}
cout<<sum_o<<endl<<sum_e;
return 0;
}
You can also do your odd and even position sum using a single loop. Your code size will reduce and look better
for(int i=0;i<l;i=i+2){
if(i%2==0)// even index means odd position numbers
sum_o+=(s[i]-'0');
else
sum_e+=(s[j]-'0');
}