Converting string value to int WITHOUT using anything from the string library [duplicate]

Question

Let's say that I have a string variable called "numbers" that is defined as "12345" in string:

string numbers = "12345";

If that is the case, how do I convert that variable so that I can store those as integers instead? Now I know that there are multiple ways to do it using the string library like stoi, atoi, stringstream, and etc... but the trick here is to make it possible WITHOUT using any of those. I personally do not understand this assignment when there are MUCH efficient ways to do it but that's just my opinion.

I was thinking if I would make a function where it stores individual numbers as array of chars using loops and somehow stick those together as an int. Am I going into the right track?

Thank you

EDIT: I am talking about c++

@KenY-N I understand, but are there any ways to replicate that in c++? — user3521038
– user3521038, Commented Jan 11, 2017 at 5:20

MRB · Accepted Answer · 2017-01-11 05:31:00Z

2

You can convert every char that you read to it's equivalent int by using (ch - '0').

std::string str = "12345";
int number = 0;

for(auto ch : str)
{
    number = (number * 10) + (ch - '0');
}

return 0;

edited Jan 11, 2017 at 5:31

answered Jan 11, 2017 at 5:22

MRB

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1 Comment

MRB Over a year ago

Yes. It's unnecessary. only if str contain any other char('a', 'b') instead of numbers, it hold resulting int between 0 to 9. I'll edit.

injung · Accepted Answer · 2017-01-11 05:24:28Z

0

Try this:

string numbers = "12345";
int result = 0;
for (int i = 0; i < numbers.size(); i++) {
  result += (numbers[i] - 48) * pow(10, (numbers.size() - i - 1));
}
cout << result << endl;

Key idea is that char '0' is 48 in ASCII
You may check whether numbers[i] is in range 48 to 57 to validate input.

edited Jan 11, 2017 at 5:24

answered Jan 11, 2017 at 5:18

injung

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2 Comments

user3521038 Over a year ago

awesome. Thank you!

Ken Y-N Over a year ago

Why not use '0' instead of 48 then? And size() is a string library function (well, method) which the OP suggests they may want to avoid.

Shreevardhan · Accepted Answer · 2017-01-11 05:27:27Z

0

#include <iostream>
using namespace std;

int main() {
    string numbers = "12345";
    int result = 0;
    for (int digit : numbers)
        result = digit - 48 + result * 10;
    cout << result << endl;
    return 0;
}

Output

answered Jan 11, 2017 at 5:27

Shreevardhan

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