How to convert a character taken from a string to an int?

user_11972111 0
Viewed 97

So i have a string containing numbers, and i want to take one of those numbers and convert it to an int like that :

string s = "13245";
int a = stoi(s.at(3));

I have tried stoi :

int a = stoi(s.at(3));

I have tried atoi :

int a = atoi(s.at(3));

But none of those ways works, the only way i've found is the C way :

int a = s.at(3)-'0';

Do you know why stoi / atoi don't work ? Do you have any other way to convert a character taken from a string to an int ?

digits char c++
4 Answers

The function stoi expects an object of the type std::string. The C function atoi expects an object of the type char * that points to a string. While you are dealing with an object of the type char.

This

int a = s.at(3)-'0';

is the common approach to convert the internal representation of a digit character to the corresponding integer value because the codes of digits follow each other without gaps.

From the C++ Standard (2.3 Character sets)

  1. ... In both the source and execution basic character sets, the value of each character after 0 in the above list of decimal digits shall be one greater than the value of the previous.
user_28772410

std::string is an array of chars and .at(3) or [3] will return you single char. stoi and atoi work on strings (many chars) and will convert string like "-42" to its number representation -42.

#include <iostream>
#include <string>


int main() {
    const std::string str = "1324509";

    for(const char ch: str) {
        const int a = ch & 0x0f; // the same as  ch - '0';
        printf("%i,",a);
    }

    return 0;
}

Live Code

user_72675250

According to documentation:

Stoi parses str interpreting its content as an integral number of the specified base and returned an int value.

int stoi (const string&  str, size_t* idx = 0, int base = 10);

Parallel to your code:

#include <iostream>
#include <string>
#include <typeinfo> 

using namespace std;

string s = "13245";
cout << typeid(s.at(3)).name();
// Prints out 'c' which means 'char'

So you need to convert char to string before using stoi

#include <iostream>
#include <string>

using namespace std

string s = "13245";
string m(1,s.at(3));
int a = stoi(m);
user_132226530

atoi accepts a const char*, whereas s.at(index) returns a char, hence the compiler will return a type error, rather you can do the following:

    string s = "13245";
    auto c = s.at(3);
    int a = atoi(&c);
user_102617110
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