C++ make a string to int function - c++

Question

I usually program in python which has a string to integer converter built in but when i use C++ it doesn't seem to work so I decided to make my own.
this is what i made so far
C++:
int strtoint(string str)
{
int values [str.length()];
int return_value = 0;
for (int i=0; i < str.length(); ++i)
if(str.at(str.length()-1-i) == '1')
values[i] = 1;
else if(str.at(str.length()-1-i) == '2')
values[i] = 2;
else if(str.at(str.length()-1-i) == '3')
values[i] = 3;
else if(str.at(str.length()-1-i) == '4')
values[i] = 4;
else if(str.at(str.length()-1-i) == '5')
values[i] = 5;
else if(str.at(str.length()-1-i) == '6')
values[i] = 6;
else if(str.at(str.length()-1-i) == '7')
values[i] = 7;
else if(str.at(str.length()-1-i) == '8')
values[i] = 8;
else if(str.at(str.length()-1-i) == '9')
values[i] = 9;
for (int i=0; i < str.length(); ++i)
return_value += values[i]^(10*i);
return return_value;
}
I seem to get very weird answers like "12" returns 13 and "23" returns 11.
I know about stoi but I prefer to make my own so I can learn C++.

Answer 1

Three problems I can see on a quick look.
The first is that
int values [str.length()];
is not valid C++. It is using a feature from the 1999 C standard, which some C++ compilers support as an extension, but it is still not valid C++.
The second is lack of handling of 0s or errors (non-digit characters) in input.
The third is the statement
return_value += values[i]^(10*i);
^ is a bitwise XOR operator in C++. Not mathematical exponentiation.
A couple of other minor tips.
You can probably simplify your code a lot by using iterators.
Also, with all standard character sets, the roman digits are sequential, starting with '0'. So a simple way to convert a digit to the numeric value you want is digit - '0', which will convert '0' to 0, '1' to 1, .... '9' to 9.

Answer 2

In C++ I use std::stringstream to convert std::string to int. Simple example I found:
#include <sstream>
#include <iostream> //used for cout
int main()
{
std::string str = "12345";
std::istringstream ss(str);
int i;
ss >> i;
if (ss.fail())
{
// Error
}
else
{
std::cout << "The integer value is: " << i;
}
return 0;
}
Mistakes in your code you can find in #Peter answer.

Answer 3

You could simplify this a lot with the char conversion where you just subtract the value of '0' and the use of iterators, especially reverse iterators which make it simpler to calculate the value.
#include <cmath>
#include <string>
/*! Converts a char to an integer.
* #param c the character to convert
* #return the integer value [0-9] or -1 if not convertible
*/
int chartoint(const char c)
{
auto value = (c - '0');
if (value < 0 || value > 9)
{
value = -1;
}
return value;
}
/*! Converts a string to an integer.
* #param string the string to convert
* #return the converted integer value or 0 if not convertible
*/
int strtoint(const std::string& string)
{
auto integer = 0;
const auto begin = string.rbegin();
for (auto it = begin, end = string.rend(); it != end; ++it)
{
const auto current = chartoint(*it);
if (current == -1)
{
integer = 0;
break;
}
else
{
const auto factor = static_cast<int>(std::pow(10, (it - begin)));
integer += factor * current;
}
}
return integer;
}
Test run:
#include <cassert>
int _tmain(int argc, _TCHAR* argv[])
{
assert(strtoint("1") == 1);
assert(strtoint("1x") == 0);
assert(strtoint("12") == 12);
assert(strtoint("10") == 10);
return 0;
}