How do you implement your own string to integer conversion - c++

Question

So I want to take a string in such as 8302 and convert this into an integer by creating my own function and not using the stoi/atoi functions.
I tried so far by doing:
int stringToInt(string input)
{
int i = 0;
while(input[i] >= '0' && input[i] <= '9')
{
input[i] = input[i] * 10 + ......
i++;
}
return i;
}
I know i need to keep multiplying by 10 everytime i find an integer so i can increase it e.g 123 = 1*10*10+2*10+3. but i dont know how to code this. could anyone suggest a method?

Answer 1

It might be easiest to do it in a recursive manner. Use the following idea:
8302 = 830 * 10 + 2
That is:
If there is only one char in the string - return it; otherwise continue
Separate the last char in the string
Convert the string (without the last char) to integer - using recursion
Multiply by 10 and add the last char
There are a lot of details here:
how to convert 1 char to an integer? - subtract '0' from it
how to separate a char from the rest of the string? - use substr
when you have a working recursive solution, you might want to convert it to an iterative solution - this will make it faster, but maybe less readable
what to do with invalid strings like "aaa" or "123haha" - my algorithm doesn't handle that

Answer 2

Before you could define a char2int conversion:
inline int ctoi(char c) {
switch (c) {
case '0':
return 0;
case '1':
return 1;
case '2':
return 2;
case '3':
return 3;
case '4':
return 4;
case '5':
return 5;
case '6':
return 6;
case '7':
return 7;
case '8':
return 8;
case '9':
return 9;
default:
throw std::runtime_error("Invalid char conversion");
}
}
And use it:
int my_stoi_dec(const std::string& str) {
int rtn = 0;
int exp = 1;
for (auto cp = str.crbegin(); cp != str.crend(); ++cp) {
char c = *cp;
if (isdigit(c)) {
rtn += ctoi(c) * exp;
exp *= 10;
} else if (c == '+') {
return rtn;
} else if (c == '-') {
return rtn * -1;
} else {
throw std::runtime_error("Integer error conversion");
}
}
}

Answer 3

This one is very close to your attempt:
int toInt(const std::string& input)
{
int i = 0;
for (const auto c : input)
{
if (c < '0' || c > '9')
break;
i = i*10 + c-'0';
}
return i;
}
The only assumption is that the characters '0' to '9' lie directly next to each other in the character set. The if statement makes sure that we stop at non-digit characters. The digit-characters are converted to their integer value using c-'0'.
Please keep in mind that this only parses the first digits of a string. Strings that start with a sign + or - are not considered.

Answer 4

A good way is to find your first digit and go from there make a multiplayer variable and multiply it by ten every digit. for every char you add you have to subtract '0' from it as '0' is not equal to int 0.
example:
string s = "12346";
int multiplayer = 1;
int i = 0;
int result = 0;
while (s[i] >= '0' && s[i] <= '9')
++i;
--i;
for(i ; i >= 0 ; --i){
result += (s[i] - '0') * multiplayer;
multiplayer *= 10;
}

Answer 5

It is much better that you start your conversion from right to left.
So you will iterate your string starting from its end, and ending on its beginning. On each iteration, we will take the character, convert it to int and multiply it with its multiplier (its position in the result integer) and then add it to the final result.
This should work:
#include <iostream>
#include <string>
int stringToInt(std::string input)
{
int result = 0;
int multiplier = 1;
for (int i = input.length() - 1; i >= 0; i--) // start from right
{
if (input[i] - '0' < 0 || input[i] - '0' > 9) // if any character is not an integer, return maximum negative
{
result = INT16_MIN;
break;
}
result += (input[i] - '0') * multiplier; // convert to int, get its position and then add it to result.
multiplier *= 10; // get next position
}
return result;
}
int main()
{
std::string MyEx = "123456";
int MyInt = stringToInt(MyEx);
std::cout << MyInt;
return 0;
}

Answer 6

void enforce( bool b ) {
if ( !b ) {
throw std::range_error("Not a valid string to convert to integer");
}
}
int stringToInt(std::string) {
for ( std::size_t i( 0 ); i != ( last - first ); ++i ) {
enforce( ( '0' <= first[ i ] ) && ( first[ i ] <= '9' ) );
result += pow(10,i) * ( first[ i ] - '0' );
}
}
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