How to convert a single char into an int

Question

I have a string of digits, e.g. "123456789", and I need to extract each one of them to use them in a calculation. I can of course access each char by index, but how do I convert it into an int?

I've looked into atoi(), but it takes a string as argument. Hence I must convert each char into a string and then call atoi on it. Is there a better way?

Answer 1

If you're not worried about encodings (and as this sounds like homework I doubt you are) you can utilise the fact that the character encodings for digits in ascii are all in order from 48 (for '0') to 57 (for '9').

Therefore, the integer value for any digit is the digit less 48 or, expressed in code (and you don't even need to know that '0' is 48, also you can do arithmetic with chars, there just really small ints under the covers). .

char c = '1';
int i = c - '0'; // i is now equal to 1, not '1'

Hope this helps,

EDIT:

Me saying "If you're not worried about encodings" can (and probably should) be interpeted as "I hope to hell you are not worried about encodings, because I don't know enough about them, and I can only be sure about how decimal digits are represented in AscII".

Apologies if this has caused any confusion.

Answer 2

If you are worried about encoding, you can always use a switch statement.

Just be careful with the format you keep those large numbers in. The maximum size for an integer in some systems is as low as 65,535 (32,767 signed). Other systems, you've got 2,147,483,647 (or 4,294,967,295 unsigned)

Answer 3

Or you could use the "correct" method, similar to your original atoi approach, but with std::stringstream instead. That should work with chars as input as well as strings. (boost::lexical_cast is another option for a more convenient syntax)

(atoi is an old C function, and it's generally recommended to use the more flexible and typesafe C++ equivalents where possible. std::stringstream covers conversion to and from strings)

Answer 4

Any problems with the following way of doing it?

int CharToInt(const 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:
     return 0;
    }
}

Answer 5

#include<iostream>
#include<stdlib>
using namespace std;

viod main()
{
     char ch;
     int x;
     cin>>ch;
     x=char (ar[1]);
     cout<<x;
}

Answer 6

The answers provided are great as long as you only want to handle Arabic numerals, and are working in an encoding where those numerals are sequential, and in the same place as ASCII.

This is almost always the case.

If it isn't then you need a proper library to help you.

Let's start with ICU.

1. First convert the byte-string to a unicode string. (Left as an exercise for the reader).
2. Then use uchar.h to look at each character.
3. if we the character is UBool u_isdigit (UChar32 c)
4. then the value is int32_t u_charDigitValue ( UChar32 c )

Or maybe ICU has some function to do it for you - I haven't looked at it in detail.

Answer 7

char a='5',b='7';

int c,d;

c=atoi(&a);

d=atoi(&b);

the above code does not work properly; when a is converted in to integer the value of 'c' become 5 but after the second execution the value of 'd' become 75.... plz give me its solution......