How to convert a single char into an int

问题

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?

回答1:

You can utilize the fact that the character encodings for digits are all in order from 48 (for '0') to 57 (for '9'). This holds true for ASCII, UTF-x and practically all other encodings (see comments below for more on this).

Therefore the integer value for any digit is the digit minus '0' (or 48).

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

is synonymous to

char c = '1';
int i = c - 48; // i is now equal to 1, not '1'

However I find the first c - '0' far more readable.

回答2:

#define toDigit(c) (c-'0')

回答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)

回答4:

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.

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

回答5:

You can make use of atoi() function

#include <stdlib.h>
#include <stdio.h>
int main(int argc, char* argv[]){
    int num ;
    num = atoi(argv[1]);
    printf("\n%d", num);
}

回答6:

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

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

回答7:

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)

回答8:

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;
    }
}

回答9:

By this way You can convert char to int and int to char easily:

int charToInt(char c)
{
   int arr[]={0,1,2,3,4,5,6,7,8,9};
   return arr[c-'0'];
}

回答10:

For me the following worked quite well:

QChar c = '5';
int x = c.digitValue();
// x is now 5

Documentation: int QChar::digitValue() const which says:

Returns the numeric value of the digit, or -1 if the character is not a digit.

来源：https://stackoverflow.com/questions/439573/how-to-convert-a-single-char-into-an-int

标签

c++

char