How to convert char to int?

问题:

What is the proper way to convert a char to int? This gives 49:

int val = Convert.ToInt32('1');
//int val = Int32.Parse("1"); // Works

I don't want to convert to string and then parse it.

回答1:

I'm surprised nobody has mentioned the static method built right into System.Char...

int val = (int)Char.GetNumericValue('8');
// val == 8

回答2:

how about (for char c)

int i = (int)(c - '0');

which does substraction of the char value?

Re the API question (comments), perhaps an extension method?

public static class CharExtensions {
    public static int ParseInt32(this char value) {
        int i = (int)(value - '0');
        if (i < 0 || i > 9) throw new ArgumentOutOfRangeException("value");
        return i;
    }
}

then use int x = c.ParseInt32();

回答3:

What everyone is forgeting is explaining WHY this happens.

A Char, is basically an integer, but with a pointer in the ASCII table. All characters have a corresponding integer value as you can clearly see when trying to parse it.

Pranay has clearly a different character set, thats why HIS code doesnt work. the only way is

int val = '1' - '0';

because this looks up the integer value in the table of '0' which is then the 'base value' subtracting your number in char format from this will give you the original number.

回答4:

int i = (int)char.GetNumericValue(c);

Yet another option:

int i = c & 0x0f;

This should accomplish this as well.

回答5:

int val = '1' - '0';

This can be done using ascii codes where '0' is the lowest and the number characters count up from there

回答6:

int val = '1' - 48;

回答7:

You may use the following extension method:

public static class CharExtensions
    {
        public static int CharToInt(this char c)
        {
            if (c < '0' || c > '9')
                throw new ArgumentException("The character should be a number", "c");

            return c - '0';
        }
    }

回答8:

The most secure way to accomplish this is using Int32.TryParse method. See here: http://dotnetperls.com/int-tryparse

回答9:

int val = '1' & 15;

The binary of the ASCII charecters 0-9 is:

0   -   00110000

1   -   00110001

2   -   00110010

3   -   00110011

4   -   00110100

5   -   00110101

6   -   00110110

7   -   00110111

8   -   00111000

9   -   00111001

and if you take in each one of them the first 4 LSB(using bitwise AND with 8'b00001111 that equels to 15) you get the actual number (0000 = 0,0001=1,0010=2,... )

回答10:

You can try something like this:

int val = Convert.ToInt32("" + '1');

回答11:

An extension of some other answers that covers hexadecimal representation:

public int CharToInt(char c) 
{
    if (c >= '0' && c <= '9') 
    {
        return c - '0';
    }
    else if (c >= 'a' && c <= 'f') 
    {
        return 10 + c - 'a';
    }
    else if (c >= 'A' && c <= 'F') 
    {
        return 10 + c - 'A';
    }

    return -1;
}