How to convert char to int? [duplicate]

This question already has answers here: Convert char to int in C# (20 answers) Closed 7 months ago.

What is the pro开发者_如何学JAVAper 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.

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

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

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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();

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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.

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

Yet another option:

int i = c & 0x0f;

This should accomplish this as well.

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int val = '1' - '0';

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

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int val = '1' - 48;

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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';
        }
    }

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

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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,... )

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You can try something like this:

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

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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;
}