About atoi function-CodePudding

#include <unistd.h>

int ft_atoi(char *str)
{
    int c;
    int sign;
    int result;

    c = 0;
    sign = 1;
    result = 0;
    while ((str[c] >= '\t' && str[c] <= '\r') || str[c] == ' ')
    {
        c  ;
    }
    while (str[c] == ' ' || str[c] == '-')
    {
        if (str[c] == '-')
            sign *= -1;
        c  ;
    }
    while (str[c] >= '0' && str[c] <= '9')
    {
        result = (str[c] - '0')   (result * 10);
        c  ;
    }
    return (result * sign);
}

#include <stdio.h>

int main(void)
{
    char *s = " --- -- 1234ab567";
    printf("%d", ft_atoi(s));
}

This line: result = (str[c] - '0') (result * 10); Why do we subtract zero and multiply by 10? How its convert ascii to int with this operations? Thanks...

CodePudding user response：

Some detail before answering your question

Internally everything is a number a char is not an exception. In C char is a promoted type of integer meaning characters are integer in C. The char which is promoted type of integer are mapped to responding ASCII Value.

For example:

Capital Letter Range

65 => 'A' to 90 => 'Z'

Small Letter Range

97 => 'a' to 122 => 'z'

Number Range

48 => '0' to 57 => '9'

To answer your question

The ASCII CHARACTER '0' subtracted from any ASCII CHARACTER that is a digit(0-9) results to an actual Integer.

For Example

'9'(57) - '0'(48) = 9 (int)
'8'(56) - '0'(48) = 8 (int)

Remember char are promoted integer in C Read the detail to understand this statement.

And Also the ASCII CHARACTER '0' added to any INTEGER in the range(0-9) results to an ASCII CHARACTER.

For Example

9   '0'(48) = '9'(57) (char)
8   '0'(48) = '8' (56)(char)

CodePudding user response：

Please see ASCII table

The ASCII code for '0' is 48 - not zero. Therefore, to convert to decimal you need to subtract 48