将字符串转换为char c

Question

I am working on a project where I can to convert a api digit to a char. I have used an array of string pointers to get the conversion. However, I want to return just a single ch, as my api that I am using will only accept a char. So ap_five will return "5". But I want to five to be a single char '5'.

I thought maybe I could cast to a char in the return. However, I just get rubbish.

I am sure there is other ways to solve this, how I am just wondering how can I solve with my present program.

char *digits_conversion[][2]=
{
    {"ap_zero", "0"},
    {"ap_one", "1"},
    {"ap_two", "2"},
    {"ap_three", "3"},
    {"ap_four", "4"},
    {"ap_five", "5"},
    {"ap_six", "6"},
    {"ap_seven", "7"},
    {"ap_eight", "8"},
    {"ap_nine", "9"},
    {"ap_star", "*"},
    {"ap_hash", "#"},
    {NULL, NULL}
};

char convert_to_char(const char *digit)
{
    int i = 0;
    for(i = 0; *digits_conversion[i][1]; i++)
    {
        if(strcmp(digits_conversion[i][0], digit) == 0)
        {
            return (char) digits_conversion[i][1];
        }
    }

    return '\0';
}

int main(void)
{
    char ch;

    ch = convert_to_char("ap_five");

    printf("Converted digit: %c\n", ch);

    return 0;
}

Answer 1

The simplest fix to your existing code is just to change:

return (char) digits_conversion[i][1];

into

return digits_conversion[i][1][0];

However, you might find that changing digits_conversion into an array of structures will give you code that is easier to understand and maintain. For example:

struct digit_mapping {
    char *api_name;
    char digit;
};
struct digit_mapping conversion_table[] = {
    { "ap_zero", '0' },
    {"ap_one", '1'},
    {"ap_two", '2'},
    {"ap_three", '3'},
    {"ap_four", '4'},
    {"ap_five", '5'},
    {"ap_six", '6'},
    {"ap_seven", '7'},
    {"ap_eight", '8'},
    {"ap_nine", '9'},
    {"ap_star", '*'},
    {"ap_hash", '#'},
    {NULL, '\0'}
};

char convert_to_char(const char *digit)
{
    int i = 0;
    for(i = 0; conversion_table[i].digit; i++)
    {
            if(strcmp(conversion_table[i].api_name, digit) == 0)
            {
                    return conversion_table[i].digit;
            }
    }

    return '\0';
}

int main(void)
{
    char ch;

    ch = convert_to_char("ap_five");

    printf("Converted digit: %c\n", ch);

    return 0;
}

Answer 2

digits_conversion[i][1]

is a C-style string. Use:

digits_conversion[ i ][ 1 ][ 0 ] // if you want the character 

char convert_to_char(const char *digit)
{
    int i = 0;
    for(i = 0; *digits_conversion[i][1]; i++)
    {
            if(strcmp(digits_conversion[i][0], digit) == 0)
            {
                    return digits_conversion[i][1][0];
            }
    }

    return '\0';
}

Answer 3

return digits_conversion[i][1][0];

Answer 4

The termination of your 'for' loop looks incorrect:

//...
    {"NULL", "NULL"}
//...
//...
for(i = 0; *digits_conversion[i][1]; i++)
//...

The "NULL" you've got at the end of your array is literally, the string "NULL". If you pass a string that doesn't match any of the "ap_zero" through "ap_hash" your program will likely crash as the loop starts dereferencing garbage beyond the end of your array.

Eric fixed this in his example, but I thought I'd point this out specifically.