将基数20转换为int

Question

I am trying to convert base 20 to int . For example if I have "1A" it needs to be converted to 30 , and so on. I have developed the code but it is giving issues in running. Code is as below in C programming language:

#include <stdio.h>
#include <stdlib.h>

int main()
{
    printf("Hello world!\n");
    char converted[20] = "1A";
    itov(converted);

    return 0;
}

void itov(char vigesimalStr[])
{
    int length = 0;
    for (int i = 0; vigesimalStr[i] != '\0'; i++)
    {
        length++;
    }

    int base = 20;
    int result = 0;
    int power = 1;
    int num = 0;
    for (int j = length; j >= 0; j--)
    {
        if (val(vigesimalStr[j]) >= base)
        {
            printf("Invalid Number");
            return -1;
        }

        num += val(vigesimalStr[j]) * power;
        power = power * base;

    }

}

int val(char c)
{
    if (c >= '0' && c <= '9')
        return (int)c - '0';
    else
        return (int)c - 'A' + 10;
}

Answer 1

I will begin by stating the obvious, there is already a library function that can convert a string containing a number in any base from (I believe) 2 to 36:

printf("%ld\n", strtol("1A", NULL, 20));
// Output: 30

If, however, as part of an exercise or homework assignment the use of this and similar library functions are prohibited, I will not do your homework for you, but I will instead give you a description of the high level algorithm for reading an integer in an arbitrary base, N:

1. Initialize an accumulator variable at zero.
2. Initialize a count variable i at 0 .
3. Multiply accumulator by N.
4. Get the numeric value of the base-N digit currently in str[i] , and add that to accumulator (your val function).
5. Increment i .
6. If str[i] is '\\0' , return accumulator and exit. Otherwise, go to step 3.

Answer 2

Two key issues I see in your code:

void itov(char vigesimalStr[])
// ...
    return -1;

The itov() function can't be void if it returns -1 on error. It also fails to return the correct value on success!

for (int j = length; j >= 0; j--)

When running a sequence backward, you want to start at length - 1 as length itself is never a valid index as it's the last index plus one.

Below is a rework of your code with the above issues fixed and other lesser issues tidied up:

#include <stdio.h>
#include <stdlib.h>

#define BASE 20

int val(char c)
{
    if ('0' <= c && c <= '9')
        return c - '0';

    return c - 'A' + 10;
}

int itov(char vigesimalStr[])
{
    int length = 0;

    for (length = 0; vigesimalStr[length] != '\0'; length++)
    {
        // nothing to see here
    }

    int power = 1;
    int number = 0;

    for (int j = length - 1; j >= 0; j--)
    {
        int digit = val(vigesimalStr[j]);

        if (digit >= BASE)
        {
            fprintf(stderr, "Invalid Number\n");
            return -1;
        }

        number += digit * power;
        power *= BASE;
    }

    return number;
}

int main()
{
    char to_convert[20] = "1A";

    printf("%d\n", itov(to_convert));

    return 0;
}