C将dec解析为十六进制，并输出为char []

Question

I need some help please. I am looking to modify the DecToHex function.

For input decimalNumber = 7 :

Actual output :

sizeToReturn = 2;
hexadecimalNumber[1] = 7;
hexadecimalNumber[0] = N/A ( is garbage );

Desired output :

sizeToReturn = 3
hexadecimalNumber[2] = 0
hexadecimalNumber[1] = 7
hexadecimalNumber[0] = N/A ( is garbage )

The function :

void DecToHex(int decimalNumber, int *sizeToReturn, char* hexadecimalNumber)
{
    int quotient;
    int i = 1, temp;
    quotient = decimalNumber;
    while (quotient != 0) {
        temp = quotient % 16;
        //To convert integer into character
        if (temp < 10)
            temp = temp + 48; else
            temp = temp + 55;
        hexadecimalNumber[i++] = temp;
        quotient = quotient / 16;
    }

    (*sizeToReturn) = i;
}

This will append each u8 to an array :

for (int k = size - 1;k > 0;k--)
        AppendChar(Str_pst, toAppend[k]);

Answer 1

You are really close, you can reverse in the array and add a '0' to the beginning with little effort, or you can leave it the way you have it and take care of it in main. Where I think you are getting wound around the axle is in the indexing of hexadecimalNumber in your function. While 7 produces one hex-digit, it should be at index zero in hexadecimalNumber (except you initialize i = 1 ) That sets up confusion in handling your conversion to string indexes . Just keep the indexes straight, initializing i = 0 and using hexadecimalNumber initialized to all zeros, if you only have a single character at index 1 , pad the string with 0 at the beginning.

Here is a short example that may help:

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

#define NCHR 32

void d2h (int n, char *hex)
{
    int idx = 0, ridx = 0;      /* index & reversal index */
    char revhex[NCHR] = "";     /* buf holding hex in reverse */

    while (n) {
        int tmp = n % 16;
        if (tmp < 10)
            tmp += '0';
        else
            tmp += '7';
        revhex[idx++] = tmp;
        n /= 16;
    }
    if (idx == 1) idx++;        /* handle the zero pad on 1-char */

    while (idx--) { /* reverse & '0' pad result */
        hex[idx] = revhex[ridx] ? revhex[ridx] : '0';
        ridx++;
    }
}

int main (int argc, char **argv) {

    int n = argc > 1 ? atoi (argv[1]) : 7;
    char hbuf[NCHR] = "";

    d2h (n, hbuf);

    printf ("int : %d\nhex : 0x%s\n", n, hbuf);

    return 0;
}

The 0x prefix is just part of the formatted output above.

Example Use/Output

$ ./bin/h2d
int : 7
hex : 0x07

$ ./bin/h2d 26
int : 26
hex : 0x1A

$ ./bin/h2d 57005
int : 57005
hex : 0xDEAD

If you do want to handle the reversal in main() so you can tack on the 0x07 if the number of chars returned in hexadecimalNumber are less than two, then you can do something similar to the following:

void d2h (int n, int *sz, char *hex)
{
    int idx = 0;
    while (n) {
        int tmp = n % 16;
        if (tmp < 10)
            tmp += '0';
        else
            tmp += '7';
        hex[idx++] = tmp;
        n /= 16;
    }
    *sz = idx;
}

int main (int argc, char **argv) {

    int n = argc > 1 ? atoi (argv[1]) : 7, sz = 0;
    char hbuf[NCHR] = "";

    d2h (n, &sz, hbuf);

    printf ("int : %d\nhex : 0x", n);
    if (sz < 2)
        putchar ('0');
    while (sz--)
        putchar (hbuf[sz]);
    putchar ('\n');

    return 0;
}

Output is the same

Look it over and let me know if you have further questions.