Convert unsigned short to char - c++

Question

I know there's other posts like this but I think mine is different. I have a group of numbers I wish to display. However, it's saved as unsigned short. As it's given to me from a network buffer, all my data is unsigned short format. So for a serial number starting with "ABC-" the two unsigned shorts will be holding 0x4142 and 0x432D (Already in ASCII format). I need to convert those to type char to display using printf and %s, but for the rest of my system, they need to remain as an unsigned short. This is what I've tried so far, but the output is blank:
unsigned char * num[3];
num[0] = (unsigned char*)(SYSTEM_N >> 8);
num[1] = (unsigned char*)(SYSTEM_N & 0x00FF);
printf("System Number: %s \r\n", num);
Can anyone shed some light on this for me? Thanks!

Answer 1

There are several errors: 1) array is too short, 2) defining the array as an array of pointers and 3) not terminating the string.
#include <stdio.h>
int main (void)
{
unsigned short system_m = 0x4142;
unsigned short system_n = 0x432D;
unsigned char num[5];
num[0] = system_m >> 8;
num[1] = system_m & 0xFF;
num[2] = system_n >> 8;
num[3] = system_n & 0xFF;
num[4] = '\0';
printf("System Number: %s \r\n", num);
return 0;
}
EDIT alternatively if you don't want to keep the string, just display the information with this:
#include <stdio.h>
int main (void)
{
unsigned short system_m = 0x4142;
unsigned short system_n = 0x432D;
printf("System Number: %c%c%c%c \r\n",
system_m >> 8, system_m & 0xFF, system_n >> 8, system_n & 0xFF);
return 0;
}
Program output:
System Number: ABC-

Answer 2

You probably meant to write
unsigned char num[3];
as you have it, you declare an array holding three char* pointers.
Also don't forget to set the closing NUL character, before printing:
num[2] = '\0';

Answer 3

A general solution might be something like this, assuming ushort_arr contains the unsigned shorts in an array and ushort_arr_size indicates its size.
char *str = malloc(ushort_arr_size * 2 + 1);
// check if str == NULL
int j = 0;
for (int i = 0; i < ushort_arr_size; i++) {
str[j++] = ushort_arr[i] >> 8;
str[j++] = ushort_arr[i];
}
str[j] = '\0';
printf("string is: %s\n", str);
free(str);
Though perhaps this might be more effective without the memory management, if you only want to print it once:
for (int i = 0, j = 0; i < ushort_arr_size; i++) {
putchar(ushort_arr[i] >> 8);
putchar(ushort_arr[i] & 0xFF);
}