Problem parsing 16-bit Hex to ASCII

I am writing a program for a class and I have an array of 16 bit ints and I need to make each entry in the array print two ASCII characters. I have tried a few different ways on my own but couldn't come up with a solution. And suggestions?

As always, post your attempt(s) and use code tags to encourage responses, and you will likely get the results you desire.

Believe me, my code isn't worth posting. It didn't even begin to work. It would compile and not do what I anticipated.

Believe me, it's not worth my time to help someone who won't follow the forum rules.


Quzah.

Okay, here's what I tried:

insert

Code:

---

void disassembler(int s_adr, int e_adr)
{
        int index;
        int memLocation=1;
        char* firstDigit, secondDigit;
        int examine;

        printf("\n\nMemory Dump:\n");
        for(index=s_adr;index<=e_adr;(index+=8))
        {
                int counter;
                printf("\n%#06x: ",index);

                for(counter=8;counter>0;counter--)
                {
                        printf(" %04x",LC3Memory[memLocation]);
                        memLocation++;
                }
               
                memLocation-=8;
                printf(" ");
                for(counter=8;counter>0;counter--)
                {
                        firstDigit=strtok(("%#04x",LC3Memory[memLocation]), "");
                        secondDigit=strtok(("%#04x",LC3Memory[memLocation]), "");
                        strcat(firstDigit,secondDigit);
                        examine= (int) firstDigit;

                        if(examine>32 && examine<126)
                                printf("%c",examine);
                        else
                                printf(".");
                }
        }
}

---

Okay, the LC3 is like a hypothetical computer for which I am supposed to create a simulator. This function is supposed to be a visual disassembly of the code. So, it prints the memory address then what is at each one of the memory locations and finally it is supposed to print a "." if the 8-bit ASCII value for each memory location is not within x21 and x7e. If the 8-bit value is within x21 and x7e it is supposed to print the character. This was not my first attempt, it was my latest. Obviously, it doesn't work.

Code:

---

                for(counter=8;counter>0;counter--)
                {
                        firstDigit=strtok(("%#04x",LC3Memory[memLocation]), "");
                        secondDigit=strtok(("%#04x",LC3Memory[memLocation]), "");
                        strcat(firstDigit,secondDigit);
                        examine= (int) firstDigit;

                        if(examine>32 && examine<126)
                                printf("%c",examine);
                        else
                                printf(".");
                }

---

Not fully understanding the question (I guess), I'd say that you don't need to strtok, but just

Code:

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firstdigit = &yourshort;
seconddigit = &yourshort + 1;

---

EDIT: What is the exact thing that you are attempting to do?

Okay, there is a 16-bit hex value at each one of the array locations LC3Memory[]. I need to make each one of these 16 bit values 2 8-bit ASCII values that can be printed to the screen.

Quote:

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Originally Posted by black_spot1984

Code:

---

void disassembler(int s_adr, int e_adr)
{
        int index;
        int memLocation=1; /* Arrays start at 0 usually, unless you have some reason to do otherwise... */
        char* firstDigit, secondDigit; /* 'secondDigit' isn't a pointer. Only 'firstDigit' is. */

---

---

There are many ways to split your two bits.

1 - Bitshift.

Code:

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unsigned short int sixteen = ...whatever...;
unsigned char byte1, byte2;
byte1 = sixteen & 0xFF;
byte2 = (sixteen >> 8) & 0xFF;

---

Here you just mask off the lower byte, stick that in one variable, then shift over eight places, and mask again. Here we're assuming a few things, like 8 bit bytes. (CHAR_BIT is more accurate.)

2 - Use a union.

Code:

---

union foo
{
    unsigned short int sixteen;
    unsigned char bytes[2];
} bar;
...
bar.bytes[0] = 'A';
bar.bytes[1] = 'B';
printf( "%u", bar.sixteen );

---

3 - Fun with pointers.

Code:

---

unsigned short int sixteen;
unsigned char *byte;

byte = (unsigned char *)&sixteen;
*byte = 'A';
byte++; /* move to the next byte */
*byte = 'B';

---

There are a few more, but those should provide you something to chew on.


Quzah.

All right, I'll play a little bit of devil's advocate.

Code:

---

#include <stdio.h>

int main(void)
{
  unsigned short  value = 0x3141;
  unsigned char  *byte  = (unsigned char*)&value;
  size_t i;
  for ( i = 0; i < sizeof value; ++i )
  {
      printf("byte[%lu] = %02X = '%c'\n", (long unsigned)i, byte[i], byte[i]);
  }
  return 0;
}

/* my output
byte[0] = 41 = 'A'
byte[1] = 31 = '1'
*/

---

Quote:

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/* my output
byte[0] = 41 = 'A'
byte[1] = 31 = '1'
*/

---

That's why I prefer shifting and masking - it is more platform independent (depends only on CHAR_BIT value, and it can be solved with masks depending on it)

Quote:

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Originally Posted by vart

That's why I prefer shifting and masking - it is more platform independent (depends only on CHAR_BIT value, and it can be solved with masks depending on it)

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Oh, I know. Sometimes I just try to provoke a little thought is all.

[edit]The assignment appears predisposed to a given conclusion.

Thank you guys!!! That worked perfectly. Sorry, I had just been working on that problem for so long that I couldn't think anymore!

Quote:

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Originally Posted by Dave_Sinkula

Oh, I know.

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Have no doubt in it. :)