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?
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?
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.
Hope is the first step on the road to disappointment.
Okay, here's what I tried:
insertOkay, 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: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("."); } } }
Not fully understanding the question (I guess), I'd say that you don't need to strtok, but justCode: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("."); }Code: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.
There are many ways to split your two bits.Originally Posted by black_spot1984
1 - Bitshift.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.)Code:unsigned short int sixteen = ...whatever...; unsigned char byte1, byte2; byte1 = sixteen & 0xFF; byte2 = (sixteen >> 8) & 0xFF;
2 - Use a union.3 - Fun with pointers.Code:union foo { unsigned short int sixteen; unsigned char bytes[2]; } bar; ... bar.bytes[0] = 'A'; bar.bytes[1] = 'B'; printf( "%u", bar.sixteen );There are a few more, but those should provide you something to chew on.Code:unsigned short int sixteen; unsigned char *byte; byte = (unsigned char *)&sixteen; *byte = 'A'; byte++; /* move to the next byte */ *byte = 'B';
Quzah.
Hope is the first step on the road to disappointment.
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' */
Last edited by Dave_Sinkula; 12-05-2006 at 11:04 PM. Reason: Whoa! Pokey fingers need some sleep.
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
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)/* my output
byte[0] = 41 = 'A'
byte[1] = 31 = '1'
*/
All problems in computer science can be solved by another level of indirection,
except for the problem of too many layers of indirection.
– David J. Wheeler
Oh, I know. Sometimes I just try to provoke a little thought is all.
[edit]The assignment appears predisposed to a given conclusion.
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
Thank you guys!!! That worked perfectly. Sorry, I had just been working on that problem for so long that I couldn't think anymore!
Have no doubt in it.
All problems in computer science can be solved by another level of indirection,
except for the problem of too many layers of indirection.
– David J. Wheeler
