[SOLVED] string to hex parser

avee137 · 09-10-2011, 10:27 AM

I need to implement string to hex parser .

My requirement is as below :
1.get 8 characters from console input.
say ab12de11
2. my program should store it as
char buf5[5] = {'0xab','0x12','0xde','0x11','\0'};

any relevant suggestion would be appreciated .

grail · 09-10-2011, 10:49 AM

Well what have you tried and where are you stuck? Looks like a simple homework type question.

Ian John Locke II · 09-10-2011, 10:55 AM

Scratch that. Misunderstood the question.

dugan · 09-10-2011, 11:09 AM

Do you know to handle arrays of arrays (and arrays of strings) in C?

avee137 · 09-10-2011, 03:07 PM

i figured out a limited parser for my use ;

code goes like this:

Code:

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

void DBG(char *str)
{
#ifdef DEBUG
	printf("%s\n",str);
#endif
}

char toHex(char ch1 , char ch2)
{
	char hexByte ;
	ch1 = ch1 << 4 ;
	hexByte = (ch1 | ch2 );
	return hexByte ;
}
char getHexEq(char inChar)
{
	switch (inChar)
	{
	case '0' :
		return 0 ;
		break ;
	case '1' :
			return 1 ;
			break ;
	case '2' :
			return 2 ;
			break ;
	case '3' :
			return 3 ;
			break ;
	case '4' :
			return 4 ;
			break ;
	case '5' :
			return 5 ;
			break ;
	case '6' :
			return 6 ;
			break ;
	case '7' :
			return 7 ;
			break ;
	case '8' :
			return 8 ;
			break ;
	case '9' :
			return 9 ;
			break ;
	case 'a' :
			return 10 ;
			break ;
	case 'b' :
			return 11 ;
			break ;
	case 'c' :
			return 12 ;
			break ;
	case 'd' :
			return 13 ;
			break ;
	case 'e' :
			return 14 ;
			break ;
	case 'f' :
			return 15 ;
			break ;
	default:
			printf("invalid character format\n");
			break ;
	}
}

int main()

{

	int i ,j;
	char ch1 ,ch2 ,chHex;
	char buf9[9] = {'\0'};

	char hexBuf4[4] = {'\0'};

	printf ("enter the 8 character hex id\n");
	scanf("%s",buf9);

	for(i = 0,j=0 ; i < 8,j<4  ; i+=2,++j)
	{
		ch1 = getHexEq(buf9[i]);
		ch2 = getHexEq(buf9[i+1]);

		chHex = toHex(ch1 , ch2);
		hexBuf4[j] = chHex ;

	}

	for(i = 0 ; i < 4  ; ++i)
	{
	printf(" 0x%x\n ", (unsigned)(unsigned char)hexBuf4[i] );
	}
	return 0;
}

ta0kira · 09-10-2011, 04:41 PM

Quote:

Originally Posted by avee137

i figured out a limited parser for my use ;

code goes like this:

Here are a few tips:

char values are actually numbers, but they're printed as characters and the zero value ('\0', not '0') has a special meaning for text. If you look up "ASCII table" on the web you'll see that the digits and the letters are respectively numbered sequentially. This means if you know a char val is a digit then you can get the integer form with val-'0'. Something similar can be done with letters, but uppercase and lowercase need to be dealt with separately.
When reading strings with scanf, always specify the maximum string length. Your program could crash if someone entered a string longer than 8 characters. In more "important" programs that can potentially be exploited to crack into your system. In this case, you'd use "%8s" as your format.
Some of your variable assignments can be skipped to make your code shorter. For example, toHex could be one line, as could your first for loop.

Kevin Barry