Thread: hex to int

hi
i'm trying to make a program that can convert hex to int..
if user enters whole bunch of hex then the program read those as character string and read string by 2 characters and convet 2 digit hex to int..

Code:

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

//int sub1(char digit);
unsigned char sub2(char digit[2]);
void sub3(char *hex, unsigned char *dec);

main()
{
	char hex[] = "B22101";
	unsigned char dec;
	int len;

	len = strlen(hex);
	printf("%s\n", hex);
	printf("%d\n", len);
	sub3(hex, &dec);
	//printf("%d", dec);
}

void sub3(char *hex, unsigned char *dec)
{
	int len;
	int i;

	len = strlen(hex);
	printf("len = %d\n", len);
	printf("hex = %s\n", hex);

	for(i=0; i<len/2; i++)
	{
		dec[i] = sub2(hex[2*i]);
	}
}

unsigned char sub2(char *digit[2])
{
//	int i1, i2;
	printf("%s", digit);
return 0;
}

but my source code gives me 3 warnings..
at dec[i] = sub2(hex[2*i]);

warnings are..
warning C4047: 'function' : 'char *' differs in levels of indirection from 'char '
warning C4024: 'sub2' : different types for formal and actual parameter 1
warning C4028: formal parameter 1 different from declaration

I don't suppose you've thought about using sscanf().

Code:

char hex[] = "B22101";
int dec;

sscanf(hex, "%x", &dec);
printf("%d\n", dec);

As for the warnings, your prototype and definition for sub2() don't match:
>unsigned char sub2(char *digit[2])
unsigned char sub2(char digit[2])

And I think the call here:
dec[i] = sub2(hex[2*i]);
Would instead be:
dec[i] = sub2(&hex[2*i]);

It doesn't do either of you any good to return -1 on error, if you don't bother checking the return value of the function to make sure it isn't -1...

Quzah.

Of course Quzah, but the error must be known, either way...

Code:

#include<stdio.h>
#include<conio.h>

#define HEXSIGNED 0x01
#define HEXUNSIGNED 0x02

int hexcharToInt(char digit){
	if('0'<=digit && digit<='9')
		return digit-'0';
	if('A'<=digit && digit<='F')
		return digit-'A'+10;
	if('a'<=digit && digit<='f')
		return digit-'a'+10;
	return -1;
}

int hexToInt(char *h, int flags){
	int i, result=0, isnegative=0;
	int tmp;
	
	switch(flags){
	case HEXSIGNED:
		tmp = hexcharToInt(h[0]);
		if(tmp == -1) return 0;

		if(tmp & 0x80)//has negative signal
			isnegative=1;
		else
			isnegative=0;
		break;
	case HEXUNSIGNED:
		isnegative=0;
		break;
	default:
 		return 0;
	}
	
	if(isnegative){ //parse the number in complement to 2
		for(i=0;h[i] && h[i]=='f' && i<8;i++);//skip f's
		for(;h[i] && i<8;i++){
			tmp = hexcharToInt(h[i]);
			if(tmp == -1) return -result-1;
			result=result*0x10+0x0f-tmp;   		
		}
		result = -result-1;
	}else
		for(i=0;h[i] && i<8;i++){
			tmp = hexcharToInt(h[i]);
			if(tmp == -1) return result;
			result=result*0x10+tmp;
		}
	
	return result;
}

int main(){
	char s[9];
	int i, k;
	
	for(i=-300, k=0;i<300;i++, k=(k==20?0:k+1)){
		sprintf(s, "%x", i);	
		printf("%d\t%2s\t%d\n", i, s, hexToInt(s, HEXSIGNED));
		if(k==20)
			getch();
	}
	return 0;
}

This way negative number are also parsed, and the function break on the first non hex char. I've included a little main for testing.
NOTE: one of the for cicles termination condition is i>=8 because a 32bit integer has only 8 4-bit block, or, 8 hex chars. If you dont' want this, remove it.

Of course Quzah, but the error must be known, either way...

I'm not sure how to parse that sentence. My point was, in the example before your code, the negative number would be processed anyway:

Code:

unsigned char sub2(char digit[2])
{
	int i1, i2;
	i1 = sub1(digit[0]);
	i2 = sub1(digit[1]);
	return i1*16+i2;
}

Of course, in their example, they should have paid attention to their warnings, and they would have noticed a "control reaches end of non-void function".

Quzah.

Originally Posted by xErath

What's the fun about that???

That's true, and in addition you don't learn much about HEX. It does make for a good verification source though.