How to hex2ascii???

[Guest_imported]

HI
im a newbie
i want a source code that converts the hex into ascii, without using the console prompt...
for example: i have 2 variables:
char mychar1="96";
char mychar2;

how can i put mychar1 hex value in mychar2 (naturally in ascii character)

at the end i'll have mychar1="96" and mychar2="i"

Can anyone help me please???

Thanx a lot!

 

[keenanbr]

are you asking 'how do i convert hex to decimal ?'. If so, the answear is to write a simple function. I don't know of any library containing one. If you just want to print, all you do is 'printf("hex is %x decimal is %d\n", mychar, mychar);

 

[Guest_imported]

yes, you understood what i mean... but i have to do that, printing nothing in the console...i have to print the result in a variable, not in the console... how to?

 

[Leibnitz]

If you want to print your results into a variable,you can use the function "sprintf".

char mychar1="96";
char mychar2;
char Buffer[100]; /*this "buffer" will receive the results*/

....
.....
sprintf(Buffer,"hex is %x decimal is %d\n", mychar, mychar);/*prints the results into the "buffer"*/

 

[Guest_imported]

your code works only if i set mychar="0x69"(69, and not 96, becuse i wronged the number when i posted the question the first time)
...the number 69 is an hex value...0x69 is equal to "i"...
if i write:

char mychar1="69";
char Buffer[100];

sprintf(Buffer,"hex is %x decimal is %c\n", mychar1, mychar1);

at the and i have Buffer="hex is (the corresponding hex value of "69&quot decimal is (the corresponding decimal value of "69&quot! instead i wanna have Buffer="hex is 69 decimal is i"...because i want to keep the number 69 as an hex value...
you understand what i mean?

 

[Leibnitz]

Maybe this will do !?

void main( void )
{
int value1 = 0x69;
char Buffer[100];
sprintf(Buffer,"C hex: 0x%x\n decimal is %d\n", value1, value1);
printf("%s",Buffer);/*just for verification*/
}

 

[lordblood]

If your trying to take a character string "69" (in hex) and make a decimal representation (105 in decimal).

This is just something to think about. I have not fully tested it ... just whipped something up. I'm sure someone can do better ;-)


Try this(take a string and figure out the dec value):

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

main()
{
	int	i = 0;

	long	int dec = 0;

	double	hold = 0.0;

        char    hex[] = &quot;69\0&quot;,
		b[64];

	strcpy(b,hex);

	for(i = (strlen(hex) - 1); i >= 0; i--)
	{
		switch(hex[i])
		{
			case '0':
			case '1':
			case '2':
			case '3':
			case '4':
			case '5':
			case '6':
			case '7':
			case '8':
			case '9':
		dec += (atoi(&hex[i]) * pow(16.0,hold));
		break;

			case 'A':
			case 'a':
		dec += (10 * pow(16.0,hold));
		break;
			case 'B':
			case 'b':
		dec += (11 * pow(16.0,hold));
		break;
			case 'C':
			case 'c':
		dec += (12 * pow(16.0,hold));
		break;
					
			case 'D':
			case 'd':
		dec += (13 * pow(16.0,hold));
		break;
					
			case 'E':
			case 'e':
		dec += (14 * pow(16.0,hold));
		break;
			case 'F':
			case 'f':
		dec += (15 * pow(16.0,hold));
		break;

			default: printf(&quot;NOT HEX VALUE\n&quot;);
				 exit(1);
				 break;
		}

		hex[i] = 0;
		hold += 1;
	}

	printf(&quot;Hex is %s ... Decimal is %ld\n&quot;, b, dec);

        return 0;
}

 

[Skatanic]

Here is an overly simple way to do it:

#include <stdio.h>

int main() {
char *hexNum = &quot;69&quot;;
long intergerNum;
char c;

integerNum = strtol(hexNum, 0, 16);
printf(&quot;%d\n&quot;, integerNum); // Should print &quot;105&quot;
c = (char) integerNum;
printf(&quot;%c\n&quot;, c); // Should print &quot;i&quot;

return(1);
}

The prototype for strol is:

long strtol(char *s, char **end_ptr, int base);

The end_ptr is a pointer to where the function stops converting hex values (ie. when it find a non-hex character like 'L'). You can set that to NULL and be fine.

Hope this helps...
-Skatanic

 

[Guest_imported]

ok, thanks very much at all!!! I solved my problem, bye bye!!!

 

[Guest_imported]

Uhm... i have another question: this is the code that posted Skatanic:

#include <stdio.h>

int main() {
char *hexNum = &quot;69&quot;;
long intergerNum;
char c;

integerNum = strtol(hexNum, 0, 16);
printf(&quot;%d\n&quot;, integerNum); // Should print &quot;105&quot;
c = (char) integerNum;
printf(&quot;%c\n&quot;, c); // Should print &quot;i&quot;

return(1);
}

...how could i use it to convert a string of hex numbers???? for example if i start with char*hexNum=&quot;696F616F&quot; instead of char*hexNum=&quot;69&quot;
how could i?

 

[Skatanic]

That code should work with the hex numbers. strtol recognizes characters 0-9 and A-F when 16 is passed as the base.

 

[Guest_imported]

yes...but it recognizes a range of hex numbers too...doesn't it? if i have char *hexNum=&quot;696F616F&quot;, with integerNum = strtol(hexNum, 0, 16); i pass hexNum in integerNum (that become &quot;0x696F616F&quot...ho can i pass this number to a char variable?

 

[abbeman]

Do you mean that you want to convert &quot;696F616F&quot; into &quot;ioao&quot;?

If so you can modify the above code to:

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

int main() {
char *hexNum = &quot;696F616F&quot;;
long integerNum;
char c;
int i;

for (i = 0; i < strlen(hexNum); i += 2) {
char *tmp = &quot; &quot;;
memcpy(tmp, hexNum + i, 2);
integerNum = strtol(tmp, 0, 16);
printf(&quot;%d\n&quot;, integerNum);
c = (char) integerNum;
printf(&quot;%c\n&quot;, c);
}

return(1);
}

 

[Guest_imported]

Cool!!! It works perfectly thanks a lot at all!!!

 

[zaq888]

how to convert ascii to OCTAL CODE?

I'm Keep Studying.... please show the way...
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