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I'm having great difficulty changing a hex value into an integer.

In an unsigned char INBuffer[2], there is a hex code in the format 0x0a etc. I would like to convert this value into a long or int variable using the strtol function:

C#
`long int strtol ( const char * str, char ** endptr, int base );`

For this I'm assuming I need to get the value contained within INBuffer[2] into a string pointer, but no matter how many examples I look up i just can't seem to get my head around pointers. Could somebody please help?

M.
Posted
Updated 3-Aug-11 3:13am
v2
CPallini 3-Aug-11 9:31am
What exactly INBuffer contains? Could you be more precise (it cannot contain the string "0x0a", do you mean INBuffer[0]='0', INBuffer[1]='a'?)?

## Solution 3

That seems a little confused to me: an unsigned char buffer with two elements can't hold a hex code in the format "0x0a" since that is four characters.

If you mean that there is a value in the two unsigned chars that equates to the hex value 0A, and that it has a range between 0000 and FFFF hex, then you need to do this:
C++
`int i = ((int) INBuffer[0] & 0xFF) + (((int) INBuffer[1] << 8) & 0xFF00);`

(The ANDs are there to prevent sign extentions, which shouldn't happen anyway) You may need to swap round the array indexes depending on your data format: big- or little-endian

If you mean that INBuffer contains two bytes which are '0' and 'A' respectively, then you need to copy them to a character array big enough to hold three bytes, and make it a null terminated string:
C++
```unsigned char data[3];
data[0] = INBuffer[0];
data[1] = INBuffer[1];
data[2] = '\0';```
You can then use that to convert it to a int:
C++
```char * pEnd;
long int i = strtol(data, &pEnd, 16);```

## Solution 5

I'm confused by your question, but let's try this...

You have an unsigned char INBuffer[2]. This means it's a two character buffer. The two elements of this buffer can be accessed individually by using INBuffer[0] and INBuffer[1].

If you want the decimal value contained in INBuffer[0], you can do this
`int buf0val = (int)INBuffer[0];`

Likewise if you want the value contained in the second character of this buffer, you can do this
`int buf1val = (int)INBuffer[1]`

However if you mean that the first character contains the ASCII number 0 and the second character contains the ASCII 'A', the you have a different problem.

You can't use strtol on this, because this (two character) string is not NULL terminated - just as OriginalGriff said above.

Hope that helps.

kutalinelucas 3-Aug-11 12:25pm
i can absolutly see where you are all coming from, the thing that is onfusing me is that if I compaire the contents of INBuffer[2] with say 0x02 (if that is what I program the firmware to return) I can find a match, where I thought each slot of an array could contain only one character. The INBuffer is declaired as 'unsigned char INBuffer[64]...why is a complete mistery to me. Guess I'll just have to put in a massive switch statement to test its contents as above.
krmed 3-Aug-11 13:14pm
I think you are sadly confused. If you compare INBuffer[2] with 0x02 like this

if (INBuffer[2] == 0x02)you are comparing only a single BYTE.

Each character of INBuffer can only contain ONE character. It's only one byte long. How that charcter is represented is a different matter. Let's say INBuffer[2] contains the letter A.

Here are some of the ways that can be represented:

Decimal representation 65

ASCII representation A
Octal representation 0101

Binary representation 1000001

So you see, 0x0A is really a decimal value of 10, which is stored in a single byte. However "0x0A" (as a string) is a character string that is 4 bytes long, plus one byte for the terminating null. To show this stored in INBuffer, you would have:

INBuffer[0] = '0' or 0x30 or 48, etcINBuffer[1] = 'x' or 0x78 or 120, etc

INBuffer[2] = '0' or 0x30 or 48, etcINBuffer[3] = 'A' or 0x41 or 65, etc

INBuffer[4] = 0 (the terminating NULL)

As you can see, each individual character contains a value between 0 and 255, and this can be stored in a single byte.

Hope this helps.
kutalinelucas 3-Aug-11 16:45pm
Sorry for the late reply, I had work to go to...but so simple! thankyou sooo much for that, you've saved me hours of head scratching

## Solution 1

I think You can do like this,

C#
```char*chVal  = "0x0a";
cout<<"size:: "<<sizeof(chVal)<<endl;
long lVal = strtol(chVal, NULL, 16);
```

## Solution 2

Did you by any means notice that a `0x0a` is four characters while your input buffer is declared as `char INBuffer[2]` which can hold only 2 characters?

C++
```/* simple strtol example */
#include <stdio.h>
#include <stdlib.h>
int main ()
{
char INBuffer[] = "0x0a 0xff";
char * endOfScan;
long int l1, l1;
l1 = strtol (INBuffer,  &endOfScan, 16);
l2 = strtol (endOfScan, &endOfScan, 16);
printf ("The decimal equivalents of %s are: %ld and %ld.\n", INBuffer, l1, l2);
return 0;
}```

The endPtr parameters is used to tell where the scanner stopped recognizing a long the last time it was called. That's why it is passed as a pointer to a pointer. When you now pass the address of variable endOfScan this character pointer will point to the last char the scanner didn't recognize as belonging to the last long.

Cheers! I hope this helped.

—MRB

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