difficulties with arrays and pointers

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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 3-Aug-11 3:11am

kutalinelucas

Updated 3-Aug-11 3:13am

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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'?)?

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OriginalGriff · Answer 1 · 2011-08-03T03:33:00

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);

krmed · Answer 2 · 2011-08-03T05:17:00

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.

Posted 3-Aug-11 5:17am

krmed

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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
Hexadecimal representation 0x41

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

వేంకటనారాయణ(venkatmakam) · Answer 3 · 2011-08-03T03:23:00

Solution 1

I think You can do like this,

C#

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

Posted 3-Aug-11 3:23am

వేంకటనారాయణ(venkatmakam)

Manfred Rudolf Bihy · Answer 4 · 2011-08-03T03:32:00

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

difficulties with arrays and pointers

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