Convert a 3-byte array into a single size integer?

I have a file I am trying to decode and the size of the portion of the file I need is specified by 3 bytes in its header. How can I convert these 3 separate bytes into 1 single integer?

Thanks for any help you have.

What, exactly, does each of the bytes represent? Are they part of a single 24 bit integer?

Viggy

Quote:

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Originally Posted by MrViggy

What, exactly, does each of the bytes represent? Are they part of a single 24 bit integer?

Viggy

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Yes they are part of a single 24 bit integer, but I can't figure out how to get the program to recognize them as a single integer, and not 3 different bytes in an array...

I don't know how to create a 24-bit integer, but if it were a 32-bit integer conversion would look something like this, where buffer contains the binary representation of a 4-byte integer. I assume a 24-bit integer would be similar. I know short = 16 bit integer, long = 32 bit. I don't know what is a 24 bit. Yes, I know those sizes are not guarenteed, but that's what they are on 16 and 32 bit operating systems such as MS-DOS, MS-Windows and *nix.

Code:

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int n = *(int *)buffer;

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Is this not a possible solution

Code:

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        BYTE arr[3] = {0x23, 0x45, 0x67};
        UINT val = arr[2] << 16 | arr[1] << 8 | arr[0];

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I have assumed that arr[2] is the MSB and arr[0] is LSB.

Thanks for the help it worked perfectly!

anyone know how to do this in .net?

Quote:

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Originally Posted by DaBoonDockSaint

anyone know how to do this in .net?

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Yes, in the exact same way. What does using .Net have to do with anything if it is a c++ program? You still have the same problem and the same kind of solution.