I have a hex value stored in a two byte array:
unsigned char hex[2] = {0x02, 0x00};
How can I convert this to a decimal value?
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1It is one of those nights... Feel free to give as much offense as possible as I deserve it. I am not sure why but the closer finals get the more I have horrible horrible lapses in "thinking". For some reason I was thinking there is more to it than that. I had my self convinced that it was much more complicated...tgai– tgai2011-04-26 00:23:04 +00:00Commented Apr 26, 2011 at 0:23
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1@Mitch Wheat: Don't worry about it. You comment actually made me pause for a second and realize that I did know the answer to my question. As I get closer to starting my coop I find my self horrified of making mistakes like this and looking rather dumb. It is probably good to be reminded that making mistakes and/or not remembering something of the top of my head is doomed to happen and not the end of the world.tgai– tgai2011-04-26 00:35:29 +00:00Commented Apr 26, 2011 at 0:35
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3 Answers
You can use (bitwise operation)
int b = (hex[0] << 8) | hex[1];
or (simple math)
int b = (hex[0] * 0x100) + hex[1];
Comments
Depends on endian-ness, but something like this?
short value = (hex[0] << 16) & hex[1];
2 Comments
Mateen Ulhaq
The "depends on what kind of Indian you've got" part's true.
tgai
Some reasons as to why this is not correct: First you are shifting the first char by 16 bits instead of 8 bits. A char is typically 8 bits. Second you are ANDing the two parts, you need to OR them. By shifting hex[0] over you are leaving the first 16 bits as zeros and when ANDed with hex[1] you will effectively remove hex[1]. (Since zero anded is always zero) Thanks for the input though!
This isn't an efficient way at least I do not think it is, but it would work in all situations regardless of size of the array and will easily convert to b.
__int8 p[] = {1, 1, 1, 1, 1}; //Let's say this was the array used.
int value = sizeof(p); //Just something to store the length.
memcpy(&value, &p, value > 4 ? 4 : value); //It's okay to go over 4, but might as well limit it.
In the above example it the variable "value" will have a value of 16,843,009. Which is equivalent to if you had done the below.
int value = p[0] | p[1] << 0x8 | p[2] << 0x10 | p[3] << 0x18;
