How to convert char[] array of hex characters to a byte array of values? [duplicate]

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Hex to char array in C

I have a char[10] array that contains hex characters, and I'd like to end up with a byte[5] array of the values of those characters.

In general, how would I go from a char[2] hex value (30) to a single decimal byte (48)?

Language is actually Arduino, but basic C wou开发者_JAVA百科ld be best.

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1 byte = 8 bit = 2 x (hex digit)

What you can do is left shift the hex-digit stored in a byte by 4 places (alternatively multiply my 16) and then add the 2nd byte which has the 2nd hex digit.

So to convert 30 in hex to 48 in decimal:

1. take first hex digit, here 3; multiply it by 16 getting 3*16 = 48
2. add the second byte, here 0; getting 48+0 = 48 which is your final answer

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Char array in "ch", byte array as "out"

byte conv[23] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, -1, -1, -1, -1, -1, -1, -1, 10, 11, 12, 13, 14, 15};

// Loop over the byte array from 0 to 9, stepping by 2
int j = 0;
for (int i = 0; i < 10; i += 2) {
  out[j] = conv[ch[i]-'0'] * 16 + conv[ch[i+1]-'0'];
  j++;
}

Untested.

The trick is in the array 'conv'. It's a fragment of an ASCII table, starting with the character for 0. The -1's are for the junk between '9' and 'A'.

Oh, this is also not a safe or defensive routine. Bad input will result in crashes, glitches, etc.