C/C++ read a byte from an hexinput from stdin

Question

Can't exactly find a way on how to do the following in C/C++.

Input : hexdecimal values, for example: ffffffffff...

I've tried the following code in order to read the input :

uint16_t twoBytes;
scanf("%x",&twoBytes);

Thats works fine and all, but how do I split the 2bytes in 1bytes uint8_t values (or maybe even read the first byte only). Would like to read the first byte from the input, and store it in a byte matrix in a position of choosing.

uint8_t matrix[50][50]

Since I'm not very skilled in formating / reading from input in C/C++ (and have only used scanf so far) any other ideas on how to do this easily (and fast if it goes) is greatly appreciated .

Edit: Found even a better method by using the fread function as it lets one specify how many bytes it should read from the stream (stdin in this case) and save to a variable/array.

size_t fread ( void * ptr, size_t size, size_t count, FILE * stream );

Parameters

ptr - Pointer to a block of memory with a minimum size of (size*count) bytes.

size - Size in bytes of each element to be read.

count - Number of elements, each one with a size of size bytes.

stream - Pointer to a FILE object that specifies an input stream.

cplusplus ref

Answer 1

%x reads an unsigned int, not a uint16_t (thought they may be the same on your particular platform).

To read only one byte, try this:

uint32_t byteTmp;
scanf("%2x", &byteTmp);
uint8_t byte = byteTmp;

This reads an unsigned int, but stops after reading two characters (two hex characters equals eight bits, or one byte).

Answer 2

You should be able to split the variable like this:

uint8_t LowerByte=twoBytes & 256;
uint8_t HigherByte=twoBytes >> 8;

Answer 3

A couple of thoughts:

1) read it as characters and convert it manually - painful

2) If you know that there are a multiple of 4 hexits, you can just read in twobytes and then convert to one-byte values with high = twobytes << 8; low = twobyets & FF;

3) %2x