Putting Hex into a byte array - c

Question

I am trying to put hex values into a Byte[], Trying to achieve 78,66,1E,B5,4F,E7,67,63
#define BYTE unsigned char
int num = 1;
long long hex[]
{
0x78661EB54FE76763,
};
int main()
{
for (int c = 0; c < num; c++)
{
printf("%llx\n", hex[c]);
unsigned int number = hex[c];
unsigned int ef = number & 0xff;
unsigned int cd = (number >> 8) & 0xff;
unsigned int ab = (number >> 16) & 0xff;
printf("%x", number & 0xff);
BYTE data2[8]{
ef, cd, ab
};
}
}
Update:
Basically I have an array of 30 odd hex values. I am trying to loop through that array called hex[], then for each hex value separate it into 2's ie 78,66,1E,B5,4F,E7,67,63 then add each one into an array of type BYTE which will hold the hex value in 8 pairs so data[0] will have the value of 78 up to data[8] which will have the value of 63, So I can then pass the array of type BYTE to another method for further work

Answer 1

Here is the solution that you want:
#include <stdio.h>
typedef unsigned char BYTE;
int main()
{
int i,j;
long long hex=0x78661EB54FE76763;
BYTE val[8];
for(j=0,i=7; i>=0; i--,j++){
val[j]= (hex>>(i*8))&0xff;
}
printf("Hexadecimal bytes are:\n");
for(j=0;j<8;j++){
printf("%02X ",val[j]);
}
return 0;
}
And the output is:
Hexadecimal bytes are:
78 66 1E B5 4F E7 67 63

Answer 2

Assuming that BYTE is a type
BYTE data2[] = {ef, cd, ab, '\0'};
Use the null terminator to allow printing with "%s" and printf().
Of course, you can add the rest of the bytes.

Answer 3

This is a pretty simple task to accomplish all you need is the initialization values that you have given. long long hex[] = { 0x78661EB54FE76763 };
Next, you need a char pointer that is pointing to the hex array
unsigned char* pByte = (unsigned char*)hex; // Point to the first element in hex array
Then create an 8 byte buffer and clear it with memset:
unsigned char byteArray[8];
memset(byteArray, 0, sizeof(byteArray));
All you need to do now is dereference your char pointer and place the value into your 8 byte buffer.
// Loop through the hex array and assign the current byte
// the byte pointer is pointing to then increment the byte pointer
// to look at the next value.
// This for loop is for a system that is little endian (i.e. win32)
for (int i = 7; i >= 0; i--)
{
byteArray[i] = *pByte;
pByte++;
}
Note: Since I ran this with Visual Studio 2015 everything is little endian. This is why the first for loop is written the way it is.
See below for the full program listing I wrote & tested below demonstrating this:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
long long hex[] = { 0x78661EB54FE76763 }; // Value of an element
unsigned char* pByte = (unsigned char*)hex; // Assign a char pointer to point to the first element in hex array
unsigned char byteArray[8]; // This is the byte buffer that will be populated from the hex array
memset(byteArray, 0, sizeof(byteArray)); // Clear out the memory of byte array before filling in the indices
// Loop through the hex array and assign the current byte
// the byte pointer is pointing to then increment the byte pointer
// to look at the next value.
// This for loop is for a system that is little endian (i.e. win32)
for (int i = 7; i >= 0; i--)
{
byteArray[i] = *pByte; // Dereference the pointer and assign value to buffer
pByte++; // Point to the next byte
}
// Print the byte array
for(int i = 0; i < 8; i++)
{
printf("The value of element %i is: %X\n", i, byteArray[i]);
}
printf("Press any key to continue..."); // Hold the console window open
getchar();
return 0;
}