unsigned char* values to hexadecimal [duplicate] - c++

Question

This question already has answers here:
Convert two ASCII Hexadecimal Characters (Two ASCII bytes) in one byte
(6 answers)
Closed 6 years ago.
I have a typical use case where i need to convert the unsigned char values to hexadecimal.
For example,
unsigned char *pBuffer = (unsigned char *)pvaluefromClient //where pvaluefromclient is received from a client
The length of pBuffer is 32 bytes and it holds the value as follows,
(gdb) p pBuffer
$5 = (unsigned char *) 0x7fd4b82cead0 "EBA5F7304554DCC3702E06182AB1D487"
(gdb) n
STEP 1: I need to split this pBuffer value as follows,
{EB,A5,F7,30,45,54,DC,C3,70,2E,06,18,2A,B1,D4,87 }
STEP 2: I need to convert the above splited values to decimal as follows,
const unsigned char pConvertedBuffer[16] = {
235,165,247,48,69,84,220,195,112,46,6,24,42,177,212,135
};
Any idea on how to achieve the STEP1 and STEP2? any help on this would be highly appreciated

Answer 1

How about something like this:
unsigned char *pBuffer = (unsigned char *)pvaluefromClient //where valuefromclient is received from a client
int i, j;
unsigned char target[16]
for(i=0;i<32;i+=2)
{
sscanf((char*)&pBuffer[i], "%02X", &j);
target[i/2] = j;
}

Answer 2

You can create a function that takes two unsigned chars as parameter and returns another unsigned char. The two parameters are the chars (E and B for the first byte). The returned value would be the numerical value of the byte.
The logic would be :
unsigned char hex2byte(unsigned char uchar1, unsigned char uchar2) {
unsigned char returnValue = 0;
if((uchar1 >= '0') && (uchar1 <= '9')) {
returnValue = uchar1 - 0x30; //0x30 = '0'
}
else if((uchar1 >= 'a') && (uchar1 <= 'f')) {
returnValue = uchar1 - 0x61 + 0x0A; //0x61 = 'a'
}
else if((uchar1 >= 'A') && (uchar1 <= 'F')) {
returnValue = uchar1 - 0x41 + 0x0A; //0x41 = 'A'
}
if((uchar2 >= '0') && (uchar2 <= '9')) {
returnValue = (returnValue <<8) + (uchar2 - 0x30); //0x30 = '0'
}
else if((uchar2 >= 'a') && (uchar2 <= 'f')) {
returnValue = (returnValue <<8) + (uchar2 - 0x61 + 0x0A); //0x61 = 'a'
}
else if((uchar2 >= 'A') && (uchar1 <= 'F')) {
returnValue = (returnValue <<8) + (uchar2 - 0x41 + 0x0A); //0x41 = 'A'
}
return returnValue;
}
The basic idea is to calculate the numerical value of the chars and to reassemble a number from two chars (hence the bit shift)
I'm pretty sure there are multiple more elegant solutions than mine here and there.

Answer 3

void Conversion(char *pBuffer, int *ConvertedBuffer)
{
int j = 0;
for(int i = 0; i < 32; i += 2)
{
std::stringstream ss;
char sz[4] = {0};
sz[0] = pBuffer[i];
sz[1] = pBuffer[i+1];
sz[2] = 0;
ss << std::hex << sz;
ss >> ConvertedBuffer[j];
++j;
}
}
int main()
{
char Buffer[] = "EBA5F7304554DCC3702E06182AB1D487";
int ConvertedBuffer[16];
Conversion(Buffer, ConvertedBuffer);
for(int i = 0; i < 16; ++i)
{
cout << ConvertedBuffer[i] << " ";
}
return 0;
}
//output:
235 165 247 48 69 84 220 195 112 46 6 24 42 177 212 135