How do I convert HEXADECIMAL LPCTSTR to BYTE

Question

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I have below function

C++

void StrToByte2(LPCTSTR str, BYTE *dest)
{
	UINT count = _ttoi(str);
	BYTE buf[4] = { 0 };
	char string[10] = { 0 };
	sprintf_s(string, 10, "%04d", count);
	for (int i = 0; i < 4; ++i)
	{
		if ((string[i] >= '0') && (string[i] <= '9'))
			buf[i] = string[i] - '0';
	}
	dest[0] = (BYTE)(buf[0] << 4) | buf[1];
	dest[1] = (BYTE)(buf[2] << 4) | buf[3];
}

If i call this function on "1234" ( any digits) , dest output some 12814!

C++

struct st               
{
    byte    btID[2];
    int     nID;
};

PTR ptr(new st);
StrToByte2(strCode, ptr->btID);

but when i call this function on any hexadecimal ex A123 , it outputs 0000 always.

Below function is used to convert back the dest code to str

C++

CString Byte2ToStr(const byte* pbuf)
{
    CString str;
    str.Format(_T("%02X%02X"), pbuf[0], pbuf[1]);
    return str;
}

How can i get A123 to converted to bytes and than back to str to display A123??

Please help!!

What I have tried:

Tried below function but on convert back to str ,some i cannot see A123, but insted some random number like 1348

C++

void StrToByte2(LPCTSTR str, BYTE *dest)
{
	BYTE buf[4] = { 0 };
	char string[10] = { 0 };
	sprintf_s(string, 10, "%04d", str);
	for (int i = 0; i < 4; ++i)
	{
		if ((string[i] >= '0') && (string[i] <= 'F'))
			buf[i] = string[i] - '0';
	}
	dest[0] = (BYTE)(buf[0] << 4) | buf[1];
	dest[1] = (BYTE)(buf[2] << 4) | buf[3];
}

Posted 31-Oct-18 7:10am

Member 14039607

Updated 31-Oct-18 11:47am

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Comments

KarstenK 31-Oct-18 14:32pm

Solution 1 is correct. Because these conversion are common tasks a lot of function are available. Use them, because they are proofed and tested code and are secure even in edge cases. ;-)

Rick York 31-Oct-18 18:43pm

If your numbers have typing prefixes like 0x for hexadecimal and 0 for octal then strtoul and _tcstoul can handle those. In fact, it takes a base as an argument so you can give it binary or any other base you want to.

2 solutions

Solution 1

#include <cstdlib>

int main()
{
    char * p = NULL;
    long n = strtol( "abcd", & p, 16 );
    printf("%ld\r\n", n);
}

Posted 31-Oct-18 7:41am

TheGreatAndPowerfulOz

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This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

CPallini · Accepted Answer · 2018-10-31T11:48:00

As suggested, use the (generic-text variant of) strtol:

C++

void StrToByte2(LPCTSTR str, BYTE * dest)
{
	LONG u16 = _tcstol( str, NULL, 16);
	dest[0] = (BYTE) (u16 >> 8);
	dest[1] = (BYTE) (u16);
}

or at least, write a sensible hand-crafted one. Try

C++

void StrToByte2Alt(LPCTSTR str, BYTE dest[])
{
	for (size_t b=0; b<2; ++b)
	{
		dest[b] = 0;
		for (size_t nb=0; nb<2; nb++)
		{
			dest[b] <<= 4;
			TCHAR tc = str[2*b+nb];
			if ( tc >= _T('0') && tc <= _T('9') )
				dest[b] |= tc - _T('0');
			else if ( tc >= _T('A') && tc <= _T('F') )
				dest[b] |= tc - _T('A') + 10;
			else if ( tc >= _T('a') && tc <= _T('f') )
				dest[b] |= tc - _T('a') + 10;
			else
			{// throw all the throwable here...
			}
		}
	}
}

or, using C++ streams:

C++

#ifdef UNICODE
    typedef std::wistringstream InputStringStream;
#else
    typedef std::istringstream InputStringStream; 
#endif // UNICODE

BOOL strtobyte2(LPCTSTR str, std::array<BYTE, 2> & b)
{
	UINT16 u16;
	InputStringStream iss(str);
	iss >> hex >> u16;
	if ( ! iss) return FALSE; 
	b[0] = static_cast<BYTE>(u16 >> 8);
	b[1] = static_cast<BYTE>(u16);
	return TRUE;
}

How do I convert HEXADECIMAL LPCTSTR to BYTE

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Solution 2

Solution 1

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How do I convert HEXADECIMAL LPCTSTR to BYTE

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