Some body help me regarding to the following problem
strFixFactorSide = _T("0.5");
dFixFactorSide = atof((const char *)(LPCTSTR)strFixFactorSide);
"dFixFactorSide" takes value as 0.0000;
How I will get correct value?
Use _tstof() instead of atof(), and cast CString to LPCTSTR, and leave it as such, instead of trying to get it to const char *. Forget about const char * (LPCSTR) while you're working with unicode and use only const _TCHAR * (LPCTSTR).
int _tmain(int argc, TCHAR* argv[], TCHAR* envp[])
{
int nRetCode = 0;
CString s1 = _T("123.4");
CString s2 = _T("567.8");
double v1 = _tstof((LPCTSTR)s1);
double v2 = _tstof((LPCTSTR)s2);
_tprintf(_T("%.3f"), v1 + v2);
return nRetCode;
}
and running this correctly gives the expected answer.
I think your CString strFixFactorSide is a Unicode (UTF-16) string.
If it is, the cast (const char *) only changes the pointer type, but the string it points to still remains Unicode.
atof() doesn't work with Unicode strings. If you shove L"0.5" into it, it will fetch bytes 0x30 ('0') and 0x00 (also part of UTF-16 '0'), treat that as a NUL-terminated ASCII string "0" and convert it to 0.0.
If CString strFixFactorSide is a Unicode string, you need to either first convert it to an ASCII string and then apply atof() or use a function capable of converting Unicode strings to numbers. _wtof() can be used for Unicode strings.
Related
I am writing small application which uses com technology. I want to convert BSTR string to an unsigned char. To do this, i used W2A() Macro to convert from BSTR to String and then copied String.C_STR() to an unsigned char array. The code snippet is as follows:
Send(BSTR *packet, int length)
{
std::string strPacket = W2A(*packet);
unsigned char * pBuffer = new unsigned char [strPacket.length()+1];
memset(pBuffer,0,strPacket.length()+1);
memcpy(pBuffer,strPacket.c_str(),strPacket.length()+1);
}
This works fine when packet contains normal string. But if the packet contains a NUL character in it, the problem occurs. Some unknown characters appear after that NUL in the pBuffer i.e, after conversion.
Can anyone please let me know how to avoid that? Or is there any other way to do it correctly?
A BSTR is a Windows API type and must be managed with API macros or functions. If you cannot use W2A macro because your string may have null chars inside, you will have to use functions as WideCharToMultiByte that can convert from wide characters of BSTR to narrow chararacters for a char*. Be sure to have the SDK documentation. Alternatively, you could make you program use WCHARs
I want to know if is there any way to convert a unicode code to a string or char in C++ 11.
I've been trying with extended latin unicode letter Á (as an example) which has this codification:
letter: Á
Unicode: 0x00C1
UTF8 literal: \xc3\x81
I've been able to do so if it's hardcoded as:
const char* c = u8"\u00C1";
But if i got the byte sequence as a short, how can I do the equivalent to get the char* or std::string 'Á'?
EDIT, SOLUTION:
I was finally able to do so, here is the solution if anyone needs it:
std::wstring ws;
for(short input : inputList)
{
wchar_t wc(input);
ws += wc;
}
std::wstring_convert<std::codecvt_utf8<wchar_t>> cv;
str = cv.to_bytes(ws);
Thanks for the comments they were very helpful.
The C++11 standard contains codecvt_utf8, which converts between some internal character type (try char16_t if your compiler has it, otherwise wchar_t) and UTF-8 encoding.
The problems is that char is only one byte length, while unicode characters require a size of two bytes.
You can still treat it as char*, but you must remember that you are not dealing with an ascii string (there will be zeros).
You may have to switch to wchar_t.
BYTE name[1000];
In my visual c++ project there is a variable defined name with the BYTE data type. If i am not wrong then BYTE is equivalent to unsigned char. Now i want to convert this unsigned char * to LPCTSTR.
How should i do that?
LPCTSTR is defined as either char const* or wchar_t const* based on whether UNICODE is defined or not.
If UNICODE is defined, then you need to convert the multi-byte string to a wide-char string using MultiByteToWideChar.
If UNICODE is not defined, a simple cast will suffice: static_cast< char const* >( name ).
This assumes that name is a null-terminated c-string, in which case defining it BYTE would make no sense. You should use CHAR or TCHAR, based on how are you operating on name.
You can also assign 'name' variable to CString object directly like:
CString strName = name;
And then you can call CString's GetBuffer() or even preferably GetString() method which is more better to get LPCTSTR. The advantage is CString class will perform any conversions required automatically for you. No need to worry about Unicode settings.
LPCTSTR pszName = strName.GetString();
I have a string 2290348
I need it to display as ACF22200 (little endian)
Because the input 2290348 is passed in via a form in textbox, I have tried to read it as string (eg. this->textBox1->Text) and convert it to int (eg. Convert::ToInt32(this->textBox1->Text)).
afterwhich, i converted it to hex via ToString("x") which i manage to get 22F2AC
I appended 00 to 22F2AC and gotten 0022F2AC still as string
now i'm stuck in converting 0022F2AC to ACF22200
While still an int, you could use e.g. htonl to convert it to "network" byte order.
#include <winsock2.h>
int main()
{
unsigned int x = 0x22F2AC;
printf("x = 0x%08x\n", x);
printf("htonl(x) = 0x%08x\n", htonl(x));
return 0;
}
The program above prints:
x = 0x0022f2ac
htonl(x) = 0xacf22200
I am getting an int value from one of the analog pins on my Arduino. How do I concatenate this to a String and then convert the String to a char[]?
It was suggested that I try char msg[] = myString.getChars();, but I am receiving a message that getChars does not exist.
To convert and append an integer, use operator += (or member function concat):
String stringOne = "A long integer: ";
stringOne += 123456789;
To get the string as type char[], use toCharArray():
char charBuf[50];
stringOne.toCharArray(charBuf, 50)
In the example, there is only space for 49 characters (presuming it is terminated by null). You may want to make the size dynamic.
Overhead
The cost of bringing in String (it is not included if not used anywhere in the sketch), is approximately 1212 bytes of program memory (flash) and 48 bytes RAM.
This was measured using Arduino IDE version 1.8.10 (2019-09-13) for an Arduino Leonardo sketch.
Risk
There must be sufficient free RAM available. Otherwise, the result may be lockup/freeze of the application or other strange behaviour (UB).
Just as a reference, below is an example of how to convert between String and char[] with a dynamic length -
// Define
String str = "This is my string";
// Length (with one extra character for the null terminator)
int str_len = str.length() + 1;
// Prepare the character array (the buffer)
char char_array[str_len];
// Copy it over
str.toCharArray(char_array, str_len);
Yes, this is painfully obtuse for something as simple as a type conversion, but somehow it's the easiest way.
You can convert it to char* if you don't need a modifiable string by using:
(char*) yourString.c_str();
This would be very useful when you want to publish a String variable via MQTT in arduino.
None of that stuff worked. Here's a much simpler way .. the label str is the pointer to what IS an array...
String str = String(yourNumber, DEC); // Obviously .. get your int or byte into the string
str = str + '\r' + '\n'; // Add the required carriage return, optional line feed
byte str_len = str.length();
// Get the length of the whole lot .. C will kindly
// place a null at the end of the string which makes
// it by default an array[].
// The [0] element is the highest digit... so we
// have a separate place counter for the array...
byte arrayPointer = 0;
while (str_len)
{
// I was outputting the digits to the TX buffer
if ((UCSR0A & (1<<UDRE0))) // Is the TX buffer empty?
{
UDR0 = str[arrayPointer];
--str_len;
++arrayPointer;
}
}
With all the answers here, I'm surprised no one has brought up using itoa already built in.
It inserts the string representation of the integer into the given pointer.
int a = 4625;
char cStr[5]; // number of digits + 1 for null terminator
itoa(a, cStr, 10); // int value, pointer to string, base number
Or if you're unsure of the length of the string:
int b = 80085;
int len = String(b).length();
char cStr[len + 1]; // String.length() does not include the null terminator
itoa(b, cStr, 10); // or you could use String(b).toCharArray(cStr, len);