Convert ATL CString to Character Array

Question

I want to convert a CString to char[] . Someone told me how to do this?

My code looks like this:

 CString strCamIP1 = _T(""); char g_acCameraip[16][17]; strCamIP1 = theApp.GetProfileString(strSection, _T("IP1"), NULL); g_acCameraip[0] = strCamIP1; 

Answer 1

It looks like the right lines; http://msdn.microsoft.com/en-us/library/awkwbzyc.aspx

 CString aCString = "A string"; char myString[256]; strcpy(myString, (LPCTSTR)aString); 

which in your case will be along the lines

 strcpy(g_acCameraip[0], (LPCTSTR)strCamIP1); 

Answer 2

From the MSDN website:

 // Convert to a char* string from CStringA string // and display the result. CStringA origa("Hello, World!"); const size_t newsizea = (origa.GetLength() + 1); char *nstringa = new char[newsizea]; strcpy_s(nstringa, newsizea, origa); cout << nstringa << " (char *)" << endl; 

CString is based on TCHAR , so if you are not compiling with _UNICODE it CStringA , or if you are compiling with _UNICODE , then this is CStringW .

In case of conversion, CStringW looks a little bit (example also from MSDN ):

 // Convert to a char* string from a wide character // CStringW string. To be safe, we allocate two bytes for each // character in the original string, including the terminating // null. const size_t newsizew = (origw.GetLength() + 1)*2; char *nstringw = new char[newsizew]; size_t convertedCharsw = 0; wcstombs_s(&convertedCharsw, nstringw, newsizew, origw, _TRUNCATE ); cout << nstringw << " (char *)" << endl; 

Answer 3

You can use wcstombs_s :

 // Convert CString to Char By Quintin Immelman. // CString DummyString; // Size Can be anything, just adjust the 100 to suit. const size_t StringSize = 100; // The number of characters in the string can be // less than String Size. Null terminating character added at end. size_t CharactersConverted = 0; char DummyToChar[StringSize]; wcstombs_s(&CharactersConverted, DummyToChar, DummyString.GetLength()+1, DummyString, _TRUNCATE); //Always Enter the length as 1 greater else //the last character is Truncated 

Answer 4

If you use ATL, you can use one of the conversion macros. CString stores the data as tchar, so you should use CT2A () (C in the macro name means const):

 CString from("text"); char* pStr = CT2A((LPCTSTR)from); 

These macros are smart if tchar represents ascii (no _UNICODE is defined), they just pass a pointer and do nothing.

See the ATL String-Conversion Classs class section below for more information: http://www.369o.com/data/books/atl/index.html?page=0321159624%2Fch05.html

Answer 5

CStringA/W cheaply and implicitly converted to const char/wchar_t * . Whenever you need a C-style string, just pass the CString object itself (or the result of .GetString() , which is the same). The pointer remains valid until the string object is alive and changed.

 strcpy(g_acCameraip[0], strCamIP1); // OR strcpy(g_acCameraip[0], strCamIP1.GetString()); 

If you need a rewritable (non-constant) buffer, use .GetBuffer() with an optional argument of maximum length.

If you have CStringW , but you need const char* and vice versa, you can use a temporary CStringA object:

 strcpy(g_acCameraip[0], CStringA(strCamIP1).GetString()); 

---

But a much better way is to have an array of CString s. You can use them if you need a string with a null character, but they will also manage the string memory for you.

 std::vector<CString> g_acCameraip(16); g_acCameraip[0] = theApp.GetProfileString(strSection, _T("IP1"), NULL); 

Answer 6

Use memcpy.

 char c [25]; Cstring cstr = "123"; memcpy(c,cstr,cstr.GetLength()); 

Answer 7

Is it really possible to copy CString objects to fixed char arrays?

 enum { COUNT=16 }; CString Cameraip[COUNT]; Cameraip[0] = theApp.GetProfileString(strSection, _T("IP1"), NULL); // add more entries... 

... and then - later - when accessing elements, for example, in this way

 for (int i=0; i<COUNT; ++i) { someOp(Cameraip[i]); // the someOp function takes const CString& } 

... you can convert them if necessary.

Answer 8

fopen is a function that needs a char * parameter. so if you have a CString as an available string, you can simply use the following code. be happy :) Here is cFDlg.GetPathName().GetString(); basically returns a CString in my code.

 char*pp = (char*)cFDlg.GetPathName().GetString(); FILE *fp = ::fopen(pp,"w"); 

Answer 9

  CString str; //Do something char* pGTA = (LPTSTR)(LPCTSTR)str;//Now the cast 

Only (LPTSTR) (LPCTSTR). Hope this is what you need :)

Answer 10

 char strPass[256]; strcpy_s( strPass, CStringA(strCommand).GetString() ); 

Answer 11

It's simple

ATL CStrings allow for very simple use without having to do many conversions between types. You can most easily do:

 CString cs = "Test"; const char* str = static_cast<LPCTSTR>(cs); 

or in a UNICODE environment:

 CString cs = "Test"; const wchar_t* str = static_cast<LPCTSTR>(cs); 

How it works

static_cast (or, conversely, C-Style cast) will call CString::operator LPCTSTR , so you do not CString::operator LPCTSTR pointer yourself, but rely on ATL code!

The documentation of this operator statement says:

This useful casting operator provides an efficient method for accessing a C string with zero completion contained in a CString object. Characters are not copied; only the pointer is returned. Be careful with this operator. If you change the CString object after receiving a pointer to a character, you can cause a reallocation of memory that invalidates the pointer.

Modifiable pointers

As mentioned above, the return pointer by the translation operator is not intended to be modified . However, if you still need to use a mutable pointer for some legacy C libraries, you can use const_cast (if you are sure that the function does not change the pointer):

 void Func(char* str) // or wchar_t* in Unicode environment { // your code here } // In your calling code: CString cs = "Test"; Func(const_cast<LPTSTR>(static_cast<LPCTSTR>(test))); // Call your function with a modifiable pointer 

If you want to change the pointer, you will not get by doing some kind of copying the memory into modifiable memory, as mentioned by other answers.