converting void* pointing to char* to a std::string in c++ - c++

From the wincrypt api I am receiving a void* pointing to a char*.
This char* is pointing to the start of a char[]. I am also receiving a void* pointing to a int with the size of the char*.
Regarding pvData and cbData I have the following documentation from Microsoft.
Data type of pvData: A pointer to an array of BYTE values. The size of this array is specified in the cbData parameter. Returns a null-terminated Unicode character string that contains the display name for the certificate.
I want to convert this void* to a std::string but so far all I am getting when outputting my std::string is the first character.
I have read: Converting a void* to a std::string but since my void* is pointing to a char* instead of std::string the static_cast in the accepted answer fails and the returned std::string* triggers a null pointer exception.
So far I have the following:
// pvData = void* pointing to char*
// cbData = void* pointing to int*
std::string tempName;
tempName.assign(static_cast<char*>(pvData), static_cast<int*>(cbData));
printf("%S \n", pvData); // entire string is shown
printf("%s \n", tempName.c_str()); // only first character is shown
I have also tried
tempName = static_cast<char*>(pvData); // only single character returned
tempName.assign(static_cast<char*>(pvData)); // only single character returned
char* arr = static_cast<char*>(pvData);
std::string tempName(arr); // only single character returned empty with printf must
// use std::cout

If the char buffer isn't null-terminated, then to use the (void*)cbData length:
char* data = static_cast<char*>(pvData);
size_t len = *static_cast<int*>(cbData);
std::string tempName(data, len);
See the std::string constructor reference (#5, from buffer) and ::assign reference (#4, buffer).
EDIT: If you're trying to use the function CertGetCertificateContextProperty with dwPropId CERT_FRIENDLY_NAME_PROP_ID, here is how you should call the function:
CERT_CONTEXT ctx;
BYTE buf[100];
DWORD len = 100;
CertGetCertificateContextProperty(&ctx, CERT_FRIENDLY_NAME_PROP_ID, buf, &len);
std::string tempName(reinterpret_cast<char*>(buf), len);
No dealing with void* pointers!

The documentation specifically states that it returns a Unicode string, which in Microsoft-speak means UTF-16. Characters that are part of the ASCII range will contain a zero in their second byte, which ends a string copy prematurely. You would get better results using wstring with a cast to wchar_t*.
If copying to string appears to work, it's because those zero bytes are invisible.
Putting this in the context of your original code:
std::wstring tempName;
tempName.assign(static_cast<wchar_t*>(pvData), (*static_cast<int*>(cbData)) / sizeof(wchar_t));
printf("%S \n", tempName.c_str());
Note that this isn't the easiest way to do it, you should also follow the advice from qxz regarding the string constructor and the passing of cbData.

Related

String to const char conversion using c_str() or toCharArray()?

I want to know more about programming and after a bit of googling I found how to convert a string to a const char.
String text1;
What I do not understand is why c_str() works,
const char *text2 = text1.c_str();
contrary to toCharArray()?
const char *text2 = text1.toCharArray();
or
const char text2 = text1.toCharArray();
The latter is more logical to me as I want to convert a string to a char, and then turn it into a const char. But that doesn't work because one is a string, the other is a char. The former, as I understand, converts the string to a C-type string and then turns it into a const char. Here, the string suddenly isn't an issue anymore oO
.
a) Why does it need a C-type string conversion and why does it work only then?
b) Why is the pointer needed?
c) Why does a simple toCharArray() not work?
.
Or do I do something terribly wrong?
Thanks heaps.
I am using PlatformIO with Arduino platform.
If you need to modify the returned c-style string in any way, or have it persist after you modify the original String, you should use toCharArray.
If you only need a null-terminated c-style string to pass as a read-only parameter to a function, use c_str.
Arduino reference for String.toCharArray()
Arduino reference for String.c_str()
The interface (and implementation) of toCharArray is shown below, from source
void toCharArray(char *buf, unsigned int bufsize, unsigned int index=0) const
{ getBytes((unsigned char *)buf, bufsize, index); }
So your first issue is that you're trying to use it incorrectly. toCharArray will COPY the underlying characters of your String into a buffer that you provide. This must be extra space that you have allocated, either in a buffer on the stack, or in some other writable area of memory. You would do it like this.
String str = "I am a string!";
char buf[5];
str.toCharArray(buf, 5);
// buf is now "I am\0"
// or you can start at a later index, here index 5
str.toCharArray(buf, 5, 5);
// buf is now "a st\0"
// we can also change characters in the buffer
buf[1] = 'X';
// buf is now "aXst\0"
// modifying the original String does not invalidate the buffer
str = "Je suis une chaine!";
// buf is still "aXst\0"
This allows you to copy a string partially, or at a later index, or anything you want. Most importantly, this array you copy into is mutable. We can change it, and since it's a copy, it doesn't affect the original String we copied it from. This flexibility comes with a cost. First, we have to have a large enough buffer, which may not be known at compile time, and takes up memory. Second, that copying takes time to do.
But what if we're calling a function that just wants to read a c-style string as input? It doesn't need to modify it at all?
That's where c_str() comes in. The String object has an underlying c-string type array (yes, null terminator and all). c_str() simply returns a const char* to this array. We make it const so that we don't accidentally change it. An object's underlying data should not be changed by random functions outside of its control.
This is the ENTIRE code for c_str():
const char* c_str() const { return buffer; }
You already know how to use it, but to illustrate a difference:
String str = "I am another string!";
const char* c = str.c_str();
// c[1] = 'X'; // error, cannot modify a const object
// modifying the original string may reallocate the underlying buffer
str = "Je suis une autre chaine!";
// dereferencing c now may point to invalid memory
Since c_str() simply returns the underlying data pointer, it's fast. But we don't want other functions to be allowed to modify this data, so it's const.

copy string function to non const char buffer

I'm trying to copy a string to buffer for further processing. I used the instruction
char *buf = line.c_str();
but buf type should be const char*, However If I'm going to use
const char *buf = line.c_str();
I'll face another problem as I'm using strtok_s function for processing the buf. this function expecting char * arg. rather than const char*.
Is there another function or a way to copy this line to char * ??
One approach is to use standard library containers, such as std::vector or C++14's std::dynarray:
std::vector<char> v(line.begin(), line.end());
Bear in mind that this copies line's characters without the null termination \0. In this sense, the buffer does not represent a string. If you need null termination (which is not clear from the question), you have to add it manually:
v.push_back('\0');
Then you can use the underlying data via
char* c1 = &v[0]
char* c2 = v.data(); // c++11
Try this.
char* buf = strdup(line.c_str());
// use strtok_s however you need
free(buf);
You can't use the buffer directly without modifying it because strtok_s does actually modify the data. c_str wants you to leave the data alone.
If you don't care about the modifications, just do:
char* buf = &line[0];

Convert std::wstring to WCHAR*

I have no idea how to convert a std::wstring to a WCHAR*
std::wstring wstrProcToSearch;
WCHAR * wpProcToSearch = NULL;
std::wcin >> wstrProcToSearch; // input std::wstring
// now i need to convert the wstring to a WCHAR*
Does anyone know how to accomplish this?
If you want to convert from std::wstring to const WCHAR* (i.e. the returned pointer gives read-only access to the string content), then calling std::wstring::c_str() method is just fine:
std::wstring wstrProcToSearch;
std::wcin >> wstrProcToSearch; // input std::wstring
// Convert to const WCHAR* (read-only access)
const WCHAR * wpszProcToSearch = wstrProcToSearch.c_str();
Instead, if you want to modify std::wstring's content, things are different. You can use &wstr[0] (where wstr is a non-empty instance of std::wstring) to access the content of the std::wstring (starting from the address of its first characters, and noting that characters are stored contiguously in memory), but you must pay attention to not overrun string's pre-allocated memory.
In general, if you have a std::wstring of length L, you can access characters from index 0
to (L-1).
Before C++17, overwriting the terminating '\0' (located at index L) was undefined behavior (in practice, it's been OK on Visual C++, at least with VC9/VS2008 and VC10/VS2010).
Starting with C++17, overwriting the terminating NUL ('\0') with another NUL has been made valid and is no more undefined behavior.
If the string has not the proper size (i.e. it's not big enough for your needs), then you can call std::wstring::resize() to make room for new characters (i.e. resizing internal std::wstring's buffer), and then use &wstr[0] to read-write std::wstring's content.
If the string is already the proper length and will not need to be changed, you can get a non-const pointer by taking a pointer to the first character:
WCHAR * wpProcToSearch = &wstrProcToSearch[0];
This is guaranteed to work in C++11 and there are no known implementations of C++03 where it doesn't.
If you only need a const pointer you should use c_str:
const WCHAR * wpProcToSearch = wstrProcToSearch.c_str();
I think you can use
wpProcToSearch = wstrProcToSearch.c_str()
like you do with a normal std::string.
I recommend this approach:
wstring str = L"Hallo x y 111 2222 3333 rrr 4444 ";
wchar_t* psStr = &str[0];
It is quite simple but you can not change the length of the string at all. So moving "\0" might not be valid...

Conversion of ATL CString to character array

I want to convert a CString into a char[]. Some body tell me how to do this?
My code is like this :
CString strCamIP1 = _T("");
char g_acCameraip[16][17];
strCamIP1 = theApp.GetProfileString(strSection, _T("IP1"), NULL);
g_acCameraip[0] = strCamIP1;
This seems to be along 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 would be along the lines of
strcpy(g_acCameraip[0], (LPCTSTR)strCamIP1);
From MSDN site:
// 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 don't compile with _UNICODE it's CStringA or if you do compile with _UNICODE then it is CStringW.
In case of CStringW conversion looks little bit different (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;
You could 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
If you are using ATL you could use one of the conversion macros. CString stores data as tchar, so you would use CT2A() (C in macro name stands for const):
CString from("text");
char* pStr = CT2A((LPCTSTR)from);
Those macros are smart, if tchar represents ascii (no _UNICODE defined), they just pass the pointer over and do nothing.
More info below, under ATL String-Conversion Classes section:
http://www.369o.com/data/books/atl/index.html?page=0321159624%2Fch05.html
CStringA/W is cheaply and implicitly convertible to const char/wchar_t *. Whenever you need C-style string, just pass CString object itself (or the result of .GetString() which is the same). The pointer will stay valid as long as string object is alive and unmodified.
strcpy(g_acCameraip[0], strCamIP1);
// OR
strcpy(g_acCameraip[0], strCamIP1.GetString());
If you need writable (non-const) buffer, use .GetBuffer() with optional maximum length argument.
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 would be to have array of CStrings. You can use them whereever you need null-terminated string, but they will also manage string's memory for you.
std::vector<CString> g_acCameraip(16);
g_acCameraip[0] = theApp.GetProfileString(strSection, _T("IP1"), NULL);
Use memcpy .
char c [25];
Cstring cstr = "123";
memcpy(c,cstr,cstr.GetLength());
Do you really have to copy the CString objects into 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 the entries, for example like this
for (int i=0; i<COUNT; ++i) {
someOp(Cameraip[i]); // the someOp function takes const CString&
}
...you may convert them, if needed.
fopen is the function which needs char* param. so if you have CString as available string, you can just use bellow code.
be happy :)
Here, cFDlg.GetPathName().GetString(); basically returns CString in my code.
char*pp = (char*)cFDlg.GetPathName().GetString();
FILE *fp = ::fopen(pp,"w");
CString str;
//Do something
char* pGTA = (LPTSTR)(LPCTSTR)str;//Now the cast
Just (LPTSTR)(LPCTSTR). Hope this is what you need :)
char strPass[256];
strcpy_s( strPass, CStringA(strCommand).GetString() );
It's simple
ATL CStrings allow very simple usage without having to do a lot of conversions between types. You can most easily do:
CString cs = "Test";
const char* str = static_cast<LPCTSTR>(cs);
or in UNICODE environment:
CString cs = "Test";
const wchar_t* str = static_cast<LPCTSTR>(cs);
How it works
The static_cast (or alternatively C-Style cast) will trigger the CString::operator LPCTSTR, so you don't do any pointer reinterpretation yourself but rely on ATL code!
The documentation of this cast operator says:
This useful casting operator provides an efficient method to access the null-terminated C string contained in a CString object. No characters are copied; only a pointer is returned. Be careful with this operator. If you change a CString object after you have obtained the character pointer, you may cause a reallocation of memory that invalidates the pointer.
Modifiable Pointers
As mentioned in the above statement, the returned pointer by the cast operator is not meant to be modified. However, if you still need to use a modifiable pointer for some outdated C libraries, you can use a const_cast (if you are sure that function wont modify 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 wish to modify the pointer, you wont get around doing some kind of memory copying to modifiable memory, as mentioned by other answers.
There is a hardcoded method..
CString a = L"This is CString!";
char *dest = (char *)malloc(a.GetLength() + 1);
// +1 because of NULL char
dest[a.GetLength()] = 0; // setting null char
char *q = (char *)a.m_pszData;
//Here we cannot access the private member..
//The address of "m_pszData" private member is stored in first DWORD of &a...
//Therefore..
int address = *((int *)&a);
char *q = (char *)address;
// Now we can access the private data!, This is the real magic of C
// Size of CString's characters is 16bit...
// in cstring '1' will be stored as 0x31 0x00 (Hex)
// Here we just want even indexed chars..
for(int i = 0;i<(a.GetLength()*2);i += 2)
dest[i/2] = *(q+i);
// Now we can use it..
printf("%s", dest);

C++ how to count correctly the characters in a const char?

I have got a const char which is made by concatenation like this:
const char *fileName = "background1";
std::stringstream sstm;
sstm << fileName << "-hd.png";
fileName = sstm.str().c_str();
My problem is that the following instruction:
printf("const char = %s size = %d", fileName, sizeof(fileName));
returns:
"const char = background1-hd.png size = 4"
whereas I would expect that it returns:
"const char = background1-hd.png size = 19"
For example, the following gives the convenient result (as there is no concatenation):
const char *fileName2 = "background1-hd";
printf("const char = %s size = %d", fileName2, sizeof(fileName2));
returns:
"const char = background1-hd.png size = 19"
How to avoid this issue and guarantee that the characters will be correctly counted in my concatenated char ?
Thanks !!
sizeof() returns the number of bytes the variable occupies in memory (in this case returns the size of the pointer fileName).
strlen() returns the length of the string (which is what you need).
You could as well try something like:
#include <iostream>
#include <cstdio>
int main()
{
std::string fileName("background1");
fileName.append("-hd.png");
printf("const char = %s size = %d", fileName.c_str(), fileName.length());
return 0;
}
sizeof returns the size of the variable you give to it; it's evaluated at compile time. The "4" is the size of a pointer on your system. You want to use strlen() to determine the length of a string.
The result of sizeof(fileName) is related to fileName being a pointer, not an array. It literally returns the size of a pointer to a constant character string, and on a 32-bit system, all pointers are 32 bits (so sizeof == 4).
What you should use instead is strlen or similar, which will count the characters in the string, up to the trailing null, and return that. The results with strlen in place of sizeof will be about what you expect.
Side-related, with const char strings there is only ever one character per "cell" (actually byte). There are character sets which make for multiple bytes per character, but packing multiple characters into a single byte is quite rare, at least in C-family languages.
sizeof calculates the size of the data type in bytes and not the size of its contents (what it points to). In your example you are calculating the sizeof char* which is 4 bytes on your system. To get the length of a C string use strlen.
There is a distinction in the language between arrays and pointers, even if this distinction seems diluted both by implicit conversions (arrays tend to decay into pointers quite easily), and common statements that they are the same.
How does this even relate to your code?
Well, a string literal is actually an array of constant characters, not a pointer to character(s). In the initialization const char *fileName = "background1"; you are creating a pointer variable that points to the first element of the array ("background1" is decaying into a pointer to the first element), and from there on the variable you are managing is pointer and not the literal.
If you mix this with the fact that sizeof will tell you the size of the variable, you get that in a platform with 32bit pointers and 8 bit chars, sizeof( const char* ) is always 4, regardless of the object that is pointed by that pointer (if there is even one).
Now, if you were treating the literal as what it actually is you would be having a bit more luck there:
const char filename[] = "background1";
assert( sizeof filename == 12 ); // note: NUL character is counted!
const char *fname = filename;
assert( sizeof filename == sizeof( void* ) );
In real code, you are not a so lucky and in many cases the literals have decayed into pointers well before you get a chance of getting the compile time size of the literal, so you cannot ask the compiler to tell you the size. In that case you need to calculate the length of the C style string, which can be done by calling strlen.
strlen has been suggested a number of times already, and for this case it's probably perfectly reasonable.
There is an alternative that will let you use sizeof though:
char fileName[] = "background1";
std::cout << sizeof(fileName) << "\n";
Since you're making fileName an array, it has all the characteristics of an array -- including the fact that your later attempt at assigning to it:
fileName = sstm.str().c_str();
...would fail (won't even compile when fileName is defined as an array). I should add, however, that it seems to me that you'd be better off just using std::string throughout:
std::string fileName("background1");
std::stringstream sstm;
sstm << fileName << "-hd.png";
fileName = sstm.str();
In this case, you can use string's size() or length() member.