LStrHandle C++ String conversion

[abriggs8]

Hi, I am wiring a cluster that holds a string into a Call External Library Node and following the "Adapt to Type" article and "Calling External Code in LabVIEW" manual, I have created a .cpp file with this code in it:

typedef struct {
double Numeric;
double Numeric2;
LStrHandle String;
} TD1;

However, I am using a function that requires a C++ string to be passed into it and it returns a C++ string that needs to be assigned to LStrHandle String. Is it possible to convert the LStrHandle format into a C++ string and vice versa? I found something regarding the LStrPrintf functions, but I'm fairly new with programming and I did not follow it very well. Thank you in advance.

[rolfk]

---

@abriggs8 wrote:
Hi, I am wiring a cluster that holds a string into a Call External Library Node and following the "Adapt to Type" article and "Calling External Code in LabVIEW" manual, I have created a .cpp file with this code in it:

typedef struct {
double Numeric;
double Numeric2;
LStrHandle String;
} TD1;

However, I am using a function that requires a C++ string to be passed into it and it returns a C++ string that needs to be assigned to LStrHandle String. Is it possible to convert the LStrHandle format into a C++ string and vice versa? I found something regarding the LStrPrintf functions, but I'm fairly new with programming and I did not follow it very well. Thank you in advance.

---

I'm not familiar enough with C++ to tell you exactly what to do. I specially find the C++ string operators quite hard to comprehend. But a LabVIEW string handle is basically as follows:

typedef struct
{
int32 len;
uInt8 buf[0];
} **LStrHandle;

The len integer tells you how many characters are there while the buf member points to the actual string but without a NULL termination character.

So in C you could do something like this:

MgErr MyFunction(LStrHandle handle)
{
char *string = malloc(LStrLen(*handle) + 1);
memcpy(string, LStrBuf(*handle), LStrLen(*handle));
string[LStrLen(*handle)] = 0;

use your string in whatever way you like. Then copy back a C string into the LabVIEW string.
Either use LStrPrintf or doing it yourself:

int32 len = strlen(string);
MgErr err = NumericArrayResize(uB, 1, &handle, len);
if (err) return err;
memcpy(LStrBuf(*handle), string, len);
LStrLen(*handle) = len;
}

You will need to adapt this to C++ strings somehow.

Rolf Kalbermatter

Message Edited by rolfk on 06-20-2005 05:15 PM

[rolfk]

---

@abriggs8 wrote:
Hi, I am wiring a cluster that holds a string into a Call External Library Node and following the "Adapt to Type" article and "Calling External Code in LabVIEW" manual, I have created a .cpp file with this code in it:

typedef struct {
double Numeric;
double Numeric2;
LStrHandle String;
} TD1;

However, I am using a function that requires a C++ string to be passed into it and it returns a C++ string that needs to be assigned to LStrHandle String. Is it possible to convert the LStrHandle format into a C++ string and vice versa? I found something regarding the LStrPrintf functions, but I'm fairly new with programming and I did not follow it very well. Thank you in advance.

---

I just had a look at my C++ textbook and it shouldn't be terribly difficult to do it for C++ as well:

MgErr MyFunction(LStrHandle handle)
{
MgErr err;

string s(LStrBuf(*handle), 0, LStrLen(*handle));

Now you can pass s as a C++ string.

Then copy back a C++ string into the LabVIEW string.

err = NumericArrayResize(uB, 1, &handle, s.length);
if (err) return err;
memcpy(LStrBuf(*handle), s.cstr, s.length);
LStrLen(*handle) = s.length;
return noErr;
}

[abriggs8]

Hi rolfk, that helps me out a great deal. It was much easier to understand than most explanations I've gotten. However, the line that reads:

string s(LStrBuf(*handle), 0, LStrLen(*handle));

is giving me a compile error. It says that it cannot convert parameter 1 from 'uChar*__w64' to 'const std::basic_string<_Elem,_Traits,_Ax>::_Myt'. I'm guessing the LStrBuf function is not returning the proper data type(which appears to be a C++ string) to be passed into the constructor there. I'm unfamiliar with that function so I'm not sure how to proceed to fix this. How might I proceed to do this? Again, thank you in advance for the advice.

[rolfk]

---

@abriggs8 wrote:
Hi rolfk, that helps me out a great deal. It was much easier to understand than most explanations I've gotten. However, the line that reads:

string s(LStrBuf(*handle), 0, LStrLen(*handle));

is giving me a compile error. It says that it cannot convert parameter 1 from 'uChar*__w64' to 'const std::basic_string<_Elem,_Traits,_Ax>::_Myt'. I'm guessing the LStrBuf function is not returning the proper data type(which appears to be a C++ string) to be passed into the constructor there. I'm unfamiliar with that function so I'm not sure how to proceed to fix this. How might I proceed to do this? Again, thank you in advance for the advice.

---

Well LStrBuf is only a macro accessing the string buffer in the handle but it is a little bit complicated so that the C++ compiler might get into trouble here seeing a type he doesn't know to interpret as C string.

Maybe you could try to use something like

s((*handle)->str, 0, LStrLen(*handle));

instead, or you might even have to cast the buffer explicitedly to a (char *) to make it work such as:

s((char *)LStrBuf(*handle), 0, LStrLen(*handle));
s((char *)(*handle)->str, 0, LStrLen(*handle));

Rolf Kalbermatter