Conversion from LPTSTR* to STL string

Well  LPTSTR is infact a LPSTR if u are not using unicode characters
And LPSTR is actually a pointer to a char string
Hence this should work (I haven't tried it)

string stlStr ( szArglist ) ;

Amit

That's the way to do it. It does work. "string" does have a constructor that takes a char *.

I tried in the way what Amit has posted, but it is not working. It throws me error:
f:\Viz\Viz_src\DataVizGUI\DataVizGUI.cpp(36): error C2664: 'std::basic_string<_Elem,_Traits,_Ax>::basic_string(const std::basic_string<_Elem,_Traits,_Ax>::_Alloc &)' : cannot convert parameter 1 from 'LPTSTR * ' to 'const std::basic_string<_Elem,_Traits,_Ax>::_Alloc &'
        with
        [
            _Elem=char,
            _Traits=std::char_traits<char>,
            _Ax=std::allocator<char>
        ]
        and
        [
            _Elem=char,
            _Traits=std::char_traits<char>,
            _Ax=std::allocator<char>
        ]

Please provide me some sample code to convert the char*/LPTSTR* to STL string.

Thanks in advance,
Pampa

This works for me:

            LPTSTR lptstr = "this is the lptstr";
            string the_string(lptstr);

            cout << the_string << endl;

If this still does not help, show your source code.

Try this:

for(int i=0;i<nArgs;i++)
{
  if(!i)
    the_string +=' ';
  the_string += szArglist[i];
}

if you want all arguments to be palced in a string.

...my mistake, if(!i) should be if(i)

> string stlStr ( szArglist ) ;

LPTSTR * is not the same as LPTSTR. The first is an array of pointers, the second is one pointer.

Since szArglist is an array of pointers, you would need to do

string stlStr1(szArglist[0]);
string stlStr2(szArglist[1]);

etc.

If you are using unicode, use wstring instead of string

wayside is absolutely correct
All of us didn't notice that ur szArglist is a pointer of type LPTSTR

Amit

I did ;)

Sorry nonubik,

I should have had a meticulous look at your code

Amit

np, maybe it was too obscure.

Hi guys,

int WINAPI WinMain(HINSTANCE hInstance,
                       HINSTANCE hPrevInstance,
                       LPSTR lpCmdLine,
                       int nCmdShow)
{
     LPTSTR *szArglist;
     int nArgs;
     szArglist = CommandLineToArgvW(GetCommandLineW(), &nArgs);
     string the_string;
//I would like to convert the szArglist to STL string object. I tried this:
          for( i=1; i<=nArgs-1; i++)
          {
      the_string = the_string + szArglist[i];
      MessageBox(NULL,szArglist[i],0,0);
         }
}

But i got this error:
f:\Viz\Viz_src\DataVizGUI\DataVizGUI.cpp(63): error C2678: binary '+' : no operator found which takes a left-hand operand of type 'std::string' (or there is no acceptable conversion).

I am unable to solve this coz I am pretty new to VC++. PLease help me out in solving this problem.



Thanks in advance,
Pampa

pampa_analytica,

std::string does have a binary + operator
and the above code works fine for me

Amit

I got this error while I was trying to do this:
      for( i=1; i<=nArgs-1; i++)
      {
                  the_string += szArglist[i];
              MessageBox(NULL,szArglist[i],0,0);
      }


f:\Viz\Viz_src\DataVizGUI\DataVizGUI.cpp(64): error C2679: binary '+=' : no operator found which takes a right-hand operand of type 'LPTSTR' (or there is no acceptable conversion)

#include <string>
using namespace std;

These are my includes in the project

I think u are using VC++6.0
But I tried on it & it works fine for me

Amit

No Amit I am using VC++7.0. I tried but still Iam getting the same error.

try a cast:
the_string += (char *)szArglist[i];