Conversion from LPTSTR* to STL string

Hi,
I would like to know how to convert the string from LPTSTR* to STL string object.

Sample code would be of great help.
Thanks in advance.

nt WINAPI WinMain(HINSTANCE hInstance,
                           HINSTANCE hPrevInstance,
                           LPSTR lpCmdLine,
                           int nCmdShow)
{
      LPTSTR *szArglist;
      int nArgs;
      szArglist = CommandLineToArgvW(GetCommandLineW(), &nArgs);
//I would like to convert the szArglist to STL string object

}

Regards,
Pampa
pampa_analyticaAsked:
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Sys_ProgCommented:
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
0
Karl Heinz KremerCommented:
That's the way to do it. It does work. "string" does have a constructor that takes a char *.
0
pampa_analyticaAuthor Commented:
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
0
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Karl Heinz KremerCommented:
This works for me:

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

            cout << the_string << endl;
0
Karl Heinz KremerCommented:
If this still does not help, show your source code.
0
nonubikCommented:
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.
0
nonubikCommented:
...my mistake, if(!i) should be if(i)
0
waysideCommented:
> 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.
0
cupCommented:
If you are using unicode, use wstring instead of string
0
Sys_ProgCommented:
wayside is absolutely correct
All of us didn't notice that ur szArglist is a pointer of type LPTSTR

Amit

0
nonubikCommented:
I did ;)
0
Sys_ProgCommented:
Sorry nonubik,

I should have had a meticulous look at your code

Amit
0
nonubikCommented:
np, maybe it was too obscure.
0
pampa_analyticaAuthor Commented:
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
0
Sys_ProgCommented:
pampa_analytica,

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

Amit
0
pampa_analyticaAuthor Commented:
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)
0
pampa_analyticaAuthor Commented:
#include <string>
using namespace std;

These are my includes in the project
0
Sys_ProgCommented:
I think u are using VC++6.0
But I tried on it & it works fine for me

Amit
0
pampa_analyticaAuthor Commented:
No Amit I am using VC++7.0. I tried but still Iam getting the same error.
0
Sys_ProgCommented:
I am not sure but I have heard that VC++7.0 is less compliant compiler as compared to VC++7.1
I am not sure since I haven't used both
However, the code should work
The std::string does have a += as well as a + operator defined
Here is the quick reference

http://www.cppreference.com/cppstring_details.html#Operators

Amit

0

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nonubikCommented:
try a cast:
the_string += (char *)szArglist[i];
0
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