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Conversion from LPTSTR* to STL string

Posted on 2004-04-14
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Last Modified: 2007-12-19
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
0
Comment
Question by:pampa_analytica
  • 6
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24 Comments
 
LVL 10

Expert Comment

by:Sys_Prog
ID: 10822031
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
 
LVL 44

Expert Comment

by:Karl Heinz Kremer
ID: 10822090
That's the way to do it. It does work. "string" does have a constructor that takes a char *.
0
 

Author Comment

by:pampa_analytica
ID: 10822315
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
 
LVL 44

Expert Comment

by:Karl Heinz Kremer
ID: 10822441
This works for me:

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

            cout << the_string << endl;
0
 
LVL 44

Expert Comment

by:Karl Heinz Kremer
ID: 10822449
If this still does not help, show your source code.
0
 
LVL 16

Expert Comment

by:nonubik
ID: 10822576
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
 
LVL 16

Expert Comment

by:nonubik
ID: 10822577
...my mistake, if(!i) should be if(i)
0
 
LVL 14

Expert Comment

by:wayside
ID: 10822627
> 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
 
LVL 11

Expert Comment

by:cup
ID: 10822810
If you are using unicode, use wstring instead of string
0
 
LVL 10

Expert Comment

by:Sys_Prog
ID: 10822861
wayside is absolutely correct
All of us didn't notice that ur szArglist is a pointer of type LPTSTR

Amit

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LVL 16

Expert Comment

by:nonubik
ID: 10822889
I did ;)
0
 
LVL 10

Expert Comment

by:Sys_Prog
ID: 10822934
Sorry nonubik,

I should have had a meticulous look at your code

Amit
0
 
LVL 16

Expert Comment

by:nonubik
ID: 10822964
np, maybe it was too obscure.
0
 

Author Comment

by:pampa_analytica
ID: 10830040
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
 
LVL 10

Expert Comment

by:Sys_Prog
ID: 10830058
pampa_analytica,

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

Amit
0
 

Author Comment

by:pampa_analytica
ID: 10830078
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
 

Author Comment

by:pampa_analytica
ID: 10830080
#include <string>
using namespace std;

These are my includes in the project
0
 
LVL 10

Expert Comment

by:Sys_Prog
ID: 10830127
I think u are using VC++6.0
But I tried on it & it works fine for me

Amit
0
 

Author Comment

by:pampa_analytica
ID: 10830157
No Amit I am using VC++7.0. I tried but still Iam getting the same error.
0
 
LVL 10

Accepted Solution

by:
Sys_Prog earned 50 total points
ID: 10830177
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
 
LVL 16

Expert Comment

by:nonubik
ID: 10830883
try a cast:
the_string += (char *)szArglist[i];
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