>I have not found any similar methods for the
>string class. I have tried the copy method but it
>does not add terminating 0. As the standard
>includes lots of generic stuff it is hard to know
>if there is some algorithm, iterator, etc. than
>can be used as replacement for Left, Right and
>Mid.


Use the std::string::substr and str::string::size member functions.

CString mstr;
std::string str;

mstr.Left(ct)          becomes    str.substr(0, ct)
mstr.Mid(first,ct)    becomes    str.substr(first, ct)
mstr.Right(ct)        becomes    str.substr(ct, str.size() - ct)




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Look up substr().

sedlar@my-deja.com wrote in message <81g613$as4$1@nnrp1.deja.com>...
>I am changing an old application and have then
>choosen to use the string class available in
>Standard C++.

I have not found any similar methods for the
>string class.




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On 27 Nov 1999 08:08:47 GMT, "Al Stevens" <alstevens@midifitz.com>
wrote:

: Use the std::string::substr and str::string::size member functions.
:
: CString mstr;
: std::string str;
:
: mstr.Left(ct)          becomes    str.substr(0, ct)
: mstr.Mid(first,ct)    becomes    str.substr(first, ct)
: mstr.Right(ct)        becomes    str.substr(ct, str.size() - ct)

Since length defaults to string::npos, the last one can be

                                   str.substr(ct)

Now the only change in parameters is the 0 for left.

John


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Al Stevens <alstevens@midifitz.com> wrote:

>> mstr.Right(ct)        becomes    str.substr(ct, str.size() - ct)

That's wrong.

   mstr.Right(ct)        becomes    str.substr(str.size() - ct)

Right is supposed to get you "ct" characters from the end of the string. Al
Stevens' suggested implementation for Right gets the string after removing
the first "ct" characters, which is not the same thing.

    -- Darin



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