Removing characters from a string

[WirelessDice]

Okay, what I basically want to do is to remove the two last characters from my string witch is a
tab and a newline character, but I think that doesn't matter, because they are always at the end of
the string. I got a code already, but it doesn't work correctly.

    string::size_type pos = 0;
    while ((pos = txt.find_first_of("\n\t", pos) != string::npos))
        txt.erase(pos, 1);

    return txt;

It seems like that code removes everything except the first character.
Thanks in advance

[potatoCode]

Hello WirelessDice

check the while condition and see what value it returns on each iteration
bye~

[WirelessDice]

Hello WirelessDice

check the while condition and see what value it returns on each iteration
bye~

I edited the code so I could see the return values.

    cout<<"TEST!!!!\n";
    string::size_type pos = 0;
    while ((pos = txt.find_first_of("\n", pos) != string::npos))
    {
        cout<<pos<<"\n";
        txt.erase(pos, 1);
    }
    cout<<"END TEST!!!\n";

The return values I got was something like this.

TEST!!!
1
1
1
END TEST!!!

And instead of getting "-1" I'll only get "-", pleas help!
Thanks in advance.

[potatoCode]

I edited the code so I could see the return values.

    cout<<"TEST!!!!\n";
    string::size_type pos = 0;
    while ((pos = txt.find_first_of("\n", pos) != string::npos))
    {
        cout<<pos<<"\n";
        txt.erase(pos, 1);
    }
    cout<<"END TEST!!!\n";

The return values I got was something like this.

TEST!!!
1
1
1
END TEST!!!

And instead of getting "-1" I'll only get "-", pleas help!
Thanks in advance.

There goes your answer~
Since the return value is always 1,
the erase() always starts at that index position!

[WirelessDice]

There goes your answer~
Since the return value is always 1,
the erase() always starts at that index position!

Well, okay.. So how do I fix it? Or else, just change the whole code to remove the two last
characters!

[potatoCode]

Well, okay.. So how do I fix it? Or else, just change the whole code to remove the two last
characters!

That's a find solution if only the last two strings need to be removed unconditionally

or do something along the line of

    while(1)
    {
        pos = txt.find_first_of("\n\t");
        if(pos == string::npos)
            break;
        txt.erase(pos, 1);
    }

xD

[Amleto]

give *complete compilable* code. e.g. what is in txt ??

also, why use find_FIRST_of, when you want to find the \t\n at the end of the string? just use find_last_of.

most importantly:

pos = txt.find_first_of("\n", pos) != string::npos;

is the same as

pos = ( txt.find_first_of("\n", pos) != string::npos );

therefore, for what you have coded, pos will only ever be 1 (or whatever bool(true) gets cast to as a string::size_type) inside that while loop.


try this:

  while ( string::npos != (pos = txt.find_first_of("\n")) )
  {
    ...
  }

[WirelessDice]

That's a find solution if only the last two strings need to be removed unconditionally

or do something along the line of

    while(1)
    {
        pos = txt.find_first_of("\n\t");
        if(pos == string::npos)
            break;
        txt.erase(pos, 1);
    }

xD

Wow, Thanks! That actually worked!
Thread solved!

[Plasmator]

Thread solved!

You should prefer to use standard algorithms wherever possible. All of that crap can be replaced by a one-liner involving std::remove_if and std::isspace:

txt.erase(std::remove_if(txt.begin(), txt.end(), &std::isspace), txt.end());

Boost's string algorithms library provides more elegant solutions, if you're interested.

[Amleto]

You should prefer to use standard algorithms wherever possible. All of that crap can be replaced by a one-liner involving std::remove_if and std::isspace:

txt.erase(std::remove_if(txt.begin(), txt.end(), &std::isspace), txt.end());

Boost's string algorithms library provides more elegant solutions, if you're interested.

cant get it to compile

_FwdIt std::remove_if(_FwdIt,_FwdIt,_Pr)' : cannot use function template 'bool std::isspace(_Elem,const std::locale &)' as a function argument

I included <locale>, not a c library though :s looks like that version isnt a predicate.

I got it to work using isspace from #include <ctype.h> which isnt a member of std namespace. i probably shouldnt be including that file though?

[laserlight]

I included <locale>, not a c library though :s looks like that version isnt a predicate.

I got it to work using isspace from #include <ctype.h> which isnt a member of std namespace. i probably shouldnt be including that file though?

The problem is that std::isspace is a function template. ::isspace from <ctype.h> is not a function template. If you want to stick with <cctype>, a simple solution is to implement your own function, possibly by overloading, that forwards the job to std::isspace.

[potatoCode]

You should prefer to use standard algorithms wherever possible.