const_cast in c++ -getting errors

Question

include "stdafx.h"
#include <iostream>
using namespace std;

class Foo{
public:
    void func()
    {
        cout<<"Hello!!"<<endl;
    }
};

void some_func(const Foo &f)
{
    //f.func();
    Foo &fr=const_cast<Foo&>(f);
    fr.func();
}
int main()
{
    some_func(Foo &f); //if const declared will add the no of errors from 2 to 3
    return 0;
}

How to invoke the some_func(const Foo &f)...If i declare the const before Foo parameter in main it shows me error... But if i'm using the code above i'm getting 2 errors..

output:

1>------ Build started: Project: const_cast, Configuration: Debug Win32 ------
1>Compiling...
1>const_cast.cpp
1>c:\documents and settings\beata\my documents\visual studio 2008\projects\const_cast\const_cast\const_cast.cpp(24) : error C2065: 'f' : undeclared identifier
1>c:\documents and settings\beata\my documents\visual studio 2008\projects\const_cast\const_cast\const_cast.cpp(24) : error C2275: 'Foo' : illegal use of this type as an expression
1>        c:\documents and settings\beata\my documents\visual studio 2008\projects\const_cast\const_cast\const_cast.cpp(8) : see declaration of 'Foo'
1>Build log was saved at "file://c:\Documents and Settings\beata\My Documents\Visual Studio 2008\Projects\const_cast\const_cast\Debug\BuildLog.htm"
1>const_cast - 2 error(s), 0 warning(s)
========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========

ta.speot.is · Accepted Answer · 2011-04-24 13:25:45Z

2

Your problem is probably with main()

int main()
{
    Foo f;
    some_func(f);
    return 0;
}

You need to declare f before you can use it.

answered Apr 24, 2011 at 13:25

ta.speot.is

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sehe · Accepted Answer · 2011-04-24 13:25:52Z

2

int main()
{
    Foo f;
    some_func(f);
    return 0;
}

answered Apr 24, 2011 at 13:25

sehe

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Comments

Lara Bailey · Accepted Answer · 2011-04-24 13:26:00Z

2

some_func(Foo &f); looks something like a declaration and something like a function call. If you meant a function call you just pass an object of the appropriate type to the function. E.g.

Foo f;
some_func(f);

or if you want to pass an unnamed temporary (legal because the function takes a const reference):

some_func(Foo());

answered Apr 24, 2011 at 13:26

Lara Bailey

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wheaties · Accepted Answer · 2011-04-24 13:26:00Z

1

The problen you're seeing is that you haven't labled the func function call as const to indicate to the compiler that it does not modify visible state. That is,

class Foo{
public:
    void func() const{
        std::cout << "Hello World!" << std::end;
    }
};

will work fine. You put the const at the end of the function calls when they do not modify state (not completely true but more advanced for this post.)

So if you want to pass an object by const ref, you will only ever be able to call methods on it that have been declared non-state modifying. Please don't use const_cast unless you absolutely have to.

Also, don't forget to declare a variable of type Foo in your main body.

answered Apr 24, 2011 at 13:26

wheaties

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Sarfaraz Nawaz · Accepted Answer · 2011-04-24 13:27:12Z

1

some_func(Foo &f); //if const declared will add the no of errors from 2 to 3

Wrong syntax.

Here is what you should be doing:

Foo f; //f is an object of type Foo
some_func(f); //pass the object to the function.

answered Apr 24, 2011 at 13:27

Sarfaraz Nawaz

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