From: krishna_sicsr [mailto:[EMAIL PROTECTED]
>Hello friends,
> I want to know why the statement class a *a; valid in main
>and what is the meaning of this in c++
It should not be valid - your compiler should be complaining about it -
That is if you mean using a variable name the same as a class name:
$ type cpp.cpp
#include <iostream>
using std::cout;
class foo{
public:
foo(){cout << "foo::foo()\n"; }
~foo(){cout << "foo::~foo()\n"; }
void bar(){cout << "foo::bar()\n";}
};
int main(){
class foo* foo = new foo;
foo->bar();
delete foo;
return 0;
}
$ make
cl cpp.cpp /EHs /Zc:forScope /Wall /W4 /WL /wd4820 /wd4619 /wd4217
/wd4710 /wd4668 /c
Microsoft (R) 32-bit C/C++ Optimizing Compiler Version 13.10.3077 for
80x86
Copyright (C) Microsoft Corporation 1984-2002. All rights reserved.
cpp.cpp
cpp.cpp(11) : error C2061: syntax error : identifier 'foo'
make: *** [cpp.obj] Error 2
$
If you mean using the specifier 'class ...' in main then the word class
is redundant in this context, i.e. you can remove it and it will make no
difference. It's like declaring a 'long int' - the int is redundant:
$ type cpp.cpp
#include <iostream>
using std::cout;
class foo{
public:
foo(){cout << "foo::foo()\n"; }
~foo(){cout << "foo::~foo()\n"; }
void bar(){cout << "foo::bar()\n";}
};
int main(){
class foo* f = new foo;
f->bar();
delete f;
return 0;
}
$ make
cl cpp.cpp /EHs /Zc:forScope /Wall /W4 /WL /wd4820 /wd4619 /wd4217
/wd4710 /wd4668 /c
Microsoft (R) 32-bit C/C++ Optimizing Compiler Version 13.10.3077 for
80x86
Copyright (C) Microsoft Corporation 1984-2002. All rights reserved.
cpp.cpp
link cpp.obj /OUT:cpp.exe /nologo
$ cpp
foo::foo()
foo::bar()
foo::~foo()
$
--
PJH
"Consistently separating words by spaces became a general custom about
the tenth century A.D., and lasted until about 1957, when FORTRAN
abandoned the practice." -- Sun FORTRAN Reference Manual
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