S.i considered a variable?

[cbowler]

struct S {
int i;
};

int main() {
S myS;
myS.i; // is this a variable according to spec?
return 0;
}

 

[tom_usenet]

struct S {
int i;
};

int main() {
S myS;
myS.i; // is this a variable according to spec?

S::i is a "member variable".

Tom

C++ FAQ: http://www.parashift.com/c++-faq-lite/
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html

 

[Victor Bazarov]

struct S {
int i;
};

int main() {
S myS;
myS.i; // is this a variable according to spec?

It is an expression of type int& (a reference to int).

return 0;
}

Victor

 

[E. Robert Tisdale]

cat main.cc
struct S {
int i; // public data member
};

#include<iostream>

int main(int argc, char* argv[]) {
S myS;
myS.i = 33;
std::cout << myS.i << " = myS.i" << std::endl;
return 0;
}

g++ -Wall -ansi -pedantic -o main main.cc
./main

33 = myS.i

Yes, myS.i *is* a variable.
No, S.i is *not* a variable
because you never *declared and instance of S named S

cat main.cc

struct S {
int i; // public data member
};

#include<iostream>

int main(int argc, char* argv[]) {
S S;
S.i = 33;
std::cout << S.i << " = S.i" << std::endl;
return 0;
}

g++ -Wall -ansi -pedantic -o main main.cc
./main

33 = S.i

 

[Eugene Alterman]

It is an expression of type int& (a reference to int).

Correction:
It is an lvalue of type int (see 5.2.5/4).

And in general, an object-expression evaluates to the type of the object it
designates - lvalue or rvalue.
References evaluate to lvalues (see 5/6) .