what is the difference between deference operator and pointer

[satyspiceoflife]

int x,a;
a=10
x=&a;
cout<<*x; //using dereference operator
this will print value 10

the other case is
int *x; //x is a pointer variable
int a=10;
x=&a;
cout<<*x; //printing value 10

what is the difference between these two cases?

[Frederick Gotham]

Richard Heathfield posted:

> int cout;

That's only a tentative definition, not a "proper" definition, unless
it's within a function.

I'm not quite sure what you mean. The following compiles for me without error
or warning, which would lead me to believe that it's a fully-fledged
definition:

int cout;

int main(void)
{
cout = 7;
return 0;
}

--

Frederick Gotham

[Ben Pfaff]

Frederick Gotham <fg*******@SPAM .comwrites:

Richard Heathfield posted:

>> int cout;

That's only a tentative definition, not a "proper" definition, unless
it's within a function.

I'm not quite sure what you mean. The following compiles for me without error
or warning, which would lead me to believe that it's a fully-fledged
definition:

From C99 (6.9.2):

If a translation unit contains one or more tentative
definitions for an identifier, and the translation unit
contains no external definition for that identifier, then
the behavior is exactly as if the translation unit contains
a file scope declaration of that identifier, with the
composite type as of the end of the translation unit, with
an initializer equal to 0.

--
"You call this a *C* question? What the hell are you smoking?" --Kaz