[tordek]

As said elsewhere: "Declaration follows use".

    int *foo;

means "foo is a pointer to int". But that's explained different from how it's written. More clearly, you can say that "foo" is an int. Then, doing

    int *foo;
    foo = 50;

is obviously wrong because "foo" is not an int; however

    int *foo;
    *foo = 50;

is correct.

Similarly for arrays:

    int *foo[50];

means "foo is an array of pointers to int", or "foo[5] is an int".

From there follows that & is the antithesis of , and they negate each other:

    int *foo; //foo is a pointer to int.
    &*foo; //the address of the contents of foo
    foo; // same as above, but shorter.

[ewjordan]

Sorry, that didn't make much sense to me - in your first example, you can't say "foo is an int", because it's not, it's a pointer to an int. And in your last example, you can't say "foo[5] is an int", because it's not, it's a pointer to int.

In response to the original question, I think the answer is that

  int *foo;

is a common way of writing it because that's the way the compiler resolves it. As mentioned elsewhere,

  int* foo, bar;

is equivalent to

  int *foo; int bar

so treating the star as part of the type can cause problems.

But I agree fully - it makes a lot more sense to me to consider the pointer star as a flag on the type, not a modifier to the name. Just one of the many warts on C and C++ that make me happy I have to use them so infrequently...

[Edit: formatting, stars were getting swallowed when put inline]

[lmkg]

> [Edit: formatting, stars were getting swallowed when put inline]

The same thing happened to the post you're responding to, and that's why it's not making sense to you. The author really meant to say star-foo and star-foo[5], but instead italicized a bunch of text in between.

[ewjordan]

D'oh.

Don't know how I didn't realize that, given that my response got messed up so that my "corrected" version read exactly like the one I was confused about...

Please disregard everything I said, in that case. :)