Thread: declaration Vs Definition

Originally Posted by Meldreth

Code:

int q; // definition
int q; // declaration/tentative definition

That won't compile!

Code:

error: redeclaration of 'q' with no linkage

A declaration tells the compiler "watch out, there's this symbol you should know about. I'll tell you more later".
A definition tells the compiler "so this is everything you need to know about this symbol"
You don't need a declaration, but you do need to have a definition (depending on where you need to use the symbol).

QuantumPete

Originally Posted by QuantumPete

That won't compile!

Actually, Prelude once showed me that it can:

Code:

int q;
int q;

int main(void)
{
    return 0;
}

Unfortunately, I am not conversant enough in this area to tell you exactly what the C standard says, much less why C has such a feature.

Originally Posted by Meldreth

you're compiling as c++. c++ doesn't allow tentative definitions.

The code is in a file called test.c and I'm running gcc test.c (gcc version 3.4.6)
With g++ I get:

Code:

error: redeclaration of `int q'

QuantumPete

Okay, back from a shower, and now I get what the C standard says

Originally Posted by C Standard, 1999 edition, Section 6.2.2, Paragraph 2

In the set of translation units and libraries that constitutes an entire program, each declaration of a particular identifier with external linkage denotes the same object or function. Within one translation unit, each declaration of an identifier with internal linkage denotes the same object or function. Each declaration of an identifier with no linkage denotes a unique entity.

Originally Posted by C Standard, 1999 edition, Section 6.2.2, Paragraph 6

The following identifiers have no linkage: an identifier declared to be anything other than an object or a function; an identifier declared to be a function parameter; a block scope identifier for an object declared without the storage-class specifier extern.

So, in my adaptation of Prelude's example, q has internal linkage, so the two declarations denote the same object, hence there is only one definition of q, and that is perfectly fine. If we make q into a local variable, it has no linkage, which means that each declaration of q denotes a unique entity, thus the error.

I still do not understand why it is the way that it is, but at least I can definitely say that it really is part of the core language and not a compiler extension.

Originally Posted by Meldreth

then gcc is wrong

I find that hard to believe!
But yeah, laserlight's code compiles fine. It seems to work only for globals.

QuantumPete

Originally Posted by QuantumPete

It seems to work only for globals.

To be precise, it only works for objects with linkage. You probably did this, in which case q has no linkage, thus gcc is correct to report an error:

Code:

int main(void)
{
    int q;
    int q;
    return 0;
}

Originally Posted by laserlight

To be precise, it only works for objects with linkage. You probably did this, in which case q has no linkage, thus gcc is correct to report an error:

Code:

int main(void)
{
    int q;
    int q;
    return 0;
}

Yes, that's precisely what I did. I'm intrigued now though. Why would internal linkage allow multiple declarations of the same symbol? What's the use? I've tried googling, but it doesn't seem to be a widely discussed issue.

QuantumPete

Originally Posted by matsp

I always thought that the typedef works the other way around: B is a synonym of A - meaning that B didn't exist before as a name, and is now the equivalent of A.

Yeah, indeed. I wrote quickly, and mixed up whether A or B is the synonym. The basic message, however, is correct.