declaration Vs Definition

I did a search on google but didn't find satisfactory answer.
please tell clarify it to me.

what is the difference between variable declaration and definition?

A declaration says "there is a variable, somewhere, with this name". So

Code:

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extern int x;

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is a declaration.

A definition does that, plus sets aside the requisite memory location for the variable to live in. So

Code:

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int q;
float f = 6.5f;

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are definitions.

thank you.

would you please explain what does it mean by extern?

I could have searched for this but I think you would rather clarify it to me in a simple language and example.

"extern", in this case, essentially means defined elsewhere (i.e. externally) .

That distinction is used only in the context of external variables and functions. A declaration tells the type of the variable while its definition reserves storage for it.

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Originally Posted by itCbitC

That distinction is used only in the context of external variables and functions.

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Hence my usage of the words "in this case".

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Originally Posted by itCbitC

A declaration tells the type of the variable while its definition reserves storage for it.

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That's not true. Variable definitions require the type of the variable to be specified as well.

The standard is a little more specific, but a declaration specifies the interpretation, by the compiler, of a set of identifiers. A definition is a particular form of declaration. If the identifier is the name of an object, a definition causes storage to be reserved for that object (this is a variable definition). If the identifier is a function, the definition includes the function body. If the identifier is an enumeration constant or a typedef name, the definition is the only declaration allowed.

I was addressing the o/p when I posted that; and yes the definition declares the type and allocates storage for the variable, atleast that's what was implied.

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Originally Posted by grumpy

If the identifier is an enumeration constant or a typedef name, the definition is the only declaration allowed.

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Can you please give reason for why for typedef definition is the only declaration allowed?

Isn't typedef just for the compiler to replace with original declaration (sort of macro type functionality) ? or typedef's have some other functionality too?

I found the following explanation on some online C book.

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Although typedef is thought of as being a storage class, it isn't really. It allows you to introduce synonyms for types which could have been declared some other way. The new name becomes equivalent to the type that you wanted

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Originally Posted by vkaushal21

Isn't typedef just for the compiler to replace with original declaration (sort of macro type functionality) ? or typedef's have some other functionality too?

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There is no replacement since the typedef defines the name as a synonym for another type name. Unlike macros, typedefs obey the rules of scope.

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Originally Posted by vkaushal21

Can you please give reason for why for typedef definition is the only declaration allowed?

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Having a declaration that is not a definition only makes sense if whatever is declared can be used in some context without knowledge of the definition.

"typedef A B;" makes A a synonym (an alternative name) of B. No usage of A is possible without knowing exactly that it is a synonym of B.

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Originally Posted by vkaushal21

Isn't typedef just for the compiler to replace with original declaration (sort of macro type functionality) ? or typedef's have some other functionality too?

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To understand the answer to that, you need to realise there are various phases of compilation for any source file. The phases are specified in the standard and the net effect is as if they occur sequentially, with output from one phase being fed as input to the next. [Particular tool chains may implement things differently, but the net effect is required to be as if the phases were executed sequentially].

The preprocessor is one of the early phases, and the compiler is a later phase.

Macros are executed as text substitutions by the preprocessor, before the code ever reaches the compiler. A typedef is an instruction that is ignored (or allowed to pass through unmolested) by the preprocessor and acted on by the compiler.

That sequence is the reason that macros disobey scope but typedefs do (scoping rules are exercised by the compiler, not the preprocessor).

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Originally Posted by grumpy

"typedef A B;" makes A a synonym (an alternative name) of B. No usage of A is possible without knowing exactly that it is a synonym of B.

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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.

--
Mats

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Originally Posted by matsp

I always thought that the typedef works the other way around: B is a synonym of A

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That's correct, I often write

Code:

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typedef unsigned long long ulong

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QuantumPete

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Originally Posted by QuantumPete

That's correct, I often write

Code:

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typedef unsigned long long ulong

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QuantumPete

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Yes, I expressed it in a not so concrete way because in practice, BOTH sides are, after the typedef, synonyms - they mean the same thing and can be used in the same way. But it CREATES the new name from B in the original example (ulong in your example).

--
Mats

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Yes, I expressed it in a not so concrete way because in practice, BOTH sides are, after the typedef, synonyms - they mean the same thing and can be used in the same way.

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That's true if you're using typedef to assign a new name to an already existing type with an already existing name, as in "typedef int foo". But that's not the purpose of the typedef statement, because a macro would do as well.

The purpose of typedef is to introduce identifiers for types which don't already have a name.

Consider the following code snippet:

Code:

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int main()
{
        struct { int x; } *a;
        struct { int x; } b;

        a = &b;
}

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This won't compile. gcc tells us that it's an "assignment from incompatible pointer type". No macro can help us here, we need typedef:

Code:

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typedef struct {int x; } foo;

int main()
{
        foo *a;
        foo b;

        a = &b;
}

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Now it compiles just fine. Note that we didn't merely introduce a synonym for "struct { int x; }". Instead, we defined a new type called "foo" which has the same range and representation as "struct {int x;}". But these are not just different names for the same thing, these are different types, the most obvious difference being that "foo" has a name while the other hasn't. Also, they cannot be used interchangeably.

Greets,
Philip

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Originally Posted by Snafuist

That's true if you're using typedef to assign a new name to an already existing type with an already existing name, as in "typedef int foo". But that's not the purpose of the typedef statement, because a macro would do as well.

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Macros do not obey the rules of scope.

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Originally Posted by Snafuist

The purpose of typedef is to introduce identifiers for types which don't already have a name.

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The general purpose of typedef is to provide a synonym for another type. Consequently, you would right to say that that is a purpose of typedef.

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Originally Posted by Snafuist

But these are not just different names for the same thing, these are different types, the most obvious difference being that "foo" has a name while the other hasn't.

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No, the typedef makes foo a synonym for the anonymous type introduced by the struct definition. By itself, a typedef does not create a new type, at least according to the C standard.

EDIT:
I'd say that the reason why your first example gave the "assignment from incompatible pointer type" error is that the types of *a and b are different. If they were the same, then it would break the one definition rule since the same struct type would have been defined twice.

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Originally Posted by tabstop

A declaration says "there is a variable, somewhere, with this name". So

Code:

---

extern int x;

---

is a declaration.

A definition does that, plus sets aside the requisite memory location for the variable to live in. So

Code:

---

int q;
float f = 6.5f;

---

are definitions.

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in your example f is a definition but q is a tentative definition. it's only a definition if there isn't another definition for q somewhere else. if there is another definition then q is a declaration.

Code:

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

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the comments might not be right. i'm not sure if the order of tenativeness is defined so it could be the other way around.

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Originally Posted by Meldreth

Code:

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

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That won't compile!

Code:

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error: redeclaration of 'q' with no linkage

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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

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Originally Posted by QuantumPete

That won't compile!

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Actually, Prelude once showed me that it can:

Code:

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int q;
int q;

int main(void)
{
    return 0;
}

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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.

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Originally Posted by QuantumPete

That won't compile!

Code:

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error: redeclaration of 'q' with no linkage

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

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Originally Posted by Meldreth

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

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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:

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error: redeclaration of `int q'

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QuantumPete

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

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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.

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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.

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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.

Quote:

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Originally Posted by QuantumPete

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:

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error: redeclaration of `int q'

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QuantumPete

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then gcc is wrong, or your code is wrong. laserlight posted a program that uses tentative definitions. do you get the same error with that one?

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Originally Posted by Meldreth

then gcc is wrong

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

QuantumPete

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It seems to work only for globals.

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your code was wrong.

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Originally Posted by QuantumPete

It seems to work only for globals.

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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:

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

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Quote:

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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:

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

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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

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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.

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