Code goes first:
#include <stdio.h>
void foo()
{
static int bar;
}
int main()
{
bar++;
return 0;
}
The compiler(Clang) complains:
static.c:10:2: error: use of undeclared identifier 'bar'
Shouldn't the statement static int bar; in foo() give bar static storage duration, which makes it declared and initialized prior to main function?
You are confusing the scope of a variable with the storage duration.
As mentioned in the C11 standard, chapter §6.2.1, Scopes of identifiers,
for file scope
[...] If the declarator or type specifier that declares the identifier
appears outside of any block or list of parameters, the identifier has file scope, which
terminates at the end of the translation unit. [...]
and for function (or block) scope
[...] If the declarator or type specifier that
declares the identifier appears inside a block or within the list of parameter declarations in
a function definition, the identifier has block scope, which terminates at the end of the
associated block. [...]
In your case, bar has the file scope in foo(). So this is not visible in main().
OTOH, for the storage duration part,
An object whose identifier is declared without the storage-class specifier
_Thread_local, and either with external or internal linkage or with the storage-class
specifier static, has static storage duration. Its lifetime is the entire execution of the
program and its stored value is initialized only once, prior to program startup.
So, to summarize, bar has static storage duration, but the scope is limited to the foo() function. So, it is
declared and initialized prior to main() function
(before main() starts, to be exact) but not visible and accessible in main().
Marking something as static within a function relocates its storage off of the stack and allows its value to persist across multiple calls.
Marking something static, however, does nothing to change the scope of the variable. While you certainly could create a pointer that is aimed at bar and manipulate it from main, the compiler will view bar as undefined within main because of scoping.
Related
A simple question;
I know that function are extern by default in C. How about variables?
int a = 0;
int main(){
int b = 1;
return a+b;
}
Are a and b static/extern by default, or they don't have any storage specifiers if nothing is specified.
The variable a is defined outside the function. It is visible outside the current translation unit (TU) if the other TU contains extern int a;.
The variable b is defined inside the function. It is an automatic variable that is not visible outside the function.
See C11 §6.2.1 Scopes of identifiers for information about scopes. The variable a has file scope; the variable b has block scope.
See also §6.2.2 Linkages of identifiers for information about linkage. The variable a has external linkage. The variable b has no linkage.
See also §6.2.4 Storage durations of objects. The variable a has static storage duration. The variable b has automatic storage duration.
This question already has answers here:
What is the difference between a definition and a declaration?
(27 answers)
Closed 5 years ago.
Can you please someone explain me the flow of below problem,
#include <stdio.h>
int main(){
extern int a;
printf("%d\n",a);
return 0;
}
int a = 20;
and the output is 20. I am not sure where is the variable a getting defined and where is it getting declared?
The variable a is declared and defined as a global variable in the line:
int a = 20;
The extern line just tells the main() function scope that a is defined in another place.
In this case, the use of extern is not really necessary. You could just declare and define a before the main() function, and then main() would be familiar with it.
Usually, you would use extern when you want to use a variable or a function that was defined in another source file (and not just later in the same source file).
The C programming language has been designed to be one-pass, so that the compiler could process each line only once from top to bottom. So considering your program:
#include <stdio.h>
int main(){
extern int a;
printf("%d\n",a);
return 0;
}
int a = 20;
The identifier a is declared twice, and defined once.
Before the 4th line extern int a;, the compiler doesn't know anything about the identifier a. The declaration extern int a; has the block scope within the function main, and it declares the identifier a as an int and that its storage duration is static and linkage is external. So the compiler can write code that access a global identifier by name a as an int variable that could be defined in another module (external linkage). This is what the compiler does on line 5 when it is used in printf.
Finally at line 9, int a = 20; is another declaration and definition. This declares and defines a as a int with static storage duration, and external linkage.
If you'd put the int a = 20; before the main, the declaration extern int a; would be useless, because it doesn't add anything. I tend to put my main and other depending functions last in my source code, so that minimal amount of extra declarations are needed.
extern is syntactically a "storage class" keyword. But there is no such storage class. C has "static storage", "dynamic storage (malloc, etc) and "automatic storage" (local variables, usually represented using a stack).
If an identifier is declared extern inside a block scope, it means that the declaration refers to an external definition. If the entity being declared is an object, then it has static storage, simply because external objects have static storage. It can be a function too; functions aren't said to have storage.
In C, there is a concept called "linkage". Objects declared outside of any function at file scope, and functions, can have "external" or "internal" linkage.
If we have extern in a block scope, as you have in the example program, there can be a prior declaration of the same name at file scope, or in a nested scope, like this:
static int x;
/* ... */
{
extern int x;
}
Here, the inner x refers to the outer x, and, in spite of being "extern", it has internal linkage because of the "static".
In a nutshell, extern usually means "refer to the earlier declaration, and if there isn't one, declare this as an identifier with external linkage".
The word "external" refers to two separate concepts: the aforementioned "external linkage" and also to the meaning "outside of any function", as in "external declaration". Confusingly, "external declarations", like the static int x above, can have "internal linkage"!
In your program, things are correct because the block scope extern declaration of a and the later int a = 20, which are in separate scopes, happen to independently agree with each other.
The int a = 20; is an external declaration, which is also an external definition (because of the initializer). Since in that scope, no prior declaration of a is visible, it gets external linkage.
So, where is a defined? It is defined as an object with external linkage, in the entire translation unit as a whole. That translation unit is what defines a. a is declared in every place of the program where a declaration appears; and its definition is also a declaration. It is declared in main and also in the last line of the translation unit's source code.
A "declaration" is syntax which makes a name known in some scope. It's a concept that is active during the translation of a program. A "definition" is the fact that some object or function is provided in some translation unit. Translated units still provide definitions, but need not retain information about declarations. (Which is why when we make libraries, we provide header files with declarations in them!)
From the point of view of your main function, that function doesn't "care" where a is defined. It has declared a in such a way that if a is used, then an external definition of a, with external linkage, must exist. That definition could come from anywhere: it could be in the same translation unit, or in another translation unit.
When ever you declare a variable as extern then It means that Variable is declared as global and you cannot initialize the variable there.Because no memory is allocated for that variable It is just declared as a Variable
you can define it some where in your code.
Let us take an example ..consider the code
int main()
{
extern int i;
i=10;
printf("%d",sizeof(i));
}
here you get an error that int 'i' is not defined
therefore you need to write it as:
int main()
{
extern int i;
int i=10;
printf("%d",sizeof(i));
}
In case of your code:
This is declaration
extern int a;
This is definition:
int a = 20;
The storage class extern specifies storage duration and linkage of the object the identifier refers to:
The storage duration is set to static, which means the variable is alive for the whole time the program runs. As you declare that variable inside a function scope, this matters in your example, because in function scope, the default storage duration would be automatic.
The linkage is set to external, this just means different translation units of the same program can share the object. It has the side effect that a definition in another (file) scope is acceptable, as shown in your example. This is a logical consequence of the shared nature, you would typically declare a variable with external linkage in all translation units using it, but define it only in one.
In C, (GCC Compiler), what is the default storage class of global variables?
Let me share a code for example:
int i;
void f()
{ --statements--}
main()
{ --- set of statements--- }
Am I correct if I say that variable i would be static and not extern ?
In this code, i has static storage duration, and external linkage.
Storage duration refers to the lifetime of the variable's storage. Static storage duration means that the variable exists for the entire lifetime of the program.
Linkage refers to the relationship between names and objects. External linkage means that all instances of a name with external linkage denote the same object. Your int i; will match a declaration extern int i; from another translation unit.
The keyword static is used in different contexts as a storage duration specifier and as a linkage specifier, so it is not clear to ask whether a variable is static. Instead, the variable's storage duration and linkage should be considered.
You have the following cases for variables declared outside of any function:
static int variableWithfileScope; // Not exported. Only visible in current file.
int globalVariableThatIsExported; // Exported.
// Forward declaration. No variable defined.
extern int globalVariableThatIsDefinedSomewhereElse;
This is defined by the standard and is applicable to all C compilers.
I am trying a small example to know about the static external variable and its uses. The static variable is of local scope and the external variable is of global scope.
static5.c
#include<stdio.h>
#include "static5.h"
static int m = 25;
int main(){
func(10);
return 0;
}
static5.h
#include<stdio.h>
int func(val){
extern int m;
m = m + val;
printf("\n value is : %d \n",m);
}
gcc static5.c static5.h
o/p :
static5.c:3: error: static declaration of m follows non-static declaration
static5.h:3: note: previous declaration of m was here
EDITED
The correct program :
a.c:
#include<stdio.h>
#include "a1_1.h"
int main(){
func(20);
return 0;
}
a1.h:
static int i = 20;
a1_1.h:
#include "a1.h"
int func(val){
extern int i;
i = i + val;
printf("\n i : %d \n",i);
}
This works fine perfectly fine. But this is compiled into a single compilation unit. Hence could able to access the static variable . Across the compilation unit we cannot use the static variable by using the extern variable.
static has a very simple logic to it. If a variable is static, it means that it is a global variable, but it's scope is limited to where it is defined (i.e. only visible there). For example:
Outside a function: global variable but visible only within the file (actually, the compilation unit)
Inside a function: global variable but visible only within the function
(C++) Inside a class: global variable but visible only to the class
Now let's see what the C11 standard says regarding static and extern (emphasis mine):
6.2.2.3
If the declaration of a file scope identifier for an object or a function contains the storage-class specifier static, the identifier has internal linkage.
6.2.2.4
For an identifier declared with the storage-class specifier extern in a scope in which a prior declaration of that identifier is visible, if the prior declaration specifies internal or external linkage, the linkage of the identifier at the later declaration is the same as the linkage specified at the prior declaration. If no prior declaration is visible, or if the prior declaration specifies no linkage, then the identifier has external linkage.
6.2.2.7
If, within a translation unit, the same identifier appears with both internal and external linkage, the behavior is undefined.
So the standard says that first, if you have:
static int m;
extern int m;
then the second declaration (with extern) would regard the first one and in the end m would still be static.
However, in any other case, if there are declarations with both internal and external linkage, the behavior is undefined. This actually leaves us with only one option:
extern int m;
static int m;
i.e., extern declaration before static declaration. gcc was nice enough to give you error in this case of undefined behavior.
Remember this (quoting Eli Bendersky):
A static variable inside a function keeps its value between
invocations.
A static global variable or a function is "seen" only in
the file it's declared in
In your code, static int m = 25; means that m's scope is limited only to that file, that is, it is only visible inside static5.c and nowhere else.
If you would like to make use of m outside of static5.c make sure to remove the keyword static from the declaration of the variable.
For a more canonical explanation, along with an example, see this answer by Eli Bendersky
EDIT: (according to Klas' recommendation) **The actual scope is a compilation unit, not the source file. The compilation unit is the way the file looks after the preprocessor step
The problem is exactly as stated in the error message. m is declared a normal int but is later defined as a static int.
extern tells the compiler/linker to look for the variable in the global table of variables.
static (outside a functon) tells the compiler to exclude the variable from the global table of variables.
Do you see the conflict?
To fix the problem, either remove the static keyword from the definition or move the definition above the inclusion of static5.h.
It should be noted that the way you have designed your files is not considered best practice. Include files don't usually contain functions.
remove the keyword static while declaring m and the errors will be removed and you will be able to get the answer as 50. The static keyword makes the scope to restrict within the file.
#include <stdio.h>
static i = 5;
int main()
{
extern int i;
printf("%d\n",i);
return 0;
}
Can someone give any use-case for declaring a static variable as extern inside a function block?
NEW:
Why is this not allowed?
int main()
{
static i = 5;
extern int i;
printf("%d\n",i);
return 0;
}
this is useful when you need to access a variable that resides within another translation unit, without exposing the external variable globally (for a few reasons, like name collision, or that the the variable shouldn't be directly accessed, so static was used to limit its scope, but that TU's header still needs access).
As an example, lets say we have a translation unit foo.c, it contains:
//foo.c
static int i = 0;
i shouldn't be changed or directly accessed outside foo.c, however, foo.h comes along requiring access to i for an inline function, but i shouldn't be exposed to any translation unit using foo.h, so we can use extern at functional level, to expose it only during the scope of IncI, the inline function requiring the use of i:
//foo.h
inline void IncI(int val)
{
extern int i;
i += val;
}
Your second example is 'disallowed' because the compiler thinks you are trying to bind two different variables to the same symbol name, ie: it creates the static i at local scope, but searches for the extern int i at global scope, but doesn't find it, because static i as at the function scope. a more clever compiler would just fix the linkage to the static i, whether or not this follows standards I wouldn't know.
Now that I have a C standards document to work from (shame on me I know...), we can see what the official stance is (in C99):
6.2.2 Linkages of identifiers
Section 3:
If the declaration of a file scope identifier for an object or a function contains the storageclass
specifier static, the identifier has internal linkage.
Section 4:
For an identifier declared with the storage-class specifier extern in
a scope in which a prior declaration of that identifier is visible,
if the prior declaration specifies internal or external linkage, the
linkage of the identifier at the later declaration is the same as the
linkage specified at the prior declaration. If no prior declaration is
visible, or if the prior declaration specifies no linkage, then the
identifier has external linkage.
thus, because static will cause internal linkage, the extern will bring that linkage into the current scope. there is also a footnote stating that this may cause hiding of variables:
23) As specified in 6.2.1, the later declaration might hide the prior declaration.
In your second case, you are telling the compiler you have a static (local) variable i and another (global) variable i that is defined somewhere else.