I was told that a function defined as static in one .c file is not accessible from other files. But in the following program, I can access the static void show() function from another file. Is my understanding of static functions in C wrong?
a.h (first file):
static void show()
{
printf("I am in static show function in a.c");
}
b.c (another file):
#include"a.h"
void main()
{
show();
}
Remember that #includes work by copy-and-pasting the content of the included file. So in your example, after the #include has been processed, you get this:
static void show()
{
printf("I am in static show function in a.c");
}
void main()
{
show();
}
So clearly main can see show.1
The solution is to not #include .c files. In general, you should only #include header (.h) files. Your static functions shouldn't be declared or defined in the header file, so main will not be able to see it.
1. However, you now actually have two definitions of the show function, one in a.c and one in b.c. For static functions, this isn't a problem, but for non-static functions you would get a linker error.
static keyword changes the linkage specification to Internal Linkage.
A function marked as static will only be visible in that Translation Unit(TU).
Perhaps, You have same named symbols available in that particular TU, where you access the function. The how part of it can be only answered after you show us the code.
EDIT:
When you define a static function in header file, A copy of the same function gets created in every Translation Unit where you include it.Each instance of such a function is treated as a separate function(address of each function is different) and each instance of these functions have their own copies of static local variables & string literals.
Clearly, this will work but this might as well increase the size of your generated binary.
The other answers are correct, but it's not quite accurate to say that the static function is not accessible from another file. It is possible to access the function through a function pointer. It would be more accurate to say that the name of the function is not accessible in another translation unit.
Remember that converting C source code to an executable program consists of conceptual stages, including:
preprocessing (in which #include directives are replaced with the contents of the included file
compilation (which processes one translation unit at a time)
linking (in which the translation units are put together into the final program)
Suppose we have three files. foo.h:
typedef void (*void_function_p)(void);
extern void_function_p foo(void);
foo.c:
#include "foo.h"
#include <stdio.h>
static void baz(void) {
printf("worked!\n");
}
void_function_p foo(void) {
return baz;
}
bar.c:
#include "foo.h"
#include <stdio.h>
int main(void) {
(*foo())();
return 0;
}
This program compiles and prints "worked!" when it runs.
There are two translation units here. One is the code in the preprocessed foo.c (which, because of how #include works also includes the code in foo.h and stdio.h). The other is the code in the preprocessed bar.c (which, again, has its own copy of the code in foo.h and stdio.h).
By having the function foo return a pointer to the static function baz, we are able to call baz from the main function.
Now, consider what happens if we modify main to look like this:
int main(void) {
(*foo())();
baz();
return 0;
}
This code will result in a linker error because the name baz in this translation unit cannot be linked to the definition of baz in the other translation unit.
This is the first advantage of static functions: another programmer cannot accidentally access our baz function from another translation unit.
Now, consider what happens if we modify bar.c to look like this:
#include "foo.h"
#include <stdio.h>
static void baz(void) {
printf("still works!");
}
int main() {
(*foo())();
baz();
return 0;
}
This code will compile, and print "worked!" followed by "still works!"
This is the second advantage of static functions: we've defined two functions (in different translation units) with the same name.
If you try to put both static definitions in the same translation unit, you will get a compiler error about defining baz twice.
As a final note, if you take the program as it now stands and remove all the statics, it will result in a linker error because baz has been defined twice (with external linkage), which is not permitted.
Related
I am writing a C99 library that is distributed among several files, e.g.
// core.h
void my_private_fn();
void API_my_public_fn();
// core.c
#include "core.h"
void my_private_fn() {
do_something();
}
void API_my_public_fn() {
do_something_else();
}
// module_a.h
#include "core.h"
void API_useful_thing();
// module_a.c
#include "module_a.h"
void API_useful_thing() {
my_private_fn();
}
I want only the API_ prefixed functions to be visible by a program using the library, but I also need to expose my_private_fn in core.h in order to be used by module_a.c. Is there a way in C to make my_private_fn only visible within the library?
If the function had to only be visible in the compilation unit where it is defined, then you could declare it static. Because C language offers few possible scopes: a symbol can have only 3 scopes:
local to a block (the block can be a function or a block inside a function)
static scope (static declaration outside of a function): the symbol is only visible in the compilation unit where it is declared
global scope (non-static declaration outside of a function): the symbol is visible throughout the whole program.
At most, you can hide the declaration in a private include file that you do not declare in the official documented API. That way obedient users should not use it. But you cannot prevent users to put the declaration in their own code and use the function.
Put them in an internal header file that's only used inside the library and not distributed to end users—say, core_internal.h.
I found a neater way to lay out my code building upon Serge's answer that I selected, whose most merit goes to.
The key is to put the "private" functions in headers that are only included in C files, not in header files. This way the "private" symbols are available internally but not to an external caller. In a complete example:
core.h:
void my_public_fn();
core_priv.h:
void my_private_fn();
core.c:
#include <stdio.h>
#include "core.h"
#include "core_priv.h"
void my_private_fn() {
printf("Private function called.\n");
}
void my_public_fn() {
printf("Public function called.\n");
}
module_a.h:
#include "core.h"
void module_a_fn();
module_a.c:
#include "core_priv.h"
#include "module_a.h"
void module_a_fn() {
my_private_fn();
my_public_fn();
}
And if we want, we can group possibly multiple modules in a common library header.
library.h:
#include "module_a.h"
// etc.
This way, a program using the library only needs to include one file with only:
main.c:
#include "library.h"
int main() {
//my_private_fn(); // This triggers a compile warning.
my_public_fn(); // I can still reach the "core" public function.
module_a_fn(); // This calls the "private" function internally.
return 0;
}
Compiling with gcc -Wall *.c -o main.o and executing ./main.o yields:
Public function called.
Private function called.
Public function called.
I'm a beginner learning c. I know that use of word "static" makes a c function and variable local to the source file it's declared in. But consider the following...
test.h
static int n = 2;
static void f(){
printf("%d", n);
}
main.c
#include <stdio.h>
#include "test.h"
int main()
{
printf("%d", n);
f();
return 0;
}
My expected result was that an error message will throw up, since the function f and variable n is local to test.h only? Thanks.
But instead, the output was
2
2
EDIT:
If it only works for a compilation unit, what does that mean? And how do I use static the way I intended to?
static makes your function/variable local to the compilation unit, ie the whole set of source code that is read when you compile a single .c file.
#includeing a .h file is a bit like copy/paste-ing the content of this header file in your .c file. Thus, n and f in your example are considered local to your main.c compilation unit.
Example
module.h
#ifndef MODULE_H
#define MODULE_H
int fnct(void);
#endif /* MODULE_H */
module.c
#include "module.h"
static
int
detail(void)
{
return 2;
}
int
fnct(void)
{
return 3+detail();
}
main.c
#include <stdio.h>
#include "module.h"
int
main(void)
{
printf("fnct() gives %d\n", fnct());
/* printf("detail() gives %d\n", detail()); */
/* detail cannot be called because:
. it was not declared
(rejected at compilation, or at least a warning)
. even if it were, it is static to the module.c compilation unit
(rejected at link)
*/
return 0;
}
build (compile each .c then link)
gcc -c module.c
gcc -c main.c
gcc -o prog module.o main.o
You have included test.h in main.c.
Therefore static int n and static void f() will be visible inside main.c also.
When a variable or function is declared at file scope (not inside any other { } brace pair), and they are declared static, they are local to the translation unit they reside in.
Translation unit is a formal term in C and it's slightly different from a file. A translation unit is a single c file and all the h files it includes.
So in your case, the static variable is local to the translation unit consisting of test.h and main.c. You will be able to access it in main.c, but not in foo.c.
Meaning that if you have another .c file including test.h, you'll get two instances of the same variable, with the same name. That in turn can lead to all manner of crazy bugs.
This is one of many reasons why we never define variables inside header files.
(To avoid spaghetti program design, we should not declare variables in headers either, unless they are const qualified.)
I declared a global static function in one file
a.c
static void Func1(void);
void Func2(void);
void Func1(void) {
puts("Func1 Called");
}
void Func2(void) {
puts("Func2 Called");
}
and accessed it in b.c
#include <stdio.h>
#include "a.c"
void main() {
Func1();
Func2();
}
the program complied successfully, but as per provided information this should give error: undefined reference to Func1. What wrong is happening here?
You don't include a source file in another, you compile and link them together.
In you case, by saying #include "a.c", you're essentially putting the whole content of a.c into b.c and then starting the compilation, so the static functions and their calls are present in the same translation unit. Thus, there is no issue for compiler to find the called function.
Instead, if you do something like
gcc a.c b.c -o a.out //try to compile and link together
you'll see the expected error, as in this case, a.c and b.c will be two separate translation units.
You declare in header files and define in .c files. So you must use header files to represent the variables or functions you defined. Instead if you use .c files then it leads to multiple definitions.I think that is why you can access that global function.
As I was going through a huge source code, I came across some files where there is only one function declared in, let's say, "code.h" but in "code.c" there are many functions declared and defined.
What is the exact use of declaring and defining functions in .c files ?
How can we access one of those functions in a source file from other source file ?
It would be helpful if anyone can provide an example. Thanks
Splitting a complex routine in multiple smaller routines might make the code easier to understand and also would allow you to follow the single responsibility principle (What is an example of the Single Responsibility Principle?). By declaring and defining them only in a .c file (and declaring them static), you can hide the internal details from users.
If you, for instance, create a library, then your header could only show the public functions you'd like your users to use.
So only put your functions in a header if you'd like those functions available to anyone who includes the header.
If you'd like to keep a function internal to your .c file, seriously consider adding the static keyword as well: what is the advantage of static function?
---edit---
How can we access one of those functions in a source file from other source file ?
If the function isn't static, you could technically declare it anywhere* you'd like and use it there as well. Though, by the above argument, this would be bad practice.
Example:
a.c
int secret_a(int x)
{
return x+1;
}
main.c
#include <stdio.h>
//access a secret function from a.c
int secret_a(int);
int main()
{
printf("secret_a(3) gives: %d\n", secret_a(3));
return 0;
}
(*) anywhere allowed by the C language
How can we access one of those functions in a source file from other source file ?
That is the whole idea with defining some functions in the header, and some only in the c-file.
The functions defined in the header, can be called by other source files.
The functions only defined in the source file, can be hidden - it is a usual practice to mark functions, only declared in a source file, as static.
That makes their visibility limited to the file-scope (only code inside the same file can 'see' the functions).
The reason why this is done is to split the code into a more cleaner and sorted format so that functions concerning a particular operation can be grouped together in to one file and then called from other file. NOTE: this is not mandatory to do but a good programming practice.
Now there are two ways to do it :
1. including the files
2. Declaring the function as extern and then linking them at linking time.
1.
Including the file is the method that you are probably seeing in your code at hand. Functions are defined in one file and that file is included using #include"filename.[h/c]" and then calling the function as it were declared in the given file.
file_to_include.c
void my_func()
{
printf("Hello!\n");
}
file_that_include.c
#include<stdio.h>
#include"file_to_include.c"
int main()
{
my_func();
return 0;
}
to compile you just do
gcc file_that_include.c
using extern you can do the same but with out using the #include directive
file_to_include.c
void my_func()
{
printf("Hello!\n");
}
file_that_include.c
#include<stdio.h>
extern void my_func();
int main()
{
my_func();
return 0;
}
In this case while compiling you need to provide both files to the compiler
gcc file_that_include.c file_to_include.c
I would suggest reading up on the "extern" keyword.
From what I can understand in your question you have two files - "code.h" and "code.c". However, "code.h" only declares certain functions in "code.c":
code.h
int func1();
int func2();
code.c
int func1()
{
return 1;
}
int func2()
{
return 2;
}
int func3()
{
return 3;
}
main.c
#include "stdio.h"
#include "code.h"
extern int func3();
int main()
{
printf("%d ", func1());
printf("%d ", func2());
printf("%d ", func3());
}
Compiling using gcc:
gcc main.c code.c
yields:
main.c: In function ‘main’:
main.c:10:17: warning: implicit declaration of function ‘func3’ [-Wimplicit-function-declaration]
printf("%d ", func3());
and when executed:
1 2 3
The "extern" keyword basically tells the compiler "there is a definition for a function with this declaration" for compiling. During linking the linker will find the definition, and link to it.
This is however a clumsy way of doing it, since the caller ("main.c") needs to know what is going on in "code.c", and continually add more extern declaration, etc. etc.
So rather include functions you which to have global scope in "code.h". If there are functions in "code.c" which od not have global scope, consider making them "static".
Some more reading:
Effects of the extern keyword on C functions
I read that
A function with internal linkage is only visible to one compilation
unit. (...) A function declared static has internal linkage
For .c files it sorta makes sense, but I was wondering what happens with static functions in headers, which get included by multiple .c files but usually have an include guard.
I was reading this answer about static functions in headers, and the first item mentions that it doesn't create a symbol with external linkage and the second item mentions the function is available purely through the header file. Isn't that contradictory? How can the function be available and at the same time have no external symbol? So I did a little test:
/* 1.h */
#ifndef ONE_H
#define ONE_H
#include <stdio.h>
static void foo() {
printf("foo from 1.h %p\n", foo);
return;
}
void bar();
#endif
/* 1.c */
#include "1.h"
#include <stdio.h>
void bar() {
printf("foo,bar from 1.c %p,%p\n", foo, bar);
foo();
}
/* 2.c */
#include "1.h"
#include <stdio.h>
int main() {
printf("foo,bar from main %p,%p\n", foo, bar);
foo();
bar();
return 0;
}
...
debian#pc:~ gcc 2.c 1.c
debian#pc:~ ./a.out
foo,bar from main 0x400506,0x400574
foo from 1.h 0x400506
foo,bar from 1.c 0x400559,0x400574
foo from 1.h 0x400559
As expected bar is the same across all files, but shouldn't foo be too? Isn't 1.h included only once? Adding inline to foo resulted in the same behavior. I'm kinda lost.
Read here, how a header file basically works. That should clarify about your actual question.
Briefly: A header file is just inserted instead of the corresponding #include directive. So any declarations or definitions are treated by the compiler as if you actually copy/pasted the text into your file.
Anyway, you should be careful with function definitions in a header. This is deemed bad style in general. It blows the code for instance, creating redundant copies of that function. Also Function pointers cannot be compared (you never know ...). It is often better to bundle the functions into a library with just the declarations in a header (non-static then: external linkage). There are good justifications sometimes, however (no rule without exceptions). One of them are inline functions.
-static functions are functions that are only visible to other functions in the same file (more precisely the same translation unit).
Check this article for a detailed explanation on linkage: http://publications.gbdirect.co.uk/c_book/chapter4/linkage.html