C program confusing output - c

#include<stdio.h>
int main(){
int main =22;
printf("%d\n",main);
return 0;
}
output:22
I am defining main as function and variable both even though compiler is not giving error. It should give error "error: redefinition of ‘main’ " . I am not able to understand why this code is working.

It will not give you an error because main is not a keyword. but main is define 2 times - Scoping rules come into play.

The main function is in the global scope - while the variable main is defined within the function main scope. They're not at the same level, thus there is no conflict.
The int main=22; line tells the compiler to use (declare) the local variable main - there is no conflict / ambiguity.
Do
int main(){
return 0;
}
int main =22;
on the other hand and you'll get an error.

The declaration of main inside the function creates a new identifier in the scope of the function. It does not override the main function which is defined in global scope.

#include <stdio.h>
#include <string.h>
void stuff();
main()
{
int val = 10;
printf("from main: %d\n", val);
stuff();
printf("from main: %d\n", val);
stuff();
}
void stuff()
{
int val = 5;
printf("from stuff: %d\n", val);
}
It doesn't matter defining the int val many times as it matters in which scope it's defined, this will output 10 5 10 5 , no errror no bad behavior

Related

Calling a function within another function (except main())

Let's assume I have 2 functions other than the main(), respectively func1() and func2(). Is it possible for me to call func1() in func(2) without declaring it first? Or should I use a pointer to the other function and pass it as an argument? Thanks in advance.
_"Is it possible for me to call func1() in func(2) without declaring it first?"_
It depends on compiler, but generally this will not work. (see exclusion example at bottom of answer.)
Some scenarios that illustrate:
Scenario 1: Normally will not work as function is being referenced before being declared:
int main(void)
{
int ret = func1();
return 0;
}
int func1(void)
{
return 1;
}
int func2(void)
{
return func1();
}
Results:
9, 15 error: implicit declaration of function 'func1' is invalid
in C99. Make sure that you include the function prototype.
Scenario 2: As all required definitions occur in order, this will compile and run without issue:
char func1(void)
{
return 1;
}
char func2(void)
{
return func1();
}
int main(void)
{
char ret = func1();
ret = func2();
return 0;
}
Results:
Compiles and runs with no problem because both functions are defined before being called (both func2() called from main() and func1() called from func1)
Scenario 3: The best way is always to pre-declare functions using prototypes either in same file before functions are called or in a header file that is #included in any source file that uses them. This clears up any potential problems, especially for those that inherit the code for maintenance:
int func1(void);
int func2(void);
int main(void)
{
int ret = func1();
return 0;
}
int func1(void)
{
return 1;
}
int func2(void)
{
return func1();
}
Regarding your comment:
"...a statement in my book caused confusion, I thought it might be related to a difference of the version of the C compiler in the book and I am using."
Could be: Per comment below, pre-standard C function definitions are supported by some modern compilers (eg gcc) thus would compile scenarios 1 & 2 without issue if functions complied with the default function definition; eg:
int func1(void)
int func2(void)
Here is how you do it:
void func2(int code); // forward declaration
void func1(const char* str)
{
func2(str[0]); // a call to a declared function
}
void func2(int code) // the callee
{
printf("code: %d\n", code);
}

Why isn't my function invoked? I don't understand why a declaration is expected

Hello folks out there,
this is my code:
main.c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include "sqlite3.h"
#include "db_typedefs.h"
#include "operations.h"
int main(){
printf("Text\n");
int f = 3;
void add_mini(3);
}
operations.h
#ifndef ADD_OPERATIONS_H_INCLUDED
#define ADD_OPERATIONS_H_INCLUDED
void add_mini(int flag);
#endif // ADD_OPERATIONS_H_INCLUDED
operations.c
void add_mini(int flag)
{
int rc;
rc = flag;
printf("%i\n", rc);
}
Operations.c has also libraries included similar to main.c.
Compiler Error
error: expected declaration specifiers or '...' before numeric constant
regarding to void add_mini(3)
It seems like I'm unable to pass a simple integer value. While debugging it's even skipping the add_mini line.
Do you have any idea what's going on?
The whole code is embedded in a larger query to determine typed orders but this works fine. I just can't pass this simple integer value.
Thanks in advance.
When you use
void add_mini(3);
the compiler thinks it is a function declaration, not a function call. The argument 3 is not valid for a function declaration. Hence, the compiler complains.
Remove the void part to call the function.
int main(){
printf("Text\n");
int f = 3;
add_mini(3);
}
or, since you have initialized f to 3,
int main(){
printf("Text\n");
int f = 3;
add_mini(f);
}
Call the function like so: add_mini(3); rather than void add_mini(3);
Remove the word void for calling add_mini from main.c :
add_mini(3);
Or
(void)add_mini(3);

C program compiles but no output

I am trying to learn creating header file in C and including it in my main.c func() . I created a simple tut1.c file with function named call() and a tut1.h header file which externs tut1.c function named call(). Thats it, now i am using eclipse Juno for C/C++ on linux fedora. I dont get any compile error but the code wont output? I tried on console and eclipse in vain. Can you check please? Thanks
---main.c-----
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#include "tut1.h"
int main (void)
{
int tut1(void);
return 0;
}
-----tut1.c------
#include <stdio.h>
#include <stdlib.h>
#include "tut1.h"
int call (void)
{
int *ptr;
int i;
ptr = &i;
*ptr = 10;
printf ("%d we are printing the value of &i\n", &i);
printf ("%d we are printing the value of *ptr\n", *ptr);
printf ("%d we are printing the value of ptr\n", ptr);
printf ("%d we are printing the value of &ptr\n", &ptr);
getchar();
return 0;
}
----tut1.h----
#ifndef TUT1_H_
#define TUT1_H_
extern int call (void);
#endif
You're not seeing anything because you're not calling the call() function from your main() function.
The main() function is the default entry point when you run the program, i.e. the first function that gets called during execution.
To execute the function call() you would need to call this from main() as follows :
int main (void)
{
int result = call();
return 0;
}
BTW, this line int tut1(void); within your main() just declares a function, which you do not seem to have defined anywhere. So I have removed it in the above shown code.

import function in c using extern keywod

I am having problem with importing external function to a main c file.
Here is my minimal code:
/* main.c */
#include<stdio.h>
extern int func()
int main(){
extern int func();
}
/*external file with one function that I want to
import*/
#include<stdio.h>
int func(){
printf("Hello World Again\n");
}
I compile and run like this - gcc main.c and then ./a.out but nothing is happening.
Any idea ?
You have to compile the file containing func also
gcc -Wall main.c external_file.c
(Note that the -Wall in the compiler command isn't absolutely necessary but is very good practice)
As noted by others, you also need to fix your code to call func rather than just re-declaring it.
Because you only declared the function, You never called it!
extern int func();
Declares a function. To call it you must have:
int main()
{
func();
}
You are just declaring again in main function..
you need to call the function to work..#include
extern int func()
int main(){
func();
}
/*external file with one function that I want to
import*/
#include<stdio.h>
int func(){
printf("Hello World Again\n");
}
Edits: question has changed.
extern is only used for external variables. You just need a prototype for the function.
#include <stdio.h>
void func(void); /* <-- prototype */
int main(int argc, char * argv[])
{
func();
return 0;
}
void func(void){
printf("Hello World Again\n");
}
Notice a few things. A prototype of int func() means no parameter checking in C - this is different to C++. Also, you are not returning anything from the function, so I replace it with void func(void)

Can preprocessor directive #include be disabled/excluded?

For example: If I have two .h files
process1.h and process2.h
and they contain two function with different output variables.
process1.h:
function(int var)
{
return 2*var;
}
process2.h:
function(int var)
{
return 10*var;
}
Can this be done in main.c:
int main()
{
int a = 2;
#include "process1.h"
printf("%d",function(a)); //output is 4
EXCLUDE #INCLUDE "process1.h" ????? <----can this be done in any way??
#include "process2.h"
printf("%d",function(a)); //output is 20
}
No, you cannot "un-include" a file. Think of all the preprocessor directives (lines starting with #) as happening before the actual C compiler even sees the source file. They just operate on the text of the file, and the preprocessor could be implemented as a separate step that just feeds new text into the actual compiler.
The best way to modify the actions of an include depending on the caller is to use further macros inside the included files, that you can #define before including them.
Still, your overall syntax is off, you can't (typically) nest functions in C.
No, and you should not try to write a program with two functions of the same name.
In the special case that the functions are actually defined in the header file (instead of just prototypes), you can do this:
#define function function_file1
#include "file1.h"
#undef function
#define function function_file2
#include "file2.h"
#undef function
int
main (void)
{
int a = 2;
printf ("%d\n", function_file1 (a));
printf ("%d\n", function_file2 (a));
}
BUT if you rename a function prototype then you haven't actually renamed the real function, so you'll get undefined symbol error when you link.
In any case, if you have two functions defined with the same name then it won't link anyway, not matter what else you do in the sources. (In C++, it is sometimes possible to define two things with the same name, but the One-Definition-Rule means the linker is allowed to assume they are both the same thing really and just pick one.)
This is why libraries are supposed to use names that won't be used elsewhere - usually by adding a common prefix to all symbol names (e.g. my_unique_lib_initialize()).
Why not use array of function pointers. Sure you need to initialize it at the start but I think it probably solves what you want to do.
int process1_function(int var);
int process2_function(int var);
int main(void)
{
int i, a = 10;
int (* functions[2])(int);
functions[0] = process1_function;
functions[1] = process2_function;
for(i=0; i < 2; i++)
{
printf("%d", (functions[i])(a));
}
return 0;
}
If you do not need to dynamically change which function you're going to call you can also just prefix the functions:
int process1_function(int var);
int process2_function(int var);
int main(void)
{
printf("%d",process1_function(a));
printf("%d",process2_function(a));
return 0;
}

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