Nested function in C

Question

Function is a variable so declaration of a function precede function definition,SO

#include<stdio.h>
main( )
{
void m( )
{
printf ( "hi" ) ;
}
m();
}

output: hi "Works correctly"

but if

#include<stdio.h>
main( )
{
m();
void m( )
{
printf ( "hi" ) ;
}
}

throws error because:

1)we have to define a function before calling it.

2)No prototype is declared before using it,so by default compiler will assume that there will be a function in below lines with return type "int" but thats not the case so type mismatch.

but if I decalre the prototype then,

#include<stdio.h>
void m();
main( )
{
m();
void m( )
{
printf ( "hi" ) ;
}
}

againg error:undefined reference to m

WHY?

last what I tried,removed the overhead of nested function

#include<stdio.h>
main( )
{
m();
}
void m( )
{
printf ( "hi" ) ;
}

there is type mismatch so it must throw error.BUT it runs correctly Why?

Comments

In 2) and 3) both cases , function definition is after calling the function inside another function main().So, when the function m() is called , not getting their definition. So, giving error

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i think if you are using standard (ANSI) c compiler , then you must be getting error in 1) itself , because in standard c functions are always external and C does not allow functions to be defined inside other functions.  these nested functions are allowed in extension of c (gnu c)

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It's a compiler dependent question , it's better to not to take care about those issues.

The point  a better programmer should keep in mind are (standard C)

1) declare a function/variable before using it.

2) In c function is actually global variable so you can't define them inside other function. // define them separately

3)Forget default types. explicitly mention return type and argument type.

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So,@rupendra  it means in standard C compiler last code must throws error because of type mismatch right?

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Where r u running the program? If u define function before u using then there shouldnot be any type mismatch error.

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I said nothing about which one will cause error or which one not. I talked about the things we should take care during writing programs. It's better if we don't give much trouble to compiler , let it just compile.

I'm taking about GCC. code 4 won't generate error as the type matching is done b/w declaration and definition. in code 3 both matching so no error. in code 4 you didn't declare so compiler doesn't have any prototype information to match with definition so no error but yes there you give some burden to compiler by not declaring function and by doing extra work it will may warn you about this mistake.

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@Srestha in the last code I am not using nested function inside the main.But this function is void type and as there is no prototype declaration has done before so I thought it will throws error but It runs correctly just with a warning message of type mismatch.

Btw I am using Codeblock :GNU GCC compiler.

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"in code 3 both matching so no error"
 

@Rupendra In code 3 there is matching but still error. 

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code 4 will definitely throw mismatch error if you use any other return data type other than int while defining as per c compiler . But here the case of void is again not present in standard c , gcc allowed use of void as well as int

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code 3 , ok you declared the function prototype but that should match with external definition of function which is not present , the inside definition of m() doesn't matter even if you remove the definition from main , you will get the same error

try this :

 

#include <stdio.h>
void m();
void n();
main()
{
    m();
    void m()
    {
    printf ("hi");
    }
    m();
    n();
}

void n()
{
    m();
}

void m()
{
    printf("hello");
}

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I am not considering "inside definition" as valid thing.

if you put the definition outside then i said it will work fine.

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ok the best thing is if there any nested function in C then it will throw error as in Standard C.