what is output

Question

#include<stdio.h>

void main(){

m();

void m(){
    printf("hi\n");

   }
}

 

please explain ?

Comments

I think compiler will throw some error!!

---

why?? please explain

---

declaration of m() after using m is wrong and also type mismatch void and int

Answer 1

Nested function is not supported in standard C i.e. ANSI C but supported in GCC with the help of an extension.

Nested function behaves as local variables and you can not call them before defining function body.

more on nested function : https://gcc.gnu.org/onlinedocs/gcc/Nested-Functions.html

http://stackoverflow.com/questions/2608158/nested-function-in-c

If no function prototype is declared before using it, here the compiler guesses or implicitly creates one depending on the arguments passed with return type as 'int'. If later we declare with same function name and if it does not match with the guessed one (by the compiler) then, compilers throws error. Warning type conflict because of the implicit 'int'.

But calling m() after defining works fine :

Comments

In compilation error:-

what is the meaning of static declaration of 'm' follows non static declaration

---

The most probable answer to your question :

If no function prototype is declared before using it, here the compiler guesses or implicitly creates one depending on the arguments passed with return type as 'int'. If later we declare with same function name and if it does not match with the guessed one (by the compiler) then, compilers throws error. Warning type conflict because of the implicit 'int'.

**PS: by looking at the error, it seems like guessed declaration is static and we declared a non static version later. But I am not sure at this point about how compiler assumes function declaration ( if not provided ) whether as static or extern.

---

Does compiler by default also assume the type of formal arguments as 'int'?

#include<stdio.h>
main()
{
float a = 15.5 ;
char ch = 'C' ;
printit ( a, ch ) ;

}
printit (float x,char y )
{

printf ( "\n%f %c", x, y ) ;
}

error: conflicting types for 'printit'
 

but ,

#include<stdio.h>
main()
{
float a = 15.5 ;
char ch = 'C' ;
printit ( a, ch ) ;

}
printit ( x, y )
{

printf ( "\n%f %c", x, y ) ;
}

output of this is

15.500000 ╠ 
if by default 'int' is assumed then output must be 15 67

Answer 2

hello @indrajeet,

Given code is 

1. #include<stdio.h>
2. void main(){
3. m();
4. void m(){
5.       printf("hi\n");
6.    }
7. }

Let us analyse the code:

• We are calling function m() at line 3.
• We have NOT declared or defined a function m(); in previous lines.
• So by default, compiler will assume that there will be a function in below lines with return type "int".  
• But at line 4, compiler sees a function definition of m() with return type void. It causes type mismatch.

Hence compiler throws an error.

 

 

Comments

function definition also can never be inside main....

Also, it will be a linker error, as m() is not declared...

So by default, compiler will assume that there will be a function in below lines with return type "int"

why this...?

 

Answer 3

while running this program i have received the following error:

prog.cpp:4:11: error: '::main' must return 'int'
 void main(){
           ^
prog.cpp: In function 'int main()':
prog.cpp:6:3: error: 'm' was not declared in this scope
 m();

ie., if you use the void method it should return some value
   ^
prog.cpp:8:9: error: a function-definition is not allowed here before '{' token
 void m(){
         ^