differnces between declaration and definition of functions and global variables.

**sandeeppoudel** · 02-19-2018

hi,
i have confusion regarding access of global variables and functions. i know that global variable can only be accessed with extern syntax but function can be accessed from anywhere declaring its prototype. why global variable can't have a prototype? what is the process behind compiler which restricts from declaring global variables like we do with functions.

Code:

//file1.cpp
#include<iostream>
printHello();
int testInteger;
int main(){
printHello();
std::cout<<testInteger<<std::endl;

}

Code:

//file2.cpp
#include<iostream>
int testInteger=34;
printHello(){
std::cout<<testInteger<<std::endl;
}

in above two translation unit. i know we must use extern to use testInteger in file1 from file2. BUT why declaration of testInteger in file1 and definition in file2 restricted? why we can't do like above code? we can declare functions multiple times and define it once in any translation unit.
what is difference? i know we will get multiple definition error but why it happens? is declaration of integer itself is the definition of that integer?
please illustrate this explaining the process the compiler and linker works with function of this declaration and definition of functions and global variables.

**Salem** · 02-20-2018

Perhaps if you didn't omit so much stuff, things would be clearer.

Code:

//file1.cpp
#include<iostream>
extern void printHello();
extern int testInteger;
int main(){
    printHello();
    std::cout<<testInteger<<std::endl;
}

//file2.cpp
#include<iostream>
int testInteger=34;
void printHello(){
    std::cout<<testInteger<<std::endl;
}

This is also a function prototype (without the extern)

Code:

void printHello();

The ; at the end gives the compiler the clue that this isn't a function definition, which would begin with an opening brace at the position the ; occupies.
Like so.

Code:

void printHello() {
  // do stuff
}

Variables on the other hand have no such syntactic clue, so you have to use the extern keyword to indicate that a variable is declared elsewhere.

Symbols declared at the file scope are automatically visible to the whole program.

To restrict visibility, use the static keyword.

Code:

//file1.cpp
static int sum ( int a, int b ) {
  return a + b;
}
int main ( ) {
}

You couldn't call sum() directly from file2.cpp for example.

The static keyword can also be applied to variables as well.

**jackar56** · 02-20-2018

That can be a definition or a declaration, depending on what type T is:

Code:

typedef

Code:

void T();
T t;// declaration of function "t"

struct X { 
  T t;// declaration of function "t".
};

typedefint T;
T t;// definition of object "t".

In C++, when using templates, there is another edge case.

Code:

template

Code:

<typename T>
struct X { 
  staticint member;// declaration
};

template<typename T>
int X<T>::member;// definition

template<>
int X<bool>::member;// declaration!

The last declaration was not a definition. It's the declaration of an explicit specialization of the static member of X<bool>. It tells the compiler: "If it comes to instantiating

Code:

X<bool>::member

, then don't instantiate the definition of the member from the primary template, but use the definition found elsewhere". To make it a definition, you have to supply an initializer

Code:

template

Code:

<>
int X<bool>::member =1;// definition, belongs into a .cpp file.

**laserlight** · 02-20-2018

jackar56: I suggest that you post your code snippets as plain text within bbcode tags, with no additional markup. The forum software will add syntax highlighting and line numbering for you. As it stands, it looks like your code snippets are getting corrupted by the forum software choking on the parsing of the additional bbcode, resulting in what appears to be missing or incorrectly formatted (to the extent of becoming incorrect) code.

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