Simple file write function in C++ [duplicate]

Answer 1

There are two solutions to this. You can either place the method above the method that calls it:

// basic file operations
#include <iostream>
#include <fstream>
using namespace std;

int writeFile () 
{
  ofstream myfile;
  myfile.open ("example.txt");
  myfile << "Writing this to a file.\n";
  myfile << "Writing this to a file.\n";
  myfile << "Writing this to a file.\n";
  myfile << "Writing this to a file.\n";
  myfile.close();
  return 0;
}

int main()
{
    writeFile();
}

Or declare a prototype:

// basic file operations
#include <iostream>
#include <fstream>
using namespace std;

int writeFile();

int main()
{
    writeFile();
}

int writeFile () 
{
  ofstream myfile;
  myfile.open ("example.txt");
  myfile << "Writing this to a file.\n";
  myfile << "Writing this to a file.\n";
  myfile << "Writing this to a file.\n";
  myfile << "Writing this to a file.\n";
  myfile.close();
  return 0;
}

Answer 2

This is a place in which C++ has a strange rule. Before being able to compile a call to a function the compiler must know the function name, return value and all parameters. This can be done by adding a "prototype". In your case this simply means adding before main the following line:

int writeFile();

this tells the compiler that there exist a function named writeFile that will be defined somewhere, that returns an int and that accepts no parameters.

Alternatively you can define first the function writeFile and then main because in this case when the compiler gets to main already knows your function.

Note that this requirement of knowing in advance the functions being called is not always applied. For example for class members defined inline it's not required...

struct Foo {
    void bar() {
        if (baz() != 99) {
            std::cout << "Hey!";
        }
    }

    int baz() {
        return 42;
    }
};

In this case the compiler has no problem analyzing the definition of bar even if it depends on a function baz that is declared later in the source code.

Answer 3

Your main doesn't know about writeFile() and can't call it.

Move writefile to be before main, or declare a function prototype int writeFile(); before main.

Answer 4

You need to declare the prototype of your writeFile function, before actually using it:

int writeFile( void );

int main( void )
{
   ...

Answer 5

Switch the order of the functions or do a forward declaration of the writefiles function and it will work I think.

Answer 6

The function declaration int writeFile () ; seems to be missing in the code. Add int writeFile () ; before the function main()