CODEWITHSUNDEEP
c++pointersstatictypedef
Jonny Henly
Jonny Henly

Reputation: 4223

C++ Pointer Inquiry

So I've been fooling around with statics, pointers and typedefs and I've run into a brick wall and was wondering if someone could offer some insight.

I have a class Utilswith a static ofstream called theLog and I have another class Work that makes use of Utils::theLog. Here's the pertinent code:

Utils.h

#ifndef UTILS_H
#define UTILS_H

// includes

class Utils {
public:
  static ofstream theLog;

  // other code 
}
#endif

Work.cpp

 // includes

typedef ofstream* pOfStream;
static pOfStream l = &(Utils::theLog); // l is a lowercase L

// constructor and destructor

void Work::doSomeWork() {
    (*l) << "Hello, world!\n";
}
// other code

This compiles and gives no runtime errors. However the reason for defining pOfStream was to eliminate the filling of my code with asterisks, parentheses and Utils::theLog. When I try replacing (*l) with just l as in: l << "Hello, world!\n"; I get compiler errors. I'm assuming the reason being is that the operator << takes precedence over *, even though l is of the type pOfStream, which is defined as being a pointer to an ofstream object, however the code is not implicitly written *l << "Hello, world!\n";.

My question being, is there a way to use the extraction operator << without having to surround a pointer with parentheses and an asterisk? Also when you define a type like the above pOfStream and then assign a variable var to the type pOfStream, does the compiler just read through the code and replace any instance of pOfStream with ofstream*?

I've been searching SO and Google for the answer but I'm not exactly sure how to phrase the question.

Upvotes: 1

Views: 74

Answers (1)

vsoftco
vsoftco

Reputation: 56567

Why don't you just take the reference of the stream, such as

ostream &l = Utils::theLog;
l << "Hello, World!\n";

In this way, you don't have to use pointers.

Upvotes: 4

Related Questions