Re: I don't use printf enough

[Plankton]

I find it kinda of ironic that in C++ the "preferred" way of writting to STDOUT is like so:

cout << "text text " << obj->method1() << " more text " << endl;

[jryan]

I assume you mean:

cout << "text text " << obj.method1() << " more text " << endl;

And would you explain why you find this ironic?

[elusion]

Presumably, he/she finds it ironic beacuse

cout << "text text " << obj.method1() << " more text " << end1;

is like using string concatenation in Perl. The root node says to use printf sometimes. printf is a C function, yet it would not be the preferred way of doing things in C.

So to prefer printf in Perl when something like the above (preferred) C snippet exists, is ironic.

[bunnyman]

printf is a C function, yet it would not be the preferred way of doing things in C.

That's half right - printf is a C function, but it is not preferred in C++. It is preferred in C, of course.

The reason why printf is not the best way in C++ is that the compiler cannot check the variable types. If you use "%s" to print a string, and then use an integer variable instead, the program crashes at run time. With perl it would just convert the integer into a string and not crash, and with C++ style I/O, it would print the integer as a string. Also, printf cannot be extended to print objects, it is only able to print a few built in types.

[pg]

His c++ syntax is just fine, his obj can be understood as a pointer to an object:

#include <iostream.h>

class aclass {
    public:
        int first_perfect_number(void); 
};

int aclass::first_perfect_number(void) {
    return 6;
}

int main() {
    aclass * ap = new aclass();
    cout << "first perfect number is: " << ap->first_perfect_number() 
+<< endl;
}

[jryan]

Right, but why on earth would you assume a pointer to an object in this case? The original bit wasn't talking about a reference to a perl object, so I assumed it was a typo. Just because the syntax passes the compiler doesn't mean it is what he meant. :)