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Steno

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We have a process where every commit gets code reviewed and graded.  I've had some good ones come my way and these were all rated as 'F' in the American grading scale.  These are all real, so I hope someone gets a chuckle out of this.  Keep in mind that we are a purely C++ shop.

 

   void SetX(int iNewX) { iX = iNewX; }

 

Comment:  int iNewX needs to be const.

 

float fMinDistance = 1.0e10f;  // Some arbitrary large distance

 

Comment:  This value is too large for a float.  Use a double.

 

char szFilename[STR_LENGTH];

 

Comment:  szFilename needs to be szFileName!!!!!  (note: !!!!! is not exaggerated)

 

 

glPushMatrix();
...
glPopMatrix();

 

Comment:  I don't know anything about OpenGL, but even I can see this is stupid!

 

// Foo is a POD type
struct Foo {
  Foo() : x(0), y(0) {}
  ...

 

Comment:  POD types can't be structs.

Comment:  What does this even mean?

 

 

OK, I have paraphrased some of the comments, but they are real.  Feel free to troll this thread for your own code review horrors.

Edited by Steno

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Well I usually define floating-point infinity as 1e20 in prototype code if the actual value is too difficult to reach (e.g. no INFINITY macro, and you have to type a complicated templated expression that you usually get wrong the first time.. thank you type abstraction!). 1e10 is a bit low though.. depending on uses.

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Comment: szFilename needs to be szFileName!!!!! (note: !!!!! is not exaggerated)

Filename can be a single world, both variants are widely used.

Comment: I don't know anything about OpenGL, but even I can see this is stupid!

Based on those two lines?

// Foo is a POD type
struct Foo {
  Foo() : x(0), y(0) {}
  ...
Comment: POD types can't be structs.
Comment: What does this even mean?

Foo is not a POD type, but not for the reasons you mentioned.

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A default constructor does not prevent a class/struct from being a POD type.

Frankly I'm not sure anymore. Quoting from wikipedia: 'Moreover, a POD class must be an aggregate, meaning it has no user-declared constructors', I'm not sure if declaring a defualt constructor falls under user-declared constructors.

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Frankly I'm not sure anymore. Quoting from wikipedia: 'Moreover, a POD class must be an aggregate, meaning it has no user-declared constructors', I'm not sure if declaring a defualt constructor falls under user-declared constructors.

Hmm. Now I'm confused too. Did this behaviour change in C++11?
struct MyPOD {
	int x, y;

	MyPOD() : x(0), y(0) {}
};

union MyUnion {
	int a;
	MyPOD pod;
};

int main() {
}
Clang with -std=c++11 compiles this just fine, but without -std=c++11 it gives me an error.

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Hmm. Now I'm confused too. Did this behaviour change in C++11?

Seems so: http://stackoverflow.com/questions/4178175/what-are-aggregates-and-pods-and-how-why-are-they-special/7189821#7189821

 

Edit: or well I'm not sure actually :P, any experts on this?

Edited by Mussi

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WTF? It just ate the important parts of my post... editing to try and add it back in...

What I tried to write:

Not an expert, but I was just writing a response about this...
 
Consider the following:

#include <type_traits>
#include <iostream>
 
struct Foo {
  Foo() : x(0), y(0) {}
  int x;
  int y;
};
 
struct Bar {
  Bar() = default;
  int x;
  int y;
};
 
int main()
{
    if (std::is_pod<Foo>::value) std::cout << "Foo is a POD" << std::endl;
    else                         std::cout << "Foo is *not* a POD" << std::endl;
    
    if (std::is_pod<Bar>::value) std::cout << "Bar is a POD" << std::endl;
    else                         std::cout << "Bar is *not* a POD" << std::endl;
}

This outputs:

Foo is *not* a POD
Bar is a POD
 

 

Let's look at what's going on...

 

Section 12.1, paragraph 5
A default constructor for a class X is a constructor of class X that can be called without an argument. If
there is no user-declared constructor for class X, a constructor having no parameters is implicitly declared
as defaulted (8.4). An implicitly-declared default constructor is an inline public member of its class.

...

A default constructor is trivial if it is neither user-provided nor deleted and ...

(note the last line, so Foo's constructor is not trivial because it is user-provided)

So let's look at what a POD is:

Section 9, paragraph 10
A POD struct is a class that is both a trivial class and a standard-layout class, and has no non-static
data members of type non-POD struct, non-POD union (or array of such types).

 

So now we ask, is Foo a trivial class?

Section 9, paragraph 6
A trivial class is a class that has a trivial default constructor (12.1) and is trivially copyable.

Nope, looks like Foo is not a trivial class, because it does not have a trivial default constructor, therefore it cannot be a POD.

Note that this is for C++11. I'd have to read through C++03 more thoroughly to be able to comment on C++03.

Edited by Cornstalks

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The change in C++11 is to unions.

From the wikipedia page: "Unions can now contain objects that have a non-trivial constructor." and "If so, the implicit default constructor of the union is deleted, forcing a manual definition."

So, its ok to have non-POD classes in unions now.
Though, it does say you have to add a default constructor to the union if you do... so still something unexplained going on with the union... Edited by Olof Hedman

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Though, it does say you have to add a default constructor to the union if you do... so still something unexplained going on with the union...

Ok, this was bugging me, but I think it makes sense.

You can have as many user-defined constructors as you like, but you have to have a default constructor, so that the compiler can construct an instance when it needs to.

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The change in C++11 is to unions.

There was certainly a change to the definition of PODs

 

Well, my point was that the reason that swiftcoders code compiled wasn't because the struct was POD (it wasn't), but because unions now allow non-POD members.

Ah, that makes sense now. I didn't realize it was directed at his post.

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