Documentation error on shaders.

[Lupinius]

The documentation (of SFML2) claims that you can use window.draw(sprite, shader); to draw something using a shader, when in reality this function does not exist anymore (or I'm just too blind to find it).

[Laurent]

It's Window::draw(Sprite, RenderStates) + the implicit constructor RenderStates::RenderStates(Shader*) which allows to write this.

You would have seen it by reading the documentation of the sf::RenderStates class.

[Lupinius]

I did actually use this, the documentation for the shader explicitly states this though:

To apply a shader to a drawable, you must pass it as an additional parameter to the Draw function:

window.draw(sprite, shader);

[Laurent]

Do you mean that this part of the doc is misleading? Because, right below, there is:

... which is in fact just a shortcut for this:

Code: [Select]

sf::RenderStates states;
states.shader = shader;
window.draw(sprite, states);

[Groogy]

The problem is probably that he have shader defined as:

sf::Shader shader;

when the RenderStates class expects a pointer. For it to work for him he has to take the reference to the shader by adding the & operator.

[Laurent]

the reference to the shader by adding the & operator.

The address of, not the reference to

[Groogy]

the reference to the shader by adding the & operator.

The address of, not the reference to

My mistake ^^
The operator name is "Reference" but is read as "Address of"

[Nexus]

& just has multiple meanings in C++:

• Type& is a reference to Type
  
• &expr is the (unary) address-of operator
  
• expr&expr is the (binary) bitwise AND operator

[Groogy]

• &expr is the (unary) address-of operator
  

Are you sure? According to Wikipedia it's name is Reference.
http://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B#Member_and_pointer_operators

[Lupinius]

Damn, I didn't get that the shader variable in the example line has to be a pointer. Guess I was confused because a variable with the same name was used as an object. Looking at the longer variant again it makes sense though. Thanks for the clearing that up

[Nexus]

Are you sure? According to Wikipedia it's name is Reference.

Yes, the standard mentions "address-of operator". Wikipedia is not a reliable source when it comes to C++ details. It's good to get an overview (also concerning C++11), but for in-depth knowledge there are better sources.

I guess they called it "reference" as an opposite to "dereference", but this is confusing because of the other meaning of &, namely the reference declarator (when used with a type). No one understands you when you say "reference" and mean "address-of"