default-arguments

I was surprised the following code resulted in a could not deduce template argument for T error: struct foo { template <typename T> void bar(int a, T b = 0.0f) { } }; int main() { foo a; a.bar(5); return 0; } Calling a.bar<float>(5) fixes the issue. Why can't the compiler deduce the type from the default argument? In C++03, the specification explicitly prohibits the default argument from being used to deduce a template argument (C++03 §14.8.2/17): A template type-parameter cannot be deduced from the type of a function default argument. In C++11, you can provide a default template argument for

This question already has an answer here: Default value of function parameter 4 answers What's the place for the default parameter value? Just in function definition, or declaration, or both places? Default parameter values must appear on the declaration, since that is the only thing that the caller sees. EDIT: As others point out, you can have the argument on the definition, but I would advise writing all code as if that wasn't true. sharptooth You can do either, but never both. Usually you do it at function declaration and then all callers can use that default value. However you can do that

I want to pass a default argument to an instance method using the value of an attribute of the instance: class C: def __init__(self, format): self.format = format def process(self, formatting=self.format): print(formatting) When trying that, I get the following error message: NameError: name 'self' is not defined I want the method to behave like this: C("abc").process() # prints "abc" C("abc").process("xyz") # prints "xyz" What is the problem here, why does this not work? And how could I make this work? Adam Wagner You can't really define this as the default value, since the default value is