Function-Parameters | Function : : Parameters - define functions and methods

First, you have to overload for pointer to member. Pointers to member aren't actually pointers, they are separate type representing an offset within type.

Sure, but you are missing the call to the lambda:

Because this is not a current feature of Swift. Once there is an associated type, there is always an associated type. It doesn't go away just because you constrain it. And once it has an associated type, it is not concrete.

There is no way to "inherit" protocols this way. What you mean is:

You use g_object_set() to set properties on elements. Simply instantiate the element:

This piece of code won't do what you want.

$1 will be replaced with the first function argument, not with _filter.

Also, this is open to SQL injection.

This is what you want:

[class.base.init]

7 The expression-list or braced-init-list in a mem-initializer is used to initialize the designated subobject (or, in the case of a delegating constructor, the complete class object) according to the initialization rules of [dcl.init] for direct-initialization.

[dcl.init]

19 If the initializer is a parenthesized expression-list, the expressions are evaluated in the order specified for function calls.

Given the above two paragraph, I would conclude that the order of evaluation in...

A "bivariant" parameter is one that is both covariant and contravariant.

The TypeScript 2.6 release notes describe the --strictFunctionTypes compiler option well, since that's the first release that supported it.

Under --strictFunctionTypes function type parameter positions are checked contravariantly instead of bivariantly. For some background on what variance means for function types check out What are covariance and contravariance?.

This article linked in that quote describes the difference between covariance and contravariance: https://www.stephanboyer.com/post/132/what-are-covariance-and-contravariance

The title is misleading, Perhaps you want to ask the difference between {0} and =0, not why a variable should be initialized. Please clarify the question.

int x{0}; initializes the variable to 0, equal to int x = 0;. It's a feature added in C++11.

Best practises include that initialization takes places in variables. If you leave the integer without initialization and since it has no default constructor, the compiler will not initialize it.

This means that by the time you use it will have an undefined value. This may not be a problem if you first write to it, but may be a serious issue if you forget that it's uninitialized and assume that it has a defaul value.

Here is a trivial bug many times seen in code:

The difference that you didn't spot is that wait returns the MutexGuard: