z****e
发帖数: 2024 | 1 template
void apply(Seq& sq, R(T::*f)(A) const, A a) {
typename Seq::iterator it = sq.begin();
while(it != sq.end())
((*it++)->*f)(a);
}
请问
1. how does the compiler understand T::*f? a member function of T, which
returns R and take A? what does "f" represent?
2. if i change T::*f to be T::*g, all other code remain the same, is this
still alright? |
d****p
发帖数: 685 | 2
Yes
OK if you change *f(a) to *g(a). It just is a parameter name
【在 z****e 的大作中提到】
: template
: void apply(Seq& sq, R(T::*f)(A) const, A a) {
: typename Seq::iterator it = sq.begin();
: while(it != sq.end())
: ((*it++)->*f)(a);
: }
: 请问
: 1. how does the compiler understand T::*f? a member function of T, which
: returns R and take A? what does "f" represent?
: 2. if i change T::*f to be T::*g, all other code remain the same, is this
|
z****e
发帖数: 2024 | 3 这个 *f 根本就没有声明，怎么能直接就用呢？
而且原来的T里边，对应的函数，也不叫f呀。
从来没有见过这种，没有声明过的identifier。
而且如果是T里边的，至少要用typename意思一下吧。
很糊涂。
【在 d****p 的大作中提到】
:
: Yes
: OK if you change *f(a) to *g(a). It just is a parameter name
|
d****p
发帖数: 685 | 4
Aren't declaring it in the parameter list?
It is a pointer to member function, not a specific member function.
It is not a type; it is a variable.
【在 z****e 的大作中提到】
: 这个 *f 根本就没有声明，怎么能直接就用呢？
: 而且原来的T里边，对应的函数，也不叫f呀。
: 从来没有见过这种，没有声明过的identifier。
: 而且如果是T里边的，至少要用typename意思一下吧。
: 很糊涂。
|
z****e
发帖数: 2024 | 5 我貌似明白了许多。
【在 d****p 的大作中提到】
:
: Aren't declaring it in the parameter list?
: It is a pointer to member function, not a specific member function.
: It is not a type; it is a variable.
|
d****p
发帖数: 685 | 6
~~~~ :-)
There are two small issues in the code.
1. Since the member function is const, the iterator should better be const
iterator
2. It is a perfect case to illustrate use of functor. The code you showed
have two performance problems (1).
the seq.end() is called multiple times (2) the f cannot be inlined due to
use of member function pointer. It will
if we use functor to replace the function pointer.
【在 z****e 的大作中提到】
: 我貌似明白了许多。
|
