c++ - 如何类型转换为模板成员函数参数?
问题描述
我实现了类模板writer
。具有writer
模板成员变量s_
。的类型s_
是Stream
。writer
期望有一个可以使用参数和Stream
调用的成员函数。const char* buf
size_t len
这是第一个版本writer
:
// clang++ -Wconversion test.cpp
#include <cstddef>
#include <ostream>
template <class F> struct size_arg_type;
template <typename Ret, typename Cls, typename T1, typename T2>
struct size_arg_type<Ret (Cls::*)(T1, T2)> {
using type = T2;
};
template <typename Stream>
struct writer {
writer(Stream& s):s_(s) {}
void write(const char* buf, size_t len) {
// The type of 2nd parameter depends on Stream
s_.write(buf, len);
}
Stream& s_;
};
struct user_stream1 {
void write(const char*, size_t) {}
};
struct user_stream2 {
void write(const char*, std::streamsize) {}
};
struct user_stream3 {
void write(const char*, size_t) {}
void write() {}
};
#include <sstream>
int main() {
{ // size_type is size_t
user_stream1 s;
writer w(s);
char buf[] = "123";
w.write(buf, sizeof(buf));
}
{ // size_type is std::streamsize
user_stream2 s;
writer w(s);
char buf[] = "123";
w.write(buf, sizeof(buf));
}
#if 1
{ // size_type is size_t but has overloaded member function
user_stream3 s;
writer w(s);
char buf[] = "123";
w.write(buf, sizeof(buf));
}
#endif
{ // size_type is std::streamsize return type is std::ostream&
std::stringstream s;
writer w(s);
char buf[] = "123";
w.write(buf, sizeof(buf));
}
}
运行演示:https ://wandbox.org/permlink/JtEHDG3plWxe4vwB
如果我将-Wconversion
标志设置为 clang++,我会收到以下警告。
clang++ -std=c++17 -Wconversion test.cpp
test.cpp:18:23: warning: implicit conversion changes signedness: 'size_t' (aka 'unsigned long') to
'std::streamsize' (aka 'long') [-Wsign-conversion]
s_.write(buf, len);
~~~~~ ^~~
test.cpp:50:11: note: in instantiation of member function 'writer<user_stream2>::write' requested here
w.write(buf, sizeof(buf));
^
test.cpp:18:23: warning: implicit conversion changes signedness: 'size_t' (aka 'unsigned long') to
'std::streamsize' (aka 'long') [-Wsign-conversion]
s_.write(buf, len);
~~~~~ ^~~
test.cpp:64:11: note: in instantiation of member function 'writer<std::__cxx11::basic_stringstream<char>
>::write' requested here
w.write(buf, sizeof(buf));
^
2 warnings generated.
Compilation finished at Wed May 1 09:37:37
我试图找到一种方法来抑制没有编译指示的警告。我想出了static_cast
办法。为了做到这一点static_cast
,我需要知道第二个参数类型。
所以我实现了一些参数类型提取器:
// clang++ -std=c++17 -Wconversion test.cpp
#include <cstddef>
#include <ostream>
template <class F> struct size_arg_type;
template <typename Ret, typename Cls, typename T1, typename T2>
struct size_arg_type<Ret (Cls::*)(T1, T2)> {
using type = T2;
};
template <typename Stream>
struct writer {
writer(Stream& s):s_(s) {}
void write(const char* buf, size_t len) {
write_impl(&Stream::write, buf, len);
}
template <typename Write>
void write_impl(Write, const char* buf, size_t len)
{
s_.write(buf, static_cast<typename size_arg_type<Write>::type>(len));
}
Stream& s_;
};
struct user_stream1 {
void write(const char*, size_t) {}
};
struct user_stream2 {
void write(const char*, std::streamsize) {}
};
struct user_stream3 {
void write(const char*, size_t) {}
void write() {}
};
#include <sstream>
int main() {
{ // size_type is size_t
user_stream1 s;
writer w(s);
char buf[] = "123";
w.write(buf, sizeof(buf));
}
{ // size_type is std::streamsize
user_stream2 s;
writer w(s);
char buf[] = "123";
w.write(buf, sizeof(buf));
}
#if 1
{ // size_type is size_t but has overloaded member function
user_stream3 s;
writer w(s);
char buf[] = "123";
w.write(buf, sizeof(buf));
}
#endif
{ // size_type is std::streamsize return type is std::ostream&
std::stringstream s;
writer w(s);
char buf[] = "123";
w.write(buf, sizeof(buf));
}
}
它按我的预期工作。但是在成员函数user_stream3
重载的情况下,会发生编译错误。write()
运行演示:https ://wandbox.org/permlink/TDPlQ3nXzIKjSlhY
为了获得成员函数的特定重载,我需要知道完整的成员函数指针类型。然而,它是不可预测的。
clang++ -std=c++17 -Wconversion test.cpp
test.cpp:18:9: error: no matching member function for call to 'write_impl'
write_impl(&Stream::write, buf, len);
^~~~~~~~~~
test.cpp:63:11: note: in instantiation of member function 'writer<user_stream3>::write' requested here
w.write(buf, sizeof(buf));
^
test.cpp:22:10: note: candidate template ignored: couldn't infer template argument 'Write'
void write_impl(Write, const char* buf, size_t len)
^
1 error generated.
Compilation exited abnormally with code 1 at Wed May 1 09:48:42
有什么好方法可以知道大小类型,或者在没有编译指示的情况下抑制警告?
解决方案
在第二个版本中获取 write 函数的地址时,它不知道要获取哪个重载地址。您需要充分约束模板参数,以免重载不明确:
template <typename Stream>
struct writer {
writer(Stream& s):s_(s) {}
void write(const char* buf, size_t len) {
write_impl(&Stream::write, buf, len);
}
template <typename Ret, typename Cls, typename T1, typename T2>
void write_impl(Ret (Cls::*)(T1, T2), const char* buf, size_t len)
{
s_.write(buf, static_cast<T2>(len));
}
Stream& s_;
};
https://wandbox.org/permlink/7Qb6xAoUQPRz3o2u
但是,如果有多个 2 参数write
函数,这仍然会失败。在这种情况下,您将不得不找到更多约束(例如,第一个 arg 必须是const char *
)或者只需要用户代码的特定签名。
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