问题
I was playing with SFINAE and found behavior I cannot explain.
This compiles fine:
template<typename Integer,
std::enable_if_t<std::is_integral<Integer>::value>* = nullptr>
void foo(Integer) {}
template<typename Floating,
std::enable_if_t<std::is_floating_point<Floating>::value>* = nullptr>
void foo(Floating) {}
While this (nullptr
replaced with 0
):
template<typename Integer,
std::enable_if_t<std::is_integral<Integer>::value>* = 0>
void foo(Integer) {}
template<typename Floating,
std::enable_if_t<std::is_floating_point<Floating>::value>* = 0>
void foo(Floating) {}
gives me a compile error:
prog.cpp: In function ‘int main()’: prog.cpp:13:10: error: no matching function for call to ‘foo(int)’
foo(3);
^ prog.cpp:5:6: note: candidate: template<class Integer, std::enable_if_t<std::is_integral<_Tp>::value>* <anonymous> > void foo(Integer) void foo(Integer) {}
^~~ prog.cpp:5:6: note: template argument deduction/substitution failed: prog.cpp:4:64: error: could not convert template argument ‘0’ to ‘std::enable_if_t<true, void>* {aka void*}’
std::enable_if_t<std::is_integral<Integer>::value>* = 0>
^ prog.cpp:9:6: note: candidate: template<class Floating, std::enable_if_t<std::is_floating_point<_Tp>::value>* <anonymous> > void foo(Floating) void foo(Floating) {}
^~~ prog.cpp:9:6: note: template argument deduction/substitution failed: prog.cpp:8:71: note: invalid template non-type parameter
std::enable_if_t<std::is_floating_point<Floating>::value>* = 0>
^
enable_if_t
expands to void
when there are no substitution failures, so I will have something like void* = 0
among the list of template parameters. Why does it break compilation?..
回答1:
Default template arguments follow their own conversion rules, which are stricter. Conversion of 0
to a pointer type in particular, is not applied.
See [temp.arg.nontype]/5.2 (emphasis mine):
for a non-type template-parameter of type pointer to object, qualification conversions ([conv.qual]) and the array-to-pointer conversion ([conv.array]) are applied; if the template-argument is of type
std::nullptr_t
, the null pointer conversion ([conv.ptr]) is applied.[ Note: In particular, neither the null pointer conversion for a zero-valued integral constant expression ([conv.ptr]) nor the derived-to-base conversion ([conv.ptr]) are applied. Although
0
is a valid template-argument for a non-type template-parameter of integral type, it is not a valid template-argument for a non-type template-parameter of pointer type. However, both(int*)0
andnullptr
are valid template-arguments for a non-type template-parameter of type “pointer to int.” — end note ]
来源:https://stackoverflow.com/questions/50765742/why-void-0-and-void-nullptr-makes-the-difference