As the title might look very confusing, let me give you an example:
typedef bool foo[2];
typedef foo bar[4];
bar what_am_i;
So, is what_am_i a [4][2] dimensional array as I presume, or a [2][4] dimensional array?
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5 Answers
It's bool[4][2]
You can verify it by static_assert:
static_assert(std::is_same<decltype(what_am_i), bool[4][2]>::value, "");
static_assert(std::is_same<decltype(what_am_i), bool[2][4]>::value, ""); // failed
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foo is an array with 2 elements of type bool, i.e. bool[2].
bar is an array with 4 elements of type foo, i.e. foo[4], each element is a bool[2].
Then what_am_i is bool[4][2].
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In order to complete @Slardar Zhang's C++ answer for C:
It's bool[4][2].
You can verify it by either one of the following:
sizeof(what_am_i)/sizeof(*what_am_i) == 4sizeof(*what_am_i)/sizeof(**what_am_i) == 2
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After inspecting the variable through the debugger, I found that I was right - what_am_i is a [4][2] dimensional array.
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When you don't know the type of a variable, one of the easy ways is this trick:
template<class T> struct tag_type{using type=T;};
template<class T> constexpr tag_type<T> tag{};
then do this:
tag_type<void> t = tag<some_type>;
almost every compiler will spit out a reasonably readable "you cannot assign tag_type<full_unpacked_type> to tag_type<void>" or somesuch (presuming your type isn't void that is).
You can also verify you have a good guess with
tag_type<your_guess> t = tag<some_type>;
sometimes you'll want to generate some_type via decltype(some_expression).
answered May 29, 2016 at 19:27
Yakk - Adam Nevraumont
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1 Comment
Nick Matteo
or you can
cout << typeid(what_am_i).name()—after piping through c++filt, it reports bool [4][2].