According to http://en.cppreference.com/w/cpp/language/explicit_cast , C-style cast and functional cast are equivalent. However, see the following example:
#include <array>
int main() {
std::array<int, 3> arr{};
(void)arr;
//void(arr);
}
While (void)arr compiles, void(arr) does not. What have I missed?
If there are no ambiguities (eg other functions with the same name, macros..) involved, the following code declares and defines two int
variables
int a = 22;
int (b) = 33;
thus you're trying to create a void variable type (with an existing name) .
And that's wrong because:
You're trying to use an existing name for another variable in the same scope
虽然void(arr)
通常与(void)arr
相同,但在此上下文中它是一个定义,并且您尝试创建一个名为arr
的变量,该变量类型为void
,这是不允许的。
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