c++ - 为什么 typedef (typedef int* pt) 可以使 pt() 发生？

Question

我写了以下代码:

typedef int* pt;
pt p = pt();    // compile ok

但是

int* p = (int*)()   // compile fail.

我想知道为什么？它们有什么区别？

Answer 1

首先，函数式转换符号中不允许使用额外的括号。例如

int i1 =  int ();  // compiles
int i2 = (int)();  // doesn't compile

但即使省略括号，您的示例也不会编译

int* p = int*();   // doesn't compile

这是因为函数式强制转换表示法要求类型名称是简单类型说明符 或类型名称说明符。

§5.2.3 [expr.type.conv]

1   A simple-type-specifier (7.1.6.2) or typename-specifier (14.6) followed by a parenthesized expression-list constructs a value of the specified type given the expression list. ...
2   The expression T(), where T is a simple-type-specifier or typename-specifier for a non-array complete object type or the (possibly cv-qualified) void type, creates a prvalue of the specified type, whose value is that produced by value-initializing (8.5) an object of type T; no initialization is done for the void() case.

simple-type-specifier 基本上表示由单个单词组成的类型名称。由于同样的原因，下一行没有编译

unsigned int i = unsigned int(); // doesn't compile