解释使用意外声明为函数的对象后出现的 GCC 错误
以下是语言新手常见的错别字，他们认为自己在定义一个对象，但实际上是在声明一个函数：
struct T
{
void foo() {}
};
int main()
{
T obj();
obj.foo();
}
GCC 4.1.2 的错误是：
In function 'int main()':
Line 9: error: request for member 'foo' in 'obj', which is of non-class type 'T ()()'
compilation terminated due to -Wfatal-errors.
为什么消息中报告的类型是 T ()()?我本来期望的是T ()。
The following is a common typo with language newcomers, who think that they are defining an object but are actually declaring a function:
struct T
{
void foo() {}
};
int main()
{
T obj();
obj.foo();
}
GCC 4.1.2's error is:
In function 'int main()':
Line 9: error: request for member 'foo' in 'obj', which is of non-class type 'T ()()'
compilation terminated due to -Wfatal-errors.
Why is the reported type in the message T ()()? I'd have expected T ().
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IIRC 这只是一个编译器错误。 GCC 4.4 对我来说是
T()而 4.2 是T()()。IIRC this is just a compiler bug. GCC 4.4 says
T()while 4.2 saysT()()for me.当您意识到通常不会在不命名函数的情况下写出函数类型时，就可以最好地理解该错误，但对于函数指针来说，这种情况更为常见。
例如，
int (*fooPtr)()命名指针。如果省略名称，则为int (*)()。现在，从函数指针到函数类型将得到int ()()。这里没有真正的标准，因为 ISO C++ 没有为所有类型定义规范名称。例如，
const volatile int与volatile const int类型相同，并且两种形式都不是规范的。The error is best understood when you realize that you usually don't write out function types without naming at least the function, but it's a bit more common for function pointers.
For instance,
int (*fooPtr)()names the pointer. If you omit the name, you haveint (*)(). Now, going from function pointer to function type would give youint ()().There's no real standard here, because ISO C++ doesn't define canonical names for all types. For instance,
const volatile intis the same type asvolatile const int, and neither form is canonical.