在main外部的函数中使用struct [英] Use a struct in a function outside of main

问题描述

要在main（）之外的函数中使用struct，你能使用forward声明并在main（）中定义它，还是必须在块之外定义？

<如果在 main（）中定义一个结构，它将隐藏结构的全局名称。因此， main（）之外的函数将只能引用全局名称。这个例子取自C ++ 2011草案，第9.1节p2：

  struct s {int a; }; 
 
 void g（）{
 struct s; //隐藏全局结构s 
 //带有块范围声明
 s * p; //指局部结构s 
 struct s {char * p; }; // define local struct s 
 struct s; //重新声明，没有效果
} 
  

没有语法来引用本地从函数范围外部定义类型。因此，即使使用模板也会失败，因为没有办法表达模板的实例化：

  template< typename F>空栏（F * f）{f-> a = 0; } 
 
 int main（）{
 struct Foo {int a; } f = {3}; 
 bar（& f）; //在C ++ 98 / C ++ 03中失败，但在C ++中确定11 
} 
  

$ b b

实际上，这在C ++ 11中已经允许，如本 answer 中所述。

To use a struct in a function outside of main(), can you use forward declaration and define it in main(), or does it have to be defined outside of a block?

解决方案

If you define a structure inside of main(), it will hide the global name of the structure. So the function outside of main() will only be able to reference the global name. This example is taken from the C++ 2011 draft, Section 9.1 p2:

struct s { int a; };

void g() {
    struct s;              // hide global struct s
                           // with a block-scope declaration
    s* p;                  // refer to local struct s
    struct s { char* p; }; // define local struct s
    struct s;              // redeclaration, has no effect
}

There is no syntax to refer to the locally defined type from outside the function scope. Because of that, even using a template will fail, because there is no way to express the instantiation of the template:

template <typename F> void bar (F *f) { f->a = 0; }

int main () {
    struct Foo { int a; } f = { 3 };
    bar(&f);                         // fail in C++98/C++03 but ok in C++11
}

Actually, this is now allowed in C++11, as explained in this answer.

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