Question

我有代码：

typedef struct foo *bar;

struct foo {
    int stuff
    char moreStuff;
}

为什么以下内容会出现incompatible pointer type错误？

foo Foo;
bar Bar = &Foo;

据我所知，bar应该被定义为指向foo的指针，不是吗？

Answer 1

完整的代码应该是

typedef struct foo *bar;

typedef struct foo {  //notice the change
    int stuff;
    char moreStuff;
}foo;

和用法

foo Foo;
bar Bar = &Foo;

如果struct foo中没有typedef，则代码不会编译。

另外，请注意结构定义之后的; [以及int stuff 之后，尽管我认为错误 >]。

Answer 2

应该是这样的：

typedef struct foo *bar;

struct foo {
    int stuff;
    char moreStuff;
};


int main()
{
  struct foo Foo;
  bar Bar = &Foo;

  return 0;
}

Answer 3

像这样更改代码。

typedef struct foo *bar;

struct foo {
  int stuff;
  char moreStuff;
};

struct foo Foo;
bar Bar = &Foo;

否则你可以将typedef用于该结构。

typedef struct foo {
  int stuff;
  char moreStuff;
}foo;

foo Foo;
bar Bar=&Foo;

Answer 4

this code is very obfuscated/ cluttered with unnecessary, undesirable elements.

In all cases, code should be written to be clear to the human reader.
this includes:
-- not making instances of objects by just changing the capitalization.
-- not renaming objects for no purpose

suggest:

struct foo 
{
    int stuff;
    char moreStuff;
};

struct foo   myFoo;
struct foo * pMyFoo = &myFoo;

which, amongst other things, actually compiles

typedef结构中的指针

4 个答案: