指針在typedef的結構

Question

我的代碼：

typedef struct foo *bar; 

struct foo { 
    int stuff 
    char moreStuff; 
} 

爲什麼下面舉一個incompatible pointer type錯誤？

foo Foo; 
bar Bar = &Foo; 

據我所知，bar應該被定義爲一個指向foo，不是嗎？

Answer 1

完整的代碼應該看起來像

typedef struct foo *bar; 

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

和使用

foo Foo; 
bar Bar = &Foo; 

，而不必struct foo typedef的，你的代碼將無法編譯。

另外，考慮到;結構後定義[後int stuff也，但我認爲更是一個錯字的。

Answer 2

更改這樣的代碼。

typedef struct foo *bar; 

struct foo { 
    int stuff; 
    char moreStuff; 
}; 

struct foo Foo; 
bar Bar = &Foo; 

否則，您可以使用該結構的typedef。

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

foo Foo; 
bar Bar=&Foo; 

Answer 3

這是應該的：

typedef struct foo *bar; 

struct foo { 
    int stuff; 
    char moreStuff; 
}; 


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

    return 0; 
} 

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