Question

我已阅读了本书了解和使用C指针，并尝试编译下面的代码。但是在编译之后我得到了警告：从不兼容的指针类型中分配。

我检查了代码并找出函数指针fptrSet并且函数ShapeSetX不兼容，因为fptrSet的第一个参数是void *而函数ShapeSetX是Shape *。

我该如何解决这个问题？谢谢！

typedef void (*fptrSet)(void*, int);
typedef int  (*fptrGet)(void*);
typedef void (*fptrDisplay)();    

typedef struct _vfunc
{
    fptrSet     setX;
    fptrGet     getX;
    fptrSet     setY;
    fptrGet     getY;
    fptrDisplay display;    
} vFunc;    

typedef struct _shape
{
    vFunc function;
    int x;
    int y;
} Shape;

void displayShape(){
    printf("Shape\n");
}
void ShapeSetX(Shape *shape, int x){
    shape->x = x;
}
void ShapeSetY(Shape *shape, int y){
    shape->y = y;
}
int ShapeGetX(Shape *shape){
    return shape->x;
}
int ShapeGetY(Shape *shape){
    return shape->y;
}    

Shape *newShape()
{
    Shape *shape = (Shape *)malloc(sizeof(Shape));
    shape->x = 10;
    shape->y = 10;
    shape->function.setX = ShapeSetX;
    shape->function.getX = ShapeGetX;
    shape->function.setY = ShapeSetY;
    shape->function.getY = ShapeGetY;
    shape->function.display = displayShape;
    return shape;
}

Answer 1

你必须尊重指针定义：指针需要第一个参数是一个指向void的指针，所以你的函数实现应该有第一个参数为void：

void ShapeSetX(void *void_shape, int x){
    Shape *shape = (Shape*) void_shape;
    shape->x = x;
}
void ShapeSetY(void *void_shape, int y){
    Shape *shape = (Shape*) void_shape;
    shape->y = y;
}
int ShapeGetX(void *void_shape){
    Shape *shape = (Shape*) void_shape;
    return shape->x;
}
int ShapeGetY(void *void_shape){
    Shape *shape = (Shape*) void_shape;
    return shape->y;
}

Answer 2

我打算说“为什么不用void替换Shape？”，直到我意识到Shape尚未定义 - 你也不能交换这两个定义，因为Shape需要vFunc，需要typedef s。

所以，这样做：

typedef struct _shape Shape; // Define _shape and Shape later

typedef void (*fptrSet)(Shape*, int);
typedef int  (*fptrGet)(Shape*);
typedef void (*fptrDisplay)();

如果编译器不喜欢，您可能需要将其更改为：

typedef struct _shape; // Define _shape later

typedef void (*fptrSet)(struct _shape*, int);
typedef int  (*fptrGet)(struct _shape*);
typedef void (*fptrDisplay)();

由于参数，函数指针和函数不兼容

2 个答案: