[英]union of value and function pointer
I am struggling with using unions. 我正在努力使用工会。 Why am I unable to pass the function pointer to where the union would be?
为什么我无法将函数指针传递给union所在的位置? Any help would be greatly appreciated.
任何帮助将不胜感激。
Edit: removed a typedef 编辑:删除了一个typedef
#include <stdio.h>
union U {
int(*fnPtr)(int);
int i;
};
enum E {
OPTION_0 = 0,
OPTION_1 = 1
};
int multiply_by_two (int x) {
return 2 * x;
}
int f (int x, enum E e, union U u) {
switch (e) {
case OPTION_0:
/* Return the sum */
return x + u.i;
case OPTION_1:
/* Return 2 * x */
return u.fnPtr (x);
}
}
int main (void) {
int a;
scanf ("%d", &a);
int b = f (a, OPTION_1, &multiply_by_two);
printf ("%d\n", b);
return 0;
}
First, this definition is not valid: 首先,这个定义无效:
union U {
typedef int(*fnPtr)(int);
int i;
};
You can't have a typedef
inside of a struct
or union
. 您不能在
struct
或union
包含typedef
。 Removing the typedef will give you a proper definition: 删除typedef将为您提供正确的定义:
union U {
int(*fnPtr)(int);
int i;
};
The second problem is here: 第二个问题在这里:
int b = f (a, OPTION_1, &multiply_by_two);
The function f
expects a union U
, but you're passing it a int (*)(int)
. 函数
f
需要一个union U
,但是你传递的是一个int (*)(int)
。 Those types are not compatible. 这些类型不兼容。 Just because the union has a member of that type doesn't mean you can use that type wherever you would use the union.
仅仅因为union具有该类型的成员并不意味着您可以在任何使用union的地方使用该类型。 You need to create a union, set the proper field, then pass that to the function.
您需要创建一个联合,设置正确的字段,然后将其传递给该函数。
union U u;
u.fnPtr = multiply_by_two;
int b = f (a, OPTION_1, u);
In main function, try this: 在main函数中,试试这个:
int main()
{
...
union U myUnion;
myUnion.fnPtr = &multiply_by_two;
int b = f (a, OPTION_1, myUnion);
...
}
And also, the union definition is not correct, you need to remove typedef
. 而且,联合定义不正确,您需要删除
typedef
。
Just to add to other answers: this is usually called a variant data type, and it makes sense to keep the type enum in a struct
, along with the union
, since you will be passing it around all the time anyway. 只是为了添加其他答案:这通常被称为变量数据类型,并且将类型枚举保留在
struct
中以及union
是有意义的,因为无论如何你将一直传递它。
So I would recommend placing both in a struct
: 所以我建议将它们放在一个
struct
:
enum var_type
{
VAR_INT = 0,
VAR_FUNC = 1
};
struct variant
{
// contains the type of the stored value
enum var_type type;
// contains the actual value
union {
int(*fnPtr)(int);
int i;
};
};
And then you can have separate functions for creating each subtype, for simpler instantiation: 然后,您可以使用单独的函数来创建每个子类型,以便更简单地实例化:
// wrap the int value in the variant struct
struct variant variant_create_int(int i)
{
return (struct variant){ .type = VAR_INT, .i = i };
}
// wrap the functino pointer in the variant struct
struct variant variant_create_func(int(*fnPtr)(int))
{
return (struct variant){ .type = VAR_FUNC, .fnPtr = fnPtr };
}
Meaning your main
would do something like: 意思是你的
main
做的事情如下:
// create variant of type 'VAR_FUNC'
struct variant var = variant_create_func(&multiply_by_two);
and just pass the var
struct forward. 然后向前传递
var
结构。
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