函数指针的使用

Question

可能重复：
How does dereferencing of a function pointer happen?

大家好， 为什么这两个码给出相同的输出， 案例1：

#include <stdio.h> 

typedef void (*mycall) (int a ,int b); 
void addme(int a,int b); 
void mulme(int a,int b); 
void subme(int a,int b); 

main() 
{ 
    mycall x[10]; 
    x[0] = &addme; 
    x[1] = &subme; 
    x[2] = &mulme; 
    (x[0])(5,2); 
    (x[1])(5,2); 
    (x[2])(5,2); 
} 

void addme(int a, int b) { 
    printf("the value is %d\n",(a+b)); 
} 
void mulme(int a, int b) { 
    printf("the value is %d\n",(a*b)); 
} 
void subme(int a, int b) { 
    printf("the value is %d\n",(a-b)); 
} 

输出：

the value is 7 
the value is 3 
the value is 10 

案例2：

#include <stdio.h> 

typedef void (*mycall) (int a ,int b); 
void addme(int a,int b); 
void mulme(int a,int b); 
void subme(int a,int b); 

main() 
{ 
    mycall x[10]; 
    x[0] = &addme; 
    x[1] = &subme; 
    x[2] = &mulme; 
    (*x[0])(5,2); 
    (*x[1])(5,2); 
    (*x[2])(5,2); 
} 

void addme(int a, int b) { 
    printf("the value is %d\n",(a+b)); 
} 
void mulme(int a, int b) { 
    printf("the value is %d\n",(a*b)); 
} 
void subme(int a, int b) { 
    printf("the value is %d\n",(a-b)); 
} 

输出：

the value is 7 
the value is 3 
the value is 10 

Answer 1

我会简化您的问题，以显示我认为您想知道的内容。

鉴于

typedef void (*mycall)(int a, int b); 
mycall f = somefunc; 

你想知道为什么

(*f)(5, 2); 

和

f(5.2); 

做同样的事情。答案是，函数名称都代表“函数指示符”。从标准：

"A function designator is an expression that has function type. Except when it is the 
operand of the sizeof operator or the unary & operator, a function designator with 
type ‘‘function returning type’’ is converted to an expression that has type ‘‘pointer to 
function returning type’’." 

当您使用间接运算符*上一个函数指针，即取消引用也是“功能符号”。从标准：

"The unary * operator denotes indirection. If the operand points to a function, the result is 
a function designator;..." 

所以f(5,2)第一规则基本上(*f)(5,2)变。这成为第二个call to function designated by f with parms (5,2)。结果是f(5,2)和(*f)(5,2)做同样的事情。

Answer 2

因为无论你用或不用引用操作使用这些函数指针自动解决。

Answer 3

你没有之前函数名

x[0] = addme; 
x[1] = subme; 
x[2] = mulme; 

但是这两种方式都是有效的使用&。