Question

int (*foo(int(*f)(int,int)))(int,int)
{
    printf("hello");
    return f+1;
}

int fun1(int x, int y)
{
    return x+y;
}
int fun2(int x,int y)
{
    printf("hey");
    return x*y;
}

int main()
{
    int (*fun_ptr[2])(int,int);
    int x,y;
    fun_ptr[0]=&fun1;
    x=fun_ptr[0](4,5);
    printf("%d
",x);
    fun_ptr[1]=&fun2;
   
    fun_ptr[0]=foo(fun_ptr[0]);
    y=fun_ptr[0](4,5);
    
    printf("%d",y);
    return 0;

}

我试图利用回报 f+1使 f返回点到fun2。因此,(在我看来)应该指出阵列中的下一个要素,即:fun_ptr? 因此,到现在为止,这应该是20个,但没有任何产出。页: 1

Answer 1

为了与阵列合作,点人必须指出阵列的要素。在这里,你有一系列职能要点,因此,一个要素的点子是a Pointer to a pointer to a function。页: 1

因此,如果你改变禁忌的声明,

int (**foo(int(**f)(int,int)))(int,int)

它应当按照你的期望开展工作。页: 1

Answer 2

Ok I fixed it 

int (*foo(int(**f)(int,int)))(int,int)
{
    
    return *f[1];
}

int fun1(int x, int y)
{
    return x+y;
}
int fun2(int x,int y)
{
    
    return x*y;
}

int main()
{
    int (*fun_ptr[2])(int,int);
    int x,y;
    fun_ptr[0]=&fun1;
    x=fun_ptr[0](4,5);
    printf("%d
",x);
    fun_ptr[1]=&fun2;
    
    fun_ptr[0]=foo(&fun_ptr[0]);
    y=fun_ptr[0](4,5);
    
    printf("%d
",y);
    return 0;

}

My foo function previously took a pointer to a function but doing that I couldn t travel 1 step inside the array fun_ptr. So, took a step back and passed the address of the array of function pointers instead. This worked.

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