About the pointer to a function in C

c, the assignment operation requires that the lvalue is of the same type as the rvalue.

The same type means that the variable itself is of the same type and the object it points to is of the same type, and both are indispensable.

For functions, the return value is the same, and the parameter type and number must be the same.

In addition to this, C also stipulates that variables must be defined before they can be used!

In your example, 1) the return value is the same, the parameter type and number are the same, so the compilation passes!

2) The return value is the same, the parameter type and number are the same, so the compilation can also pass!

So I don't understand what kind of specific problem the landlord is compiling!

Is the problem with the function being undefined?

int cmp(const void*,const void*);This is just a statement, not a definition! You have to define it later.

Also, the problem is related to the compiler's scanning process! This is a bit complicated, but you can get a general idea of how the compilation works!

You can try adjusting p=cmp; The position of the statement, such as inside or outside of main, and see what the difference is.

Here's what it looks like in general:

typedef int (*p)(int ,int );

p max;

p is a pointer type, which is equivalent to the same effect as "int*" in "int* a".

As for your situation, I can't explain it, you should post the error code.

1. Suppose the function void f(int b).

1) There is a definition int a[15], and f(a) is called, passing the first address.

2) If you call f(&a[1]) is equivalent to call f(a+1), you will pass the address, not the entire array.

Second, as mentioned in the first one, it is still the address.

int arr[3][4];

int (*a)[4] = arr;

a points to arr[0], the address of the first line.

A+1 points to arr[1], the second line address.

1.Arrays are all passed pointers, so they are the first address.

2.It's just the address of a[1].

Address 4In relation to the number of dimensions, if it is one-dimensional, move to the second element.

...How do you look like you didn't write Jeon-eun.

When used as arguments, int* a and int a are the same, both are the first address of the array.

a[1] passes the address of the second element of the array. And just operate on the address of the second element, without opening up another space.

2 If int* a is the address of A, 3 if a is int a[const]; a++ gets the address that is &a[1] plus an offset of the type size for each move.

By the way, BS does not give points.

The C++ pointer to a function is defined as:

Return type. *pointer name) (list of function parameters), for example.

void*p)(int) is to point to a return value of void

A function whose argument is an int type.

And how do you define a function pointer that points to a class member function? Use of member function pointers.

1) Non-static member functions.

Definition method: return type.

Class name: *pointer name) (function parameter list) for example, void

a::*p)(int) is a function pointer to a member function in class a.

Assignment method: p=&a:: function name, while the general function pointer is assigned to p=function name, pay attention to the difference. (The member function must be of type public).

Invocation method: The invocation of the member function pointer must be invoked through the class object, a*p(int) to call a member function (which is of type public).

2) Static member functions.

The definition and usage of static member functions are the same as those of ordinary function pointers, except that the method of assigning function pointers to non-static members is the same.

This is because the pointer type of a static member function is the same as that of a normal function.