What do you know about the awesome C language pointer?

Today, I read the pointer in the "Interview Guide for C/C ++ programmers". I have understood the pointer problem for a while.

Next, you can perform a test. If you can answer all the following pointer questions correctly, then your C language skills are amazing.

1. int * P;

2. Int ** P;

3. int * P [10];

4. INT (* P) [10];

5. int * P (INT );

6. INT (* p) (INT );

7. INT (* P [10]) (INT );

These seven questions are not very difficult. I believe most readers can answer them.

Answer:

1. A pointer to integer data

2. A pointer pointing to an integer data

3. An array with ten pointers pointing to integer data

4. A pointer pointing to ten Integer Data Arrays

5. For a function (not a function pointer), the function has an integer parameter, and the return value is a pointer to an integer.

6. A function pointer. The function has an integer parameter and the return value is of the integer type.

7. An array with ten pointers. the pointer in this array points to a function. This function has an integer parameter and returns an integer number.

Ah, at that time, I took a sigh of relief, just like a god, but I went on to see one of the following, and I collapsed completely .......

 

The question is as follows:

Parse the meaning of (* (void (*) () 0. // How is it? Hold it!

Analyze the problem:

I can't hold it anymore. The answer in the book is,

Some microprocessor starts from 0 address, sometimes in order to simulate the situation when the boot, you need to design a C statement to execute the content of 0 address, so there is (* (void (*) () 0.

This statement seems a headache at a glance, but it is quite simple after analysis.

First, when the following function declaration is available:

Void fun (PARAM );

This function is called in the form of fun (PARAM );

The function of the question has no parameters, so it is simplified to fun ();

0 is the entry address of the function, that is, 0 is the pointer value of the function. The pointer function declaration is:

Void (* pfun) (PARAM );

The call format is: (* pfun) (PARAM );

For this question, you can write: (* 0 )();

However, the function pointer variable cannot be a constant. Therefore, we need to forcibly convert 0 to a function pointer. According to the original question, the function pointer prototype of the metafunction is void (*)();

So (* (void (*) () 0) () can be analyzed in this way. First, void (*) () is used to forcibly convert 0 to a function pointer, then, call the function pointed to by function 0.

Typedef can be used to enhance the understanding of this sentence as follows:

Typedef void (* pfun )();

(* Pfun) 0 )();

These two sentences are equivalent to (* (void (*) () 0) (), but this helps to deepen the understanding of this sentence.

CONCLUSION: (* (void (*) () 0) () is the call of the function corresponding to * (void (*) () 0.

 

Awesome C language pointer... hope to help you grasp the pointer (* ^__ ^ *)