P * P & amp; P, * P

P * P & P, * P
Zookeeper

Background Summary

 

I attended the group C ++ seminar last night, which involves many aspects, including this pointer, heavy-duty operator function, array pointer and pointer array, static member static, and so on. When we talk about the knowledge points of * P and P, it gets stuck...

 

Starting from preparing C ++, we have generally reflected that * P and P are a bit confusing. Sometimes, it may happen to be correct, and changing the data won't happen again... The pointer is required to learn C ++. Today, I want to explain my understanding to you. Although the exam is coming soon, I still hope to help students who need it, correction is not appropriate.

 

P * P & P

 

To identify these expressions, you need to understand their meanings.

Before that, let's look at the following table in detail:

 

Expression	Int	Int * P
Type	Int	Int *
Variable	A	P

 

P, pointer variable. To learn the pointer, we first understand that the pointer stores the address pointing to the bucket. Simply put, the pointer variable stores the address, so P represents the address, the house number.

 

* P: value. The actual content of the address space memory that the P Pointer Points to is obtained, that is, the person in the house with the specified house number.

 

& P is an address. It is easy to understand that it is an address, so & P gets an address and a house number.

 

The following is an example:

Int a [5] = {1, 2, 3, 4, 5}, * P = a. Now I want to get the value 4, which has the following expressions:

A) * P + 4 B) * (P + 3) C) P + 3

 

Next, let's take a look at the meaning of each expression:

 

A, * P, value, and get the storage content of the first address of the bucket pointed to by P. Here is 1, so * P + 4 = 5, the option is incorrect.

B, * P, value; P, representing the address. Now P points to 20, so P + 3 = 23, which is equivalent to moving the pointer down three storage units and now pointing to address 23, so * (P + 4) = 4.

C. I have learned from option B that P + 3 = 23. It means the address, so C is incorrect.

 

The above is my understanding of the pointer. I am about to take the test. I wish you all the tests will pass!

What are the similarities and differences between C Language * p ++ and (* p) ++ and * (p ++?

* P ++ is equivalent to * (p ++), indicating to take the value of the unit referred to by p. p points to the next unit, that is, p automatically adds 1.
(* P) ++ indicates taking the value of the unit referred to by p. The value of this unit is incremented by 1.

// The following main test program and running status.
# Include <stdio. h>
Void main ()
{
Int a [] = {1, 1, 1}, * p;
P =;
* P ++;
Printf ("p = % d * p = % d \ n", p, * p );
P =;
(* P) ++;
Printf ("p = % d * p = % d \ n", p, * p );
}

// Program running status:
P = 1439896 * p = 1
P = 1439892 * p = 2
Press any key to continue...

* P ++, (* p) ++, * ++ p, ++ * p are different?

For example

Int a [5] = {1, 2, 3, 4, 5 };
Int * p =;

* P ++ first obtains the value pointed to by the pointer p (the first element 1 of the array), and then increases the pointer p by 1;

Cout <* p ++; // The result is 1.

Cout <(* p ++); // 1

(* P) ++ first goes to the value pointed to by the pointer p (the first element 1 of the array), and then increases the value by 1 (the first element of the array is changed to 2 ).
Cout <(* p) ++; // 1
Cout <(* p) ++) // 2
* ++ P first increments the pointer p by 1 (pointing to the second element of the array at this time), and * retrieves the value.

Cout <* ++ p; // 2
Cout <(* ++ p) // 2

+ * P first obtains the value pointed to by the pointer p (the first element 1 of the array), and then increases the value by 1 (the first element of the array is changed to 2)
Cout <++ * p; // 2
Cout <(++ * p) // 2

Note that each cout output above must be output separately to obtain the subsequent results.