C ++ basic accumulation (2) explanation of the relationship between pointers and arrays, pointers and functions

Reprinted! Please indicate the source when reprint: http://blog.csdn.net/aa4790139/article/details/8144416

This article will detail these relationships...

First, let's figure out the priority of the operator:

Priority	Operator	Description	Example	Overloading	Associativity
1	::	Scope resolution (C ++ proprietary)	Class::age = 2;	No	From left to right
2	++	Increasing suffix	I ++
	--	Decrease suffix	I --
	{}	Combination	{I ++; A * = I ;}
	()	Function call or variable Initialization	C_tor (int x, int y): _ x (x), _ y (y * 10 ){}
	[]	Array access	Array [4] = 2;
	.	Access members as objects	OBJ. Age = 34;	No
	->	Access members using pointers	PTR-> age = 34;
	dynamic_cast	Runtime check type conversion (C ++ proprietary)	Y & Y = dynamic_cast <Y &> (X );	No
	static_cast	Unverified type conversion (C ++ proprietary)	Y & Y = static_cast <Y &> (X );	No
	reinterpret_cast	Redefinition type conversion (C ++ proprietary)	Int const * P = reinterpret_cast <int const *> (0x1234 );	No
	const_cast	Change the extraordinary property (C ++ proprietary)	Int * q = const_cast <int *> (P );	No
	typeid	Obtain type information (C ++ proprietary)	STD: type_info const & t = typeid (X );	No
3	++	Increasing prefix	++ I		From right to left
	--	Decrease prefix	-- I
	+	Mona1 normal number	Int I = + 1;
	-	One dollar negative number	Int I =-1;
	! not	Non-logical !Alternate spelling	If (! Done )...
	~ compl	Bitwise Inversion ~Alternate spelling	Flag1 = ~ Flag2;
	(type)	Forced type conversion	Int I = (INT) floatnum;
	*	Reference	Int DATA = * intptr;
	&	Take the address of XX (refer)	Int * intptr = & data;
	sizeof	XX size	Size_t S = sizeof (INT );	No
	new	Dynamic Memory Allocation (C ++ proprietary)	Long * pvar = new long;
	new[]	Dynamic Array Memory Allocation (C ++ proprietary)	Long * array = new long [20];
	delete	Dynamic memory release (C ++ proprietary)	Delete pvar;
	delete[]	Dynamic Array Memory release (C ++ proprietary)	Delete [] array;
4	.*	Member object selection (C ++ private)	OBJ. * Var = 24;	No	From left to right
	->*	Member pointer selection (C ++ proprietary)	PTR-> * Var = 24;
5	*	Multiplication	Int I = 2*4;
	/	Division	Float F = 10.0/3.0;
	%	Modulus (remainder)	Int REM = 4% 3;
6	+	Addition	Int I = 2 + 3;
	-	Subtraction	Int I = 5-1;
7	<<	Shifts bits left	Int flags = 33 <1;
	>>	Shift bit right	Int flags = 33> 1;
8	<	Less than link	If (I <42 )...
	<=	Less than or equal	If (I <= 42 )...
	>	Greater than link	If (I> 42 )...
	>=	Greater than or equal to the link	If (I> = 42 )...
9	== eq	Equal to link ==Alternate spelling	If (I = 42 )...
	!= not_eq	Not equal to link !=Alternate spelling	If (I! = 42 )...
10	& bitand	Bit and &Alternate spelling	Flag1 = flag2 & 42;
11	^ xor	Bit XOR (exclusive or) ^Alternate spelling	Flag1 = flag2 ^ 42;
12	| bitor	Bit or (including or) |Alternate spelling	Flag1 = flag2 | 42;
13	&& and	Logic and &&Alternate spelling	If (conditiona & conditionb )...
14	|| or	Logic or ||Alternate spelling	If (conditiona | conditionb )...
15	c?t:f	Ternary conditional operation	Int I = A> B? A: B;	No	From right to left
16	=	Direct assignment	Int A = B;
	+=	And assign values	A + = 3;
	-=	Assignment by difference	B-= 4;
	*=	Assign values by multiplication	A * = 5;
	/=	Assign values by Division	A/= 2;
	%=	Returns the remainder value.	A % = 3;
	<<=	Shifts the value to the left using bits.	Flags <= 2;
	>>=	Shifts the value by bit to the right.	Flags> = 2;
	&= and_eq	Value by bit and &=Alternate spelling	Flags & = new_flags;
	^= xor_eq	Assign values by bit XOR ^=Alternate spelling	Flags ^ = new_flags;
	|= or_eq	Value by bit or |=Alternate spelling	Flags | = new_flags;
17	throw	Throw an exception	Throw eclass ("message ");	No
18	,	Cyclic Evaluation	For (I = 0, j = 0; I <10; I ++, J ++ )...		From left to right

Remember a rough idea. Otherwise, it would be too big to remember so many heads...

Operator priority: parentheses> Arithmetic Operators> bitwise operators> value assignment operators

Pointer: Int * P points to an int type pointer variable.

Pointer Array: Int * P [] an array composed of pointers (combined with P [] And then int *)

Array pointer: Int (* P) [] refers to a pointer to a one-dimensional array (first combined with (* P), then pointing to int []).

Pointer Functions: Int * P () returns an int type pointer function.

Function pointer: Int (* p) () pointer to a function entry

Two-dimensional pointer:Int ** P; pointer variable pointing to an int type pointer

 

Why do we need pointers? For example, why does an array exist?

For example:

Char s [3] [3] = {"A", "de", "H "};

Cout <sizeof (s) <Endl;

Char * P [] = {"A", "de", "H "};
 
// Cout <sizeof (p) <Endl; I don't know how to print the size of the memory occupied by the P pointer pointing to the address .... if a friend knows how to print a pointer pointing to the address to occupy the memory, please let me know? Thanks

Running result:

9

7

In fact, they all know the truth...

S [3] [5] The first character is not five characters long enough. It is an ending character, and one byte is not used. The same is true for the subsequent character... the array pointed to by P directly points to, no waste of money...

 

Adding char occupies one byte, so this array occupies

 

I think the following examples are quite good ~ Then, ask your answers and check if they are the same as the results below. In this way, your understanding and impressions can be deepened.

1. Int X [5]; int ** P = & X; do you think this is correct?

Answer: an error is returned because the type does not match. The P type is int **, And the & X type int (*) [5] is obtained because the X address is obtained, that is to say, this address is the address of the array, not just the pointer of the first element of the array (two-dimensional pointer), but the address of the entire array. The correct method is as follows: int (* P) [5] = & X;

2. How many bytes does int X [5]; int (* P3) [5]; how many bytes does P3 ++ move forward?

Answer: P3 points to an array, which is the entire array. Therefore, when P3 moves, it regards an array as a whole. So move 4*5 bytes (INT occupies 4 bytes)

3. Can int A [2] [3]; int ** P = & A; initialize it like this?

Answer: Certainly not. The type of... & A is int (*) [2] [3], and the type of P is int **;

1: int (* P) [2] [3] = &;

Correct syntax 2: int (* P) [3] =;

The second method is explained: A [0] And a [2] are also three pointers.

A [0] ---> A [0] [0], a [0] [1], a [0] [2]

A [1] ---> A [1] [0], a [1] [1], a [1] [2]

In fact, array A has two elements: a [0] And a [1]. Then, the value of a is the address of the first element, that is, & A [0]. What type is this? We know that if we regard a [0] as a whole, for example, if we use a to replace a [0], then a [0], A [1] is equivalent to a [0] [0], a [0] [1]. In this case, a is an array of the int type, and the & A type is actually the INT (* P) [3] type.

I hope you can read it carefully...

Thank you for reading my blog and hope you can see your footprints!
If something is wrong or incorrect,I also hope you can point it out!

References:

Http://zh.wikipedia.org/wiki/C%E5%92%8CC%2B%2B%E9%81%8B%E7% AE %97%E5%AD%90

Http://www.360doc.com/content/11/0506/22/6903212_114913991.shtml