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people who use OSG know that OSG's memory management approach uses smart pointers that allow OSG to handle object destruction by means of smart pointers. There are two smart pointer types in OSG, one is the ref_ptr we are familiar with, and the other one may not always be used by many people, but it's really very good observer_ptr. Both of these types appear as smart pointers, so what's the difference between them? Why are there two things like that?
void* In fact can be understood as a context, using it can be very good to achieve C Language Object-oriented programming . I think the use of void*, not to look good, but a convention, the people who are programmed by C + + can see void*, and habitually think that-this is a private data that only the defined party has the right to interpret; and that's what we p
voidThe (nonexistent) value of avoidExpression (an expression, the has typevoid) shall not being usedinchAny , andImplicitOrExplicitConversions (except tovoid) shall not being applied to such an expression. If An expression of any other type isEvaluated asAvoidexpression, its value or designator isDiscarded. (AvoidExpression isevaluated forIts side effects.)void;/***************************************nullptr Both (int*) 0 and nullptr is valid templat
following cases: Program 1: void Mymalloc (char *s)//allocating memory in a function, and then returning { s= (char *) malloc (MB) ; void Main () { char *p=null; Mymalloc (P); The value of p actually or null,p here does not change if (p) free (p). Program 2: A void Mymalloc (char **s)//pointer
(* (Void (*) ()) 0) (), this is in the C trap and the flaw, regarding solves the computer after the boot, the hardware reads the first address to 0 the subroutine code. First Look (*FP) (), FP is a function pointer, *FP is the function that the pointer points to (*FP) () is the way that the function is invoked. We can literally write (*0) if we can, as the ques
A glance is a bit vague; note the difference: 1. function pointer: the type of the function return value.. Pointer function: the return value is a pointer (address) 2. Is int (* p) () the same as int * p? Why? Int (* p) () is a variable description, indicating that p is a pointer variable pointing to the functio
declaration of a function that returns an integer pointer.For example:void (*fptr) ();To assign a function's address to a function pointer, you can use the following two forms:fptr=function;Fptr=function;The address operator is not required because a single function identifier represents its address on a label, and if it is a function call, it must also contain a parameter table enclosed in parentheses.There are two ways to call a function by using
code, and in the dev C + + and VC + + compiled through!) ）#include void Main (){Declaring variables:int nnumber;int *ppointer;Now assign them values:Nnumber = 15;Ppointer = nNumber;Print out the value of the variable Nnumber:coutNow change the value of the Nnumber by the pointer:*ppointer = 25;Prove that Nnumber has been changed by the above procedure.Re-print the value of the Nnumber:cout}Read through the
typedef void (*sighandler_t) (int);sighandler_t signal (int signum, sighandler_t handler); Here, because a typedef has been added to customize a new type of sighandler_t, the second line of the function prototype looks much more pleasing, the form is the same as the int func (char c, int i), but if you do not understand the TypeDef statement, these two sentences is still a nightmare.To understand the TypeDef, just remember a word, that is: th
(Including the head pointer and tail pointer) implementation of various functions of the circular two-way linked list, pointer Implement the following functions for cyclic double-stranded tables: Void meau (); // menu function void Initlist (List * list); // initialize
the difference between void, NULL, and 0a detailed description of void:void literal means "no type" or "null type",void* is "needle-free pointer", that means void* can point to any type of data. As we all know, if pointers p1 and p2 are of the same type, then we can assign values to each other directly between P1 and P
Distinguish function pointer and pointer function _. There are a lot of complaints about the constantly changing handling of pointers and arrays, but it should be clear now. With the foundation of the previous lecture, the content of this lecture is relatively easy to understand. 1. pointer to the function (function pointer
Pointer (Computer programming)Pointer arithmetic cannot be performed on void pointers because the void type have no size, and thus the pointed address CA N not being added to, although GCC and other compilers would perform byte arithmetic on void* as a non-standard extension
1:goto disabled The meaning of 2:void void modifier function return value and parametersIf the function does not return a value, it should be declared as a void typeIf the function has no arguments, it should declare that its argument is voidThe void modifier returns values and parameters only to indicate that none
Reprinted from: http://blog.csdn.net/xiaofeng_yan/article/details/5248693 (it seems it is also reproduced from another) The following macro definition is displayed in #define min(x, y) ({ \ typeof(x) _min1 = (x); \ typeof(y) _min2 = (y); \ (void) (_min1 == _min2); \ _min1 This is a generic macro. The GCC extension supports the command typeof (). There a
There are two functions of void"1" No"2" any typeThere will be different interpretations of where void appears."1" void func (void)Func to the left of void, which means "no return value"The void in parentheses to the right of the
C ++ Pointer and Reference vomit blood arrangement pointer Reference, vomit blood Pointer Many people have a headache and confuse when talking about the pointer of C ++. It is often confused with variable names, references, etc. In fact, the main reason is that many programmers have problems with basic knowledge, resu
First, post a document Int p; // This is a common integer variable int * p; // It starts from P and is first combined with *, so P is a pointer and then combined with int, indicates that the Pointer Points to the int type of content. so P is a pointer to return integer data, int p [3]; // starting from P, it is first combined with [], indicating that P is an arra
See http://community.csdn.net/Expert/topic/3662/3662366.xml? Temp =. 1618006 there is a post about the excitement.The specific question is what is the macro function, # define macro1 (T, m) (INT) (void *) (T *) 0-> m ). Is to get the offset of T (structure or class) member m in the structure (or class. However, many people cannot understand (T *) 0-> M. I will analyze it. First write a program segment:# Pragma pack (1)Typedef struct {Char;Int B;};# P
example of the error, as follows:void GetMemory1 (char *p,int num){P=malloc (sizeof (int) *num);Return}void Test1 (){Char *p=null;GetMemory (P);}The above example is very common to apply a pointer as a parameter to a dynamic memory space, but this program is wrong.Cause of the error:Since *p is actually a copy of P in Test1, the compiler always makes a temporary copy of each parameter of the function. In t
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