dereference

what is return type of assignment operator?

China☆狼群 提交于 2019-11-27 01:01:36
I am just starting C++. I am a bit confused about the return type of assignment and dereference operator. I am following the book C++ Primer. At various occasions, the author says that the return type of assignment operator is reference to the type of left hand operand but later on, he says that the return type is the type of the left hand operand. I have referred C++11 Standard Sec. 5.17, where the return type is described as "lvalue referring to left hand operand". Similarly, I can't figure out whether dereference returns the pointed-to object or the reference to the object. Are these

Dereferencing a pointer to 0 in C

喜你入骨 提交于 2019-11-27 00:50:31
问题 Sometimes data at memory address 0x0 is quite valuable -- take x86 real mode IVT as a more known example: it starts at 0x0 and contains pointers to interrupt handlers: a dword at 0x00 is a pointer to division by zero error handler. However, C11 language standard prohibits dereferencing null pointers [WG14 N1570 6.5.3.2], which are defined as pointers initialized with 0 or pointers initialized with a null pointer [WG14 N1570 6.3.2.3], effectively banning the very first byte. How do people

Is a closure for dereferencing variables useful?

淺唱寂寞╮ 提交于 2019-11-26 20:47:39
I'm not sure whether or when it is useful (to improve performance) to dereference variables. var x = a.b.c.d[some_key].f; while (loop) { do_something_with(x); } seems to be better than while (loop) { do_somthing_with(a.b.c.d[some_key].f); } Is that needed or is this done automagically by smart JavaScript engines? But my actual question is whether I should do this, for example, in a library. (function() { var slice = Array.prototype.slice; Function.prototype.x = function x() { var args = slice.call(arguments, 0); ... }; })(); or just Function.prototype.x = function x() { var args = Array

Why can't I treat an array like a pointer in C?

和自甴很熟 提交于 2019-11-26 20:44:13
I see this question a lot on SO. Maybe not in so many words... but time and again there is confusion on how arrays are different from pointers. So I thought I would take a moment to Q&A a few points about this. For purposes of this Q&A we're going to assume a 32-bit system and the following have been declared: char * ptr = "hello"; char arr[10] = "hello"; int iarr[10] = {0}; Here's a list of questions that surmise the confusion I see on SO. As I see new ones I'll add to my list of Q&A (others feel free to as well, and correct me if you see any mistakes!) Isn't a pointer and an array basically

Does dereferencing a pointer make a copy of it?

巧了我就是萌 提交于 2019-11-26 20:31:24
问题 Does dereferencing a pointer and passing that to a function which takes its argument by reference create a copy of the object? 回答1: In this case the value at the pointer is copied (though this is not necessarily the case as the optimiser may optimise it out). int val = *pPtr; In this case however no copy will take place: int& rVal = *pPtr; The reason no copy takes place is because a reference is not a machine code level construct. It is a higher level construct and thus is something the

Difference between pointer to pointer and pointer to array?

回眸只為那壹抹淺笑 提交于 2019-11-26 20:26:37
问题 Given that the name of an array is actually a pointer to the first element of an array, the following code: #include <stdio.h> int main(void) { int a[3] = {0, 1, 2}; int *p; p = a; printf("%d\n", p[1]); return 0; } prints 1 , as expected. Now, given that I can create a pointer that points to a pointer, I wrote the following: #include <stdio.h> int main(void) { int *p0; int **p1; int (*p2)[3]; int a[3] = {0, 1, 2}; p0 = a; p1 = &a; p2 = &a; printf("p0[1] = %d\n(*p1)[1] = %d\n(*p2)[1] = %d\n",

Difference between dereferencing pointer and accessing array elements

江枫思渺然 提交于 2019-11-26 19:52:12
问题 I remember an example where the difference between pointers and arrays was demonstrated. An array decays to a pointer to the first element in an array when passed as a function parameter, but they are not equivalent, as demonstrated next: //file file1.c int a[2] = {800, 801}; int b[2] = {100, 101}; //file file2.c extern int a[2]; // here b is declared as pointer, // although the external unit defines it as an array extern int *b; int main() { int x1, x2; x1 = a[1]; // ok x2 = b[1]; // crash

meaning of “referencing” and “dereferencing”

老子叫甜甜 提交于 2019-11-26 19:37:54
I read different things on the internet and got confused, because every website says different things. Speaking about C. I read about * referencing operator and & dereferencing operator; or that referencing means making a pointer point to a variable and dereferencing is accessing the value of the variable that the pointer points to. So I got confused. Can I get a simple but thorough explanation about "referencing and de-referencing"? A B Referencing means taking the address of an existing variable (using &) to set a pointer variable. In order to be valid, a pointer has to be set to the address

Is dereferencing null pointer valid in sizeof operation [duplicate]

主宰稳场 提交于 2019-11-26 19:06:08
This question already has an answer here: Why doesn't my program seg fault when I dereference a NULL pointer inside of malloc? 4 answers I've come across a snippet of code that to me should crash with a segmentation fault , and yet it works without a hitch. The code in question plus relevant data structure is as follows (with associated comment found right above): typedef struct { double length; unsigned char nPlaced; unsigned char path[0]; } RouteDefinition* Alloc_RouteDefinition() { // NB: The +nBags*sizeof.. trick "expands" the path[0] array in RouteDefinition // to the path[nBags] array

Pointer Arithmetic: ++*ptr or *ptr++?

僤鯓⒐⒋嵵緔 提交于 2019-11-26 18:45:26
I am learning C language and quite confused the differences between ++*ptr and *ptr++ . For example: int x = 19; int *ptr = &x; I know ++*ptr and *ptr++ produce different results but I am not sure why is that? templatetypedef These statements produce different results because of the way in which the operators bind. In particular, the prefix ++ operator has the same precedence as * , and they associate right-to-left. Thus ++*ptr is parsed as ++(*ptr) meaning "increment the value pointed at by ptr ,". On the other hand, the postfix ++ operator has higher precedence than the dereferrence operator