Related
Diapier In this code snippet, the large lookup table is easier to access using a pointer to the VLA : #pragma GCC diagnostic warning "-Wcast-qual"
char
lookup(int a, int b, int c, char const *raw, int x, int y, int z)
{
typedef char const (*DATA_PTR)[a][b
Diapier In this code snippet, the large lookup table is easier to access using a pointer to the VLA : #pragma GCC diagnostic warning "-Wcast-qual"
char
lookup(int a, int b, int c, char const *raw, int x, int y, int z)
{
typedef char const (*DATA_PTR)[a][b
Diapier In this code snippet, the large lookup table is easier to access using a pointer to the VLA : #pragma GCC diagnostic warning "-Wcast-qual"
char
lookup(int a, int b, int c, char const *raw, int x, int y, int z)
{
typedef char const (*DATA_PTR)[a][b
Diapier In this code snippet, the large lookup table is easier to access using a pointer to the VLA : #pragma GCC diagnostic warning "-Wcast-qual"
char
lookup(int a, int b, int c, char const *raw, int x, int y, int z)
{
typedef char const (*DATA_PTR)[a][b
Diapier In this code snippet, the large lookup table is easier to access using a pointer to the VLA : #pragma GCC diagnostic warning "-Wcast-qual"
char
lookup(int a, int b, int c, char const *raw, int x, int y, int z)
{
typedef char const (*DATA_PTR)[a][b
emkey08 C++ provides implicit conversions from T *to const T *. If now used T *in container classes like in vector<T *>, then of course there is no implicit conversion anymore vector<const T *>. Using a reinterpret_castseems to cast the entire container, but i
emkey08 C++ provides implicit conversions from T *to const T *. If now used T *in container classes like in vector<T *>, then of course there is no implicit conversion anymore vector<const T *>. Using a reinterpret_castseems to cast the entire container, but i
Anton Tretyakov The good guys of the internet! Throughout my new bee journey through C's stormy land, I found this code: void * add (void * _set, const void *_element)
{ int * set = _set;
const int * element =_element;
assert(set > heap && set < heap
Algildas Predisius Imagine a person (for one reason or another) wants to implement a function that acts as a proxy when calling the function. One could implement it as a template that takes a pointer to a function and all its parameters. However, if there is a
Anton Tretyakov The good guys of the internet! Throughout my new bee journey through C's stormy land, I found this code: void * add (void * _set, const void *_element)
{ int * set = _set;
const int * element =_element;
assert(set > heap && set < heap
Anton Tretyakov The good guys of the internet! Throughout my new bee journey through C's stormy land, I found this code: void * add (void * _set, const void *_element)
{ int * set = _set;
const int * element =_element;
assert(set > heap && set < heap
Anton Tretyakov The good guys of the internet! Throughout my new bee journey through C's stormy land, I found this code: void * add (void * _set, const void *_element)
{ int * set = _set;
const int * element =_element;
assert(set > heap && set < heap
Anton Tretyakov The good guys of the internet! Throughout my new bee journey through C's stormy land, I found this code: void * add (void * _set, const void *_element)
{ int * set = _set;
const int * element =_element;
assert(set > heap && set < heap
Wengers I have the following code: #include <iostream>
int main(){
int v1 = 20;
int *p1 = &v1;
int **p2 = &p1;
return 0;
}
What I'm trying to do here is point one pointer to another and it works fine in this case. I make p1 point to a cons
MathMan Suppose we have int *p1;
const int **pp2= &p1;
const int n=13;
// This means we cannot use PP2 to modify // the value of the variable whose address is stored in *PP2. *pp2 stores the address of some variable (n in this case) The book I'm reading - Int
Acetate I have an array of integers int foo[3]and I want to pass it to another function. I want to accomplish two things: pass by reference Set it to constant as it should not be modified. The function I define is: void print_foo(const int(*const foo)[3]) {
Cerno I have the following basic code: base code int main()
{
int i = 1;
const int* p = &i;
int* q = &i;
test_ptr(p);
test_ptr(q);
}
Can anyone explain why the first and third examples work with the above basic code, but not the second? Example im
Cerno I have the following basic code: base code int main()
{
int i = 1;
const int* p = &i;
int* q = &i;
test_ptr(p);
test_ptr(q);
}
Can anyone explain why the first and third examples work with the above basic code, but not the second? Example im
Cerno I have the following basic code: base code int main()
{
int i = 1;
const int* p = &i;
int* q = &i;
test_ptr(p);
test_ptr(q);
}
Can anyone explain why the first and third examples work with the above basic code, but not the second? Example im
MathMan Suppose we have int *p1;
const int **pp2= &p1;
const int n=13;
// This means we cannot use PP2 to modify // the value of the variable whose address is stored in *PP2. *pp2 stores the address of some variable (n in this case) The book I'm reading - Int
Acetate I have an array of integers int foo[3]and I want to pass it to another function. I want to accomplish two things: pass by reference Set it to constant as it should not be modified. The function I define is: void print_foo(const int(*const foo)[3]) {
MathMan Suppose we have int *p1;
const int **pp2= &p1;
const int n=13;
// This means we cannot use PP2 to modify // the value of the variable whose address is stored in *PP2. *pp2 stores the address of some variable (n in this case) The book I'm reading - Int
Cerno I have the following basic code: base code int main()
{
int i = 1;
const int* p = &i;
int* q = &i;
test_ptr(p);
test_ptr(q);
}
Can anyone explain why the first and third examples work with the above basic code, but not the second? Example im
csguy Isn't passing pointers by const reference for optimization? predecessor. bool testHelper(const TreeNode*& p, const TreeNode*& q) {
return false;
}
TreeNode* test(TreeNode* root, TreeNode* p, TreeNode* q) {
recursiveHelper(p, q);
}
mistake: Line
mr_Alex_Nok_ I have created a menu structure in an embedded system as follows: struct DisplayMenu
{
short int MenuTitle; // Menu title text offset - the title used in the calling menu
char NoOfSubmenuItems;
Cerno I have the following basic code: base code int main()
{
int i = 1;
const int* p = &i;
int* q = &i;
test_ptr(p);
test_ptr(q);
}
Can anyone explain why the first and third examples work with the above basic code, but not the second? Example im
Cerno I have the following basic code: base code int main()
{
int i = 1;
const int* p = &i;
int* q = &i;
test_ptr(p);
test_ptr(q);
}
Can anyone explain why the first and third examples work with the above basic code, but not the second? Example im
MathMan Suppose we have int *p1;
const int **pp2= &p1;
const int n=13;
// This means we cannot use PP2 to modify // the value of the variable whose address is stored in *PP2. *pp2 stores the address of some variable (n in this case) The book I'm reading - Int
MathMan Suppose we have int *p1;
const int **pp2= &p1;
const int n=13;
// This means we cannot use PP2 to modify // the value of the variable whose address is stored in *PP2. *pp2 stores the address of some variable (n in this case) The book I'm reading - Int