Related
Jon I'm playing with arrays and passing pointers by reference functions. Take the following code as an example: #include<iostream>
void test_fn(const int* const &i){
std::cout<<*i<<std::endl;
}
int main(){
int arr_1[5] {1, 3, 6, 4, 5};
int *int_p
Jon I'm playing with arrays and passing pointers by reference functions. Take the following code as an example: #include<iostream>
void test_fn(const int* const &i){
std::cout<<*i<<std::endl;
}
int main(){
int arr_1[5] {1, 3, 6, 4, 5};
int *int_p
Jon I'm playing with arrays and passing pointers by reference functions. Take the following code as an example: #include<iostream>
void test_fn(const int* const &i){
std::cout<<*i<<std::endl;
}
int main(){
int arr_1[5] {1, 3, 6, 4, 5};
int *int_p
Jon I'm playing with arrays and passing pointers by reference functions. Take the following code as an example: #include<iostream>
void test_fn(const int* const &i){
std::cout<<*i<<std::endl;
}
int main(){
int arr_1[5] {1, 3, 6, 4, 5};
int *int_p
godfather I want to write a wrapper for read and write unix functions, but read has const void pointer parameter and write a simple void pointer as parameter. So a prototype like this won't be able to use one of the following: typedef ssize_t (*genericStreamHa
godfather I want to write a wrapper for read and write unix functions, but read has const void pointer parameter and write a simple void pointer as parameter. So a prototype like this won't be able to use one of the following: typedef ssize_t (*genericStreamHa
Dean Why does the following work? #include <iostream>
using namespace std;
class PolyLine {
public:
PolyLine() = default;
PolyLine(PolyLine * ptr) {
std::cout << "Ctor called" << std::endl;
}
};
void function(const PolyLine& pt) {
}
int main() {
Thomas B. Suppose you have a set of pointers (yes...): std::set<SomeType*> myTypeContainer;
Then suppose you want to search this collection from a const method of SomeType: bool SomeType::IsContainered() const
{
return myTypeContainer.find(this) != myType
Thomas B. Suppose you have a set of pointers (yes...): std::set<SomeType*> myTypeContainer;
Then suppose you want to search this collection from a const method of SomeType: bool SomeType::IsContainered() const
{
return myTypeContainer.find(this) != myType
Thomas B. Suppose you have a set of pointers (yes...): std::set<SomeType*> myTypeContainer;
Then suppose you want to search this collection from a const method of SomeType: bool SomeType::IsContainered() const
{
return myTypeContainer.find(this) != myType
edwardlam0328 I'm currently learning the basics of C++ and OOP at my university. I'm not 100% sure how it works when assigning a function to a function pointer. I came across the following code: void mystery7(int a, const double b) { cout << "mystery7" << endl
edwardlam0328 I'm currently learning the basics of C++ and OOP at my university. I'm not 100% sure how it works when assigning a function to a function pointer. I came across the following code: void mystery7(int a, const double b) { cout << "mystery7" << endl
Lundin It has been argued that in modern C we should always pass arrays to functions via array pointers due to their strong typing. example: void func (size_t n, int (*arr)[n]);
...
int array [3];
func(3, &array);
It sounds like this might be a good idea to
Lundin It has been argued that in modern C we should always pass arrays to functions via array pointers due to their strong typing. example: void func (size_t n, int (*arr)[n]);
...
int array [3];
func(3, &array);
It sounds like this might be a good idea to
Lundin It has been argued that in modern C we should always pass arrays to functions via array pointers due to their strong typing. example: void func (size_t n, int (*arr)[n]);
...
int array [3];
func(3, &array);
It sounds like this might be a good idea to
Lundin It has been argued that in modern C we should always pass arrays to functions via array pointers due to their strong typing. example: void func (size_t n, int (*arr)[n]);
...
int array [3];
func(3, &array);
It sounds like this might be a good idea to
Lundin It has been argued that in modern C we should always pass arrays to functions via array pointers due to their strong typing. example: void func (size_t n, int (*arr)[n]);
...
int array [3];
func(3, &array);
It sounds like this might be a good idea to
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
catch_0x16 When using 3D/2D graphics libraries, I often find myself passing a pointer to the render window to each drawable class that draws itself onto the canvas. However, every time I do this, I worry that by passing a pointer to the render window, the obje
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
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
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
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