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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
Taoist group I'm learning c++ and recently I read a book that recommends that you use references to const when possible (if the underlying object doesn't change). I have a question, you should pass a reference to a const pointer instead of a const pointer if p
Taoist group I'm learning c++ and recently I read a book that recommends that you use references to const when possible (if the underlying object doesn't change). I have a question, you should pass a reference to a const pointer instead of a const pointer if 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
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
michalt38 Why with a pointer to const I can change the value of the variable pointed to by the pointer, but not with a const reference? int a = 10;
int * const ptr = &a;
(*ptr)++; // OK
const int & ref = a;
ref++; // error
Elias In your first example, you hav
Godru int val2 = 38;
int *ptr = &val2;
const int *&ptrRef = ptr; // ERROR
int i = 92;
int &ref_i = i;
const int &ref_i2 = ref_i;
Hi, can someone explain why I can't use a const reference to refer to a non-const pointer? I thought that if you access the const
username For simple types, the position of const to the left of * or & has the same effect. That is: const type *
equivalent to type const *
(refer to). E.g: const double pi = 3.14159;
const const double * pip1 = π
const double const * pip2 = π
const d
Godru int val2 = 38;
int *ptr = &val2;
const int *&ptrRef = ptr; // ERROR
int i = 92;
int &ref_i = i;
const int &ref_i2 = ref_i;
Hi, can someone explain why I can't use a const reference to refer to a non-const pointer? I thought that if you access the const
Godru int val2 = 38;
int *ptr = &val2;
const int *&ptrRef = ptr; // ERROR
int i = 92;
int &ref_i = i;
const int &ref_i2 = ref_i;
Hi, can someone explain why I can't use a const reference to refer to a non-const pointer? I thought that if you access the const
Godru int val2 = 38;
int *ptr = &val2;
const int *&ptrRef = ptr; // ERROR
int i = 92;
int &ref_i = i;
const int &ref_i2 = ref_i;
Hi, can someone explain why I can't use a const reference to refer to a non-const pointer? I thought that if you access the const
username For simple types, the position of const to the left of * or & has the same effect. That is: const type *
equivalent to type const *
(refer to). E.g: const double pi = 3.14159;
const const double * pip1 = π
const double const * pip2 = π
const d
Godru int val2 = 38;
int *ptr = &val2;
const int *&ptrRef = ptr; // ERROR
int i = 92;
int &ref_i = i;
const int &ref_i2 = ref_i;
Hi, can someone explain why I can't use a const reference to refer to a non-const pointer? I thought that if you access the const
Godru int val2 = 38;
int *ptr = &val2;
const int *&ptrRef = ptr; // ERROR
int i = 92;
int &ref_i = i;
const int &ref_i2 = ref_i;
Hi, can someone explain why I can't use a const reference to refer to a non-const pointer? I thought that if you access the const
Godru int val2 = 38;
int *ptr = &val2;
const int *&ptrRef = ptr; // ERROR
int i = 92;
int &ref_i = i;
const int &ref_i2 = ref_i;
Hi, can someone explain why I can't use a const reference to refer to a non-const pointer? I thought that if you access the const
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
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
Ernestas I came across a similar snippet of code today, and it struck me as odd. I did a small experiment myself as shown below. Why does the first function stuffwith a shared pointer allow to modify the value but the second does not? #include <memory>
void s
Ernestas I came across a similar snippet of code today, and it struck me as odd. I did a small experiment myself as shown below. Why does the first function stuffwith a shared pointer allow to modify the value but the second does not? #include <memory>
void s
Ernestas I came across a similar snippet of code today, and it struck me as odd. I did a small experiment myself as shown below. Why does the first function stuffwith a shared pointer allow to modify the value but the second does not? #include <memory>
void s
Ernestas I came across a similar snippet of code today, and it struck me as odd. I did a small experiment myself as shown below. Why does the first function stuffwith a shared pointer allow to modify the value but the second does not? #include <memory>
void s
Ernestas I came across a similar snippet of code today, and it struck me as odd. I did a small experiment myself as shown below. Why does the first function stuffwith a shared pointer allow to modify the value but the second does not? #include <memory>
void s
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
Ultrasonic Banana In the function declaration below the second parameter, is a const pointer to const data. ti_rc_t ti_uart_write_buffer(const ti_uart_t uart, const uint8_t *const data, uint32_t len);
Below is sample code to call the function. Why is there on
Oliver Dain: Consider the following: template <class T> void Foo(const T* x) {
std::cout << "I am the pointer overload" << std::endl;
}
template <class T> void Foo(const T& x) {
std::cout << "I am the reference overload" << std::endl;
}
Given the above,