Sorry about the confusing title, did not know what to title it.
I have this function:
static void smooth5(IntVector*& v)
{
IntVector* tmp = new IntVector();
for(int i=0; i<v->size(); i+=2)
tmp->push_back(v->at(i));
delete v;
v = tmp;
}
and in the main I do this:
IntVector* v = new IntVector();
v->push_back(0);
v->push_back(1);
v->push_back(2);
v->push_back(3);
smooth5(v);
//print the contents of v
When I print the contents of v, the output is 0 2.
But I did not understand what
IntVector*& v
really means when v is a pointer to an object on the heap. Can someone please explain?
IntVector*& declares a reference to a pointer. Using this as a function argument allows smooth5 to modify the caller's copy of v.
It is similar to but more readable than passing a pointer to a pointer - IntVector**.
In your example, smooth5 deletes the caller's IntVector and points it to its tmp variable instead.
This is not a pointer to reference but a reference to a pointer. This means that instead of passing a copy of the pointer to the function you pass a reference to it. So as a result any modifications you do to the pointer in the function smooth5 will affect the argument you pass to it in this case the pointer v.
What IntVector*& v in your code means depends who's reading it.
When the compiler reads it, it means that v is a reference to
a pointer to IntVector (which is a typedef to
std::vector<int>?). When I read it, it means that the author
of the code probably didn't understand how C++ works, and is
trying to adapt Java or C# idioms that aren't appropriate. The
idiomatic way of writing this would be:
void smooth5( IntVector& v )
{
IntVector results;
for ( int i = 0; i < v->size(); i += 2 ) {
results.push_back( v[i] );
}
v.swap( results );
}
Any code which involved pointers to vectors is highly
suspicious.
Related
I am trying to create a sorting function with the parameters being a pointer of a list and I am trying to access an element of the given list. Hopefully this code speaks for the problem better than I can:
void bubbleSort(std::vector<int> *L) {
unsigned int i = 0; int temp;
while(isSorted(*L)) {
if(i==L->size()-1) {
i = 0;
}
if(i<L[i]/*<-ERROR here.*/) {
temp = L[i+1]; // ERROR HERE
L[i+1] = L[i]; // ERROR HERE
L[i] = temp; // ERROR HERE
}
}
}
You don't need to painfully dereference every individual use of L (and indeed doing so is error-prone, as you've demonstrated by missing one in your answer).
Instead, just write:
void bubbleSort(std::vector<int> *Lptr) {
auto &L = *Lptr;
and keep the rest of the code the same.
NB. It would be even better to change the function itself, to
void bubbleSort(std::vector<int> &L) {
as it should have been written in the first place, but I'm assuming there's some artificial reason you can't do that.
The function accepts a pointer to an object of type std::vector<int>.
void bubbleSort(std::vector<int> *L) {
To access the original vector using the pointer, you can write either *L or L[0]. That is, both expressions yield an lvalue reference of type std::vector<int> & to the vector.
To get the i-th element of the vector using the subscript operator through the pointer, you can write either (*L)[i] or L[0][i],
However, in this if statement:
if(i<L[i]/*<-ERROR here.*/) {
You are trying to compare the variable i of type unsigned int to the object L[i] of type std::vector<int>. When i is not equal to 0, this yields a non-existent object of the vector type.
It seems you mean something like the following instead:
if ( (*L)[i] < (*L)[i+1] ) {
or:
if ( L[0][i] < L[0][i+1] ) {
or, vice versa:
if ( L[0][i+1] < L[0][i] ) {
Depending on whether the vector is sorted in ascending or descending order.
Pay attention to the fact that there is no sense in declaring the parameter as a pointer to a std::vector<int>. The function would be much clearer and readable if it accepted the vector by reference instead:
void bubbleSort(std::vector<int> &L) {
In this case, the if statement would look like this:
if ( L[i] < L[i+1] ) {
Although I prefer to change the source code as other answer. But, for this question, you can use ->at() function to access the element in a vector pointer.
if(i<L->at(i)) {
temp = L->at(i+1);
L->at(i+1) = L->at(i);
L->at(i) = temp;
}
As the title describes, I am trying to pass the pointer to the data of a std::vector into a function expecting a double pointer. Take as an example the code below. I have an int pointer d which is passed to myfunc1 as &d (still not sure if call it the pointer's reference or what), where the function changes its reference to the beginning of an int array filled with 1,2,3,4. However, if I have a std::vector of ints and try to pass &(vec.data()) to myfunc1 the compiler throws the error lvalue required as unary ‘&’ operand. I have already tried something like (int *)&(vec.data()) as per this answer, but it does not work.
Just for reference, I know I can do something like myfunc2 where I directly pass the vector as reference and the job is done. But I want to know if it's possible to use myfunc1 with the std::vector's pointer.
Any help will be very much appreciated.
#include <iostream>
#include <vector>
using std::cout;
using std::endl;
using std::vector;
void myfunc1(int** ptr)
{
int* values = new int[4];
// Fill all the with data
for(auto& i:{0,1,2,3})
{
values[i] = i+1;
}
*ptr = values;
}
void myfunc2(vector<int> &vec)
{
int* values = new int[4];
// Fill all the with data
for(auto& i:{0,1,2,3})
{
values[i] = i+1;
}
vec.assign(values,values+4);
delete values;
}
int main()
{
// Create int pointer
int* d;
// This works. Reference of d pointing to the array
myfunc1(&d);
// Print values
for(auto& i:{0,1,2,3})
{
cout << d[i] << " ";
}
cout << endl;
// Creates the vector
vector<int> vec;
// This works. Data pointer of std::vector pointing to the array
myfunc2(vec);
// Print values
for (const auto &element : vec) cout << element << " ";
cout << endl;
// This does not work
vector<int> vec2;
vec2.resize(4);
myfunc1(&(vec2.data()));
// Print values
for (const auto &element : vec2) cout << element << " ";
cout << endl;
return 0;
}
EDIT: What my actual code does is to read some binary files from disk, and load parts of the buffer into the vector. I was having troubles getting the modified vector out of a read function, and this is what I came up with that allowed me to solve it.
When you write:
myfunc1(&(vec2.data()));
You are getting the address of a rvalue. The pointed int* is so a temporary that is destroyed right after the call.
This is why you get this error.
But, as #molbdnilo said, in your myfunc1() function, you are reassigning the pointer (without caring to destroy previously allocated memory by the way).
But the std::vector already manages its data memory on its own. You cannot and you must not put your hands on it.
What my actual code does is to read some binary files from disk, and load parts of the buffer into the vector.
A solution could be to construct your std::vector by passing the iterator to the beginning and the iterator to the end of the desired part to extract in the constructor's parameters.
For example:
int * buffer = readAll("path/to/my/file"); // Let's assume the readAll() function exists for this example
// If you want to extract from element 5 to element 9 of the buffer
std::vector<int> vec(buffer+5, buffer+9);
If the std::vector already exists, you can use the assign() member function as you already did in myfunc2():
vec.assign(buffer+5, buffer+9);
Of course in both cases, you have to ensure that you are not trying to access an out of bounds element when accessing the buffer.
The problem is that you cannot take the address of data(), since it is only a temporary copy of the pointer, so writing to a pointer to it makes not that much sense. And that is good that way. You DO NOT want to pass data() to this function since it would overwrite the pointer with a new array and that would break the vector. You can remove one * from the function and only assign to it and not allocate the memory there. This will work, but make sure to allocate the memory in the caller (with resize, just reserve will result un undefined behavior, since data() is only a pointer to the beginning of the valid range [data(), data() + size()). The range [data(), data() + capacity ()) is not necessary valid.
I am still perfecting the art of posting here so bear with me, I will edit and fix anything suggested!
I have a homework that requires me to create functions that manipulate vectors. The "catch" is that all the data passed to the function is passed by reference to this struct:
struct Vector { // this struct must stay as is
int sze = 0; // "size" took out i for compatability
int capacity = 0;
int * data = nullptr ;
}a,b,c;
i.e.
void construct_vector ( Vector& v, int size= 0, int initVal= 0);
The problem that I am having is in the function construct_vector() I have to, you guessed it, construct a vector and use int* data to point to the vector on the heap? I am not positive about that last part). I just know I have to use the int pointer to point to the vector created within the construct function, and cannot for the life of me figure out how to do that.
An example of what I am trying:
void construct_vector ( Vector &v, int size, int initVal){
std::vector<int> t(size,initVal);
*v.data = &t ; // ERROR: Assigning to 'int' from incompatible type 'std::vector<int> *'
v.capacity = size; //
v.sze = size;
for (int i=0; i < t.size(); i++){
/* I originally tried to implement a
dynamic int pointer here but I cannot change int* data
to int*data[sze] within the struct*/
}
}
The reason int * data must point to the vector is because the data is passed to the subsequent functions by reference to struct member v:
void destroy_vector ( Vector & v );
void copy_data ( Vector & v );
Edit: My problem was that I misunderstood the objective of my assignment but I think the answers I received can really help people understand dynamic memory and how it should be used within functions. So I am going to leave everything as is!
You have two problems here:
std::vector<int> t(size,initVal);
*v.data = &t ; // ERROR: Assigning to 'int' from
First, *v.data and &t are different types, one is int, the other is a pointer to a vector of ints.
You can get it compile with (but you SHOULD NOT, see the second problem)
v.data = t.data();
The other problem is the vector is local to the function. As soon as the function returns, your pointer will dangle.
So the right solution for your problem is using a dynamic array:
v.data = new int[size];
Don't forget to delete[] it in the struct's destructor when you are done using it:
delete [] data;
Instead of
std::vector<int> t(size,initVal);
*v.data = &t ;
You need
v.data = new int[size];
To fill up the object with the input value, use
for ( int i = 0; i < size; ++i )
{
v.data[i] = initVal;
}
You can use std::fill to make your code a bit simpler.
std::fill(v.data, v.data+size, initVal);
Make sure to follow The Rule of Three when you manage dynamic memory yourself.
I was playing through c++ and trying to understand vector and its signature .
In below method printPrimes I need to use pointer with address of why ?
Is vector<int> &primes not enough as from main method printPrimes is already sending address .
void printPrimes(long long l, long long r, vector<int>* &primes) {
// some code
}
vector<int>* sieve() {
vector<int> *prime = new vector<int>();
return prime;
}
int main() {
vector<int> *primes = sieve();
printPrimes(l, r, primes);
return 0;
}
I need to use pointer with address of
Here, & does not mean "address of"; it means the type "reference to".
It's clearer if you write it not like this:
vector<int>* &primes
but like this:
vector<int>*& primes
Though the choice of whitespace is artificial, that better documents that this & is "part of the type".
Have some types:
std::vector<T> = A vector of Ts
std::vector<T>& = A reference to a vector of Ts
std::vector<T>* = A pointer to a vector of Ts
std::vector<T>*& = A reference to a pointer to a vector of Ts
std::vector<T>*** = A pointer to a pointer to a pointer to a vector of Ts
std::vector<T>**& = A reference to a pointer to a pointer to a vector of Ts
…and so forth.
As for why you need a vector<int>*& for printPrimes to do its job, we could not tell you without actually being able to see it. I will say that it seems unlikely it needs a pointer at all, and that if it wants to modify that pointer it's going to cause problems with the new and delete in the calling scope.
In fact, all that dynamic allocation is completely pointless and only complicates things.
The following was likely intended instead:
void printPrimes(long long l, long long r, vector<int>& primes) {
// some code
}
vector<int> sieve() {
vector<int> prime;
return prime;
}
int main() {
vector<int> primes = sieve();
printPrimes(l, r, primes);
}
vector<int>* &primes parameter has to be read this way:
Reference to a pointer of vector of int
and not
Address of a pointer of vector of int (which, you are right, would be useless)
Passing by reference allows to directly manipulate any instance outside of scope (like with pointers, but a safer way since a reference cannot be nullptr, and its existence is auto-managed (no need to delete)).
In c++ & in function parameter used to pass parameter by reference. vector<int>* &primes declares primes to be a reference to a pointer to vector<int>.
If printPrimes means to print only the vector passed to the function then the signature
void printPrimes(long long l, long long r, vector<int> &primes);
can also do the job.
Reference to a pointer is needed when the pointer passed to the function is need to be modified and it's effect is expected to seen in the caller function.
void foo(int*& p){
p = new int[10];
// rest of the code
}
if a function bar is calling foo like
void bar(/* some parameters */){
// ...
int *p;
foo(p);
// rest of the code
}
foo is modifying the pointer itself and this modification will be seen to bar also and memory allocated to p can be accessed from bar.
I apologize if this question has already been answered (I tried searching around, but couldn't find anything quite the same, and similar questions' solutions didn't work), but how do I pass an object (in this case, a vector of objects) and have the function edit those values without returning anything?
Example:
void incVector(std::vector<int> vec)
{
for (auto l = 0; l < int(vec.size()); l++)
{
vec[l]++;
}
}
int main()
{
std::vector<int> vec;
vec.push_back(1);
vec.push_back(2);
for (auto l = 0; l < int(vec.size()); l++)
{
std::cout << vec[l]; //Output: "12"
}
incVector(vec);
for (auto l = 0; l < int(vec.size()); l++)
{
std::cout << vec[l]; //Output: "12"
//This output should be "23"
}
}
Obviously, what I'm actually using this for is much more complex, but this is enough of an example to get the point of what I'm trying to do across. In the actual project, it's a rabbit hole of different functions, some of which return things while others don't, so having it simply return the vector isn't an option.
I have tried making incVector accept a reference to a vector, a pointer to a vector, a pointer to a reference to a vector, and a pointer to a pointer to a vector (which are solutions that seemed to work for other similar questions) but none of those are working for me.
EDIT:
God, I feel stupid. I swear I'd tried using a reference before and it didn't work. Yet now, trying it again works just fine. Sorry! ^^;
you pass the vector by value, thus modifications are purely local:
change prototype of your function to:
void incVector(std::vector<int> &vec)
to pass by reference and get the one from main modified
Take the argument by reference. You can also use a modern for-loop.
void incVector(std::vector<int>& vec)
{
for (auto& l : vec)
{
++l;
}
}
But you don't actually need to do any of this. Applying an operation on each element of the vector can be done easily using a standard algorithm (std::for_each) and a lambda function that takes a reference to the vector's element:
#include <algorithm>
// ...
std::for_each(vec.begin(), vec.end(), [](int& l){ ++l; });
for_each is going to call the lambda for each element in the vector and pass it as the argument. Since you take the argument by reference (int&), incrementing it will increment the actual element contained in the vector.
The key point to take away from this however, is that when you want to give a new name to an object that already exists, you use a reference. Those are declared with & before their identifier:
int i = 0;
int &i_ref = i;
Here, i_ref is a reference to i, meaning it's just another name for i. Modifying i_ref is the same as modifying i.
The same applies to function arguments. If a function argument is a reference, it means it's another name for the object that was passed to the function. Modifying the reference is the same as modifying the object that was passed to the function.