Queue array pointers (C++)

Question

I'm new to C++. And thus also the idea of using pointers. I have some arrays that I want to queue. I understand that my code does not work as I'm using pointers. However I do not know how I should fix this? I need the elements from the queue to be a float* type for later use with BLAS.

#include <iostream>
#include <queue>
using namespace std;


float* initialize(float* vec, int len){
    for(int i = 0; i < len; ++i){
        vec[i] = 0;     // Initialize
    }
    return vec;
}

void printVector(float* vec, int len){
    for(int i=0;i<len;i++)
        cout << vec[i] << endl;
    cout << "--" << endl;
}

int main ()
{
    queue<float*> q;
    int len = 3;
    float* k = new float[len];
    k = initialize(k,len);

    for(int t=0;t<len;t++){
        k[t] = 1;   
        printVector(k, len);
        q.push(k);
        k[t] = 0;
    }

    // I would like the one below to give same output as above
    cout << "Should have been the same:" << endl;

    while (!q.empty()){
        printVector(q.front(), len);
        q.pop();
    }

    return 0;
}

Bonus question does these types of "pointer arrays" have a special name? float* k = new float[len];

You could use std::vector<float>. If you have to pass C arrays to BLAS functions you could extract them using std::vector::data() (or &std::vector::operator[0]). The size (number of elements) (which is probably required as well) is available with std::vector::size(). — Scheff's Cat
– Scheff's Cat, Commented Jun 11, 2018 at 14:03
k is just a pointer to the first element of the new array. It is indistinguishable from a pointer to any other float, except in what you can/must do with it. — François Andrieux
– François Andrieux, Commented Jun 11, 2018 at 14:03
Thanks. And yes, I understand that. However, I do not know how to come around this if I want q.front() to be a float*. — Christian
– Christian, Commented Jun 11, 2018 at 14:07
@Christian you probably don't need that. If you used a std::queue<std::vector<float>> you could just as easily do q.front().data(). — François Andrieux
– François Andrieux, Commented Jun 11, 2018 at 14:09
In general, you can do it the way you tried BUT: you allocate storage only once. To make it working you have to allocate storage in the loop (the one with the q.push();). The other loop should delete[] the popped pointer resp. (Otherwise you might get memory leaks.) As you did it, all queue entries share the same storage (and hence the side effects). => Assigning a pointer does not copy the pointed contents. — Scheff's Cat
– Scheff's Cat, Commented Jun 11, 2018 at 14:09

Scheff's Cat · Accepted Answer · 2018-06-11 14:21:13Z

The essential message is: Storing a pointer (to an array or what ever) does not copy the contents. It just copies the address of it.

Re-ordering your statements a bit I got OP's sample running (and doing what's probably intended):

#include <iostream>
#include <queue>
using namespace std;


float* initialize(float* vec, int len){
    for(int i = 0; i < len; ++i){
        vec[i] = 0;     // Initialize
    }
    return vec;
}

void printVector(float* vec, int len){
    for(int i=0;i<len;i++)
        cout << vec[i] << endl;
    cout << "--" << endl;
}

int main ()
{
    queue<float*> q;
    int len = 3;

    for(int t=0;t<len;t++){
        float* k = new float[len];
        k = initialize(k,len);
        k[t] = 1;   
        printVector(k, len);
        q.push(k);
    }

    // I would like the one below to give same output as above
    cout << "Should have been the same:" << endl;

    while (!q.empty()){
        printVector(q.front(), len);
        delete[] q.front();
        q.pop();
    }

    return 0;
}

Output:

1
0
0
--
0
1
0
--
0
0
1
--
Should have been the same:
1
0
0
--
0
1
0
--
0
0
1
--

Using new in C++ is error prone and can be prevented in many cases. Doing the same with std::vector<float>, it could look like this:

#include <iostream>
#include <queue>
#include <vector>

using namespace std;

void printVector(const float* vec, int len){
    for(int i=0;i<len;i++)
        cout << vec[i] << endl;
    cout << "--" << endl;
}

int main ()
{
    queue<std::vector<float> > q;
    int len = 3;

    for(int t=0;t<len;t++){
        q.push(std::vector<float>(3, 0.0f));
        q.back()[t] = 1.0f;
        printVector(q.back().data(), q.back().size());
    }

    // I would like the one below to give same output as above
    cout << "Should have been the same:" << endl;

    while (!q.empty()){
        printVector(q.front().data(), q.front().size());
        q.pop();
    }

    return 0;
}

Output is identical.

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