Does it need to initialise the variable after allocating memory?

Question

I am trying to implement matrix multiplication in c++. I found a sample code using a class that writes in .h and .cpp files. This is just a part of the code that related to my question:

#include "Matrix.h"

// Constructor - using an initialisation list here
Matrix::Matrix(int rows, int cols, bool preallocate): rows(rows), cols(cols), size_of_values(rows * cols), preallocated(preallocate)
{
   // If we want to handle memory ourselves
   if (this->preallocated)
   {
      // Must remember to delete this in the destructor
      this->values = new double[size_of_values];
   }
}

void Matrix::matMatMult(Matrix& mat_left, Matrix& output)
{
   // The output hasn't been preallocated, so we are going to do that

      output.values = new double[this->rows * mat_left.cols];

   // Set values to zero before hand
   for (int i = 0; i < output.size_of_values; i++)
   {
      output.values[i] = 0;
   }

I wonder why they initialised using the output matrix with 0s output.values[i] = 0; while it has been allocated memory before?

Answer 1

From cppreference on new expression:

The object created by a new-expression is initialized according to the following rules:

• [...]
• If type is an array type, an array of objects is initialized.
  • If initializer is absent, each element is default-initialized
  • If initializer is an empty pair of parentheses, each element is value-initialized.

"default-initialized" ints are colloquially not initialized. They have indeterminate values. The empty pair of parantheses refers to what Ted mentioned in a comment:

output.values = new double[this->rows * mat_left.cols]{};

Value initialization is described here. The case that applies here is

• otherwise, the object is zero-initialized.

---

I wonder why they initialised using the output matrix with 0s output.values[i] = 0; while it has been allocated memory before?

Allocating memory and initializing an object are two seperate steps. Yes, the elements have to be initialzed, allocating memory is not sufficient.