sizeof in D language

Question

import std.stdio;

void main() {
    int[] a = [1,2,3,4,5,6,7,8,9,10];
    write(a.sizeof);
}

In following code sizeof of static array is equals to 8 byte. I use x86 Windows 8, so pointer is equals to 4 byte. Why I get 8 byte size of array?

Answer 1

Because int[] is a dynamic array, not a pointer. Arrays in D are not pointers. What they are is essentially

struct(T)
{
    T* ptr;
    size_t length;
}

So, if you want the underlying pointer, you need to use the array's ptr member, though that's usually only needed when interacting with C/C++ code (since dynamic arrays in C/C++ are just pointers). However, the length member is used all the time and helps make arrays in D far more powerful and pleasant to work with than arrays in C/C++ are. If you want to know more about arrays in D, then you should read this article. It goes into a fair bit of detail about them, and I would consider it a must-read for all D programmers.

Regardless, what sizeof is giving you is the size of ptr and length together, which would be 8 on 32-bit systems, and 16 on 64-bit systems.

Answer 2

A dynamic array (what you have) is behind the scenes actually a struct with a pointer and a size_t length both being 4 on your CPU.

This allows D to carry along the length of the array to avoid out of bounds reading and writing (if you have the check enabled) and a O(1) slice operation.