question on stl fill function in C++

Question

Let's say I have an array like this:

string x[2][55];

If I want to fill it with "-1", is this the correct way:

fill(&x[0][0],&x[2][55],"-1");

That crashed when I tried to run it. If I change x[2][55] to x[1][54] it works but it doesn't init the last element of the array.

Here's an example to prove my point:

  string x[2][55]; 
  x[1][54] = "x";
  fill(&x[0][0],&x[1][54],"-1");
  cout<<x[1][54]<<endl; // this print's "x"

Answer 1

Because when you have a multi-dimensional array, the address beyond the first element is a little confusing to calculate. The simple answer is you do this:

&x[1][55]

Let's consider what a 2d array x[N][M] is laid out in memory

[0][0] [0][1] ... [0][M-1] [1][0] [1][1] ... [1][M-1] [N-1][0] .. [N-1][M-1]

So, the very last element is [N-1][M-1] and the first element beyond is [N-1][M]. If you take the address of [N][M] then you go very far past the end and you overwrite lots of memory.

Another way to calculate the first address beyond the end is to use sizeof.

&x[0][0] + sizeof(x) / sizeof(std::string);

Answer 2

From the formal and pedantic point of view, this is illegal. Reinterpreting a 2D array as a 1D array results in undefined behavior, since you are literally attempting to access 1D x[0] array beyond its boundary.

In practice, this will work (although some code analysis tools might catch an report this as a violation). But in order to specify the pointer to the "element beyond the last" correctly, you have to be careful. It can be specified as &x[1][55] or as &x[2][0] (both are the same address).