Question

This following snippet of code from the Chromium source caught my eye (see line 155 here):

std::string PrintPreviewUI::GetPrintPreviewUIAddress() const {
  // Store the PrintPreviewUIAddress as a string.
  // "0x" + deadc0de + '\0' = 2 + 2 * sizeof(this) + 1;
  char preview_ui_addr[2 + (2 * sizeof(this)) + 1];
  base::snprintf(preview_ui_addr, sizeof(preview_ui_addr), "%p", this);
  return preview_ui_addr;
}

Doesn’t 2 + (2 * sizeof(this)) + 1 evaluate to 3 + 2 * sizeof(this)? Why did the authors choose to write the expression this way?

Answer 1

Yes, it does evaluate the same way.

Presumably the authors wrote it that way to make it clearer how their array was laid out — i.e. that it contained 2 bytes for one thing, then 2 pointers, and then 1 more byte after that. (Actually I’m not sure why they chose to use the sizeof() operator in this case, since the length of a string representation of a pointer isn’t the same as the pointer’s in-memory width)

The compiler will optimize away the math at compile time, so performance isn’t effected; it’s just to keep other programmers from having to figure out where the 3 came from.