Vals are initialized in the order they are declared (well, precisely, non-lazy vals are), so properties is getting initialized before loadedProps. Or in other words, loadedProps is still null when propertiesis getting initialized. The simplest solution here is to define loadedProps before properties: class Config { private val loadedProps = { val p = new … Read more
TL;DR Answer: No, this is well-defined behaviour. Explanation: As per the C11 standard document, chapter 6.7.9, initalization, If there are fewer initializers in a brace-enclosed list than there are elements or members of an aggregate, or fewer characters in a string literal used to initialize an array of known size than there are elements in … Read more
I’m just curious to know why this sudden change in the behavior of uninitialized bool? Disassemble the code and see what the compiler’s doing. My guess: since the value is now only used locally, the compiler optimizes it away completely. Since the behaviour is undefined anyway, the compiler can safely just assume any value, e.g. … Read more
You can in C++0x: int* p = new int[3] { 1, 2, 3 }; … delete[] p; But I like vectors better: std::vector<int> v { 1, 2, 3 }; If you don’t have a C++0x compiler, boost can help you: #include <boost/assign/list_of.hpp> using boost::assign::list_of; vector<int> v = list_of(1)(2)(3);
Before we start: Let me point out that a lot of the confusion around this syntax comes because in C and C++ you can use the = {0} syntax to initialize all members of a C-style array to zero! See here: https://en.cppreference.com/w/c/language/array_initialization. This works: // z has type int[3] and holds all zeroes, as: `{0, … Read more
Visual Basic 9.0 doesn’t support this yet. However, Visual Basic 10.0 will.
Initialization of flexible array member in this way is not allowed as per C standard. C11: 6.7.2.1 Structure and union specifiers (p20-21): 21 EXAMPLE 2 After the declaration: struct s { int n; double d[]; }; the structure struct s has a flexible array member d. […] 22 Following the above declaration: struct s t1 … Read more
Looks like you forgot to return value from factory: export function init(config: ConfigService) { return () => { return config.load(); // add return }; } or the same code can be written a bit shortly: export function init(config: ConfigService) { return () => config.load(); }
Short version Initialization via {..} is list-initialization, which prohibits narrowing conversions. For example, if LLONG_MAX is the maximum value of an long long int, and your int cannot represent that: int x = LLONG_MAX; // probably accepted with a warning int x {LLONG_MAX}; // error Similarly: long long y = /*something*/; int x = y; … Read more
No, there is no way to do this. According to the language specs, instance variables aren’t even initialized until a super() call has been made. These are the steps performed during the constructor step of class instance creation, taken from the link: Assign the arguments for the constructor to newly created parameter variables for this … Read more