- C++ Middleware Writer - 1 Update
- Arrays of strings - 2 Updates
| Mr Flibble <flibbleREMOVETHISBIT@i42.co.uk>: May 19 11:55PM +0100 On 19/05/2016 22:43, David Brown wrote: > So it is a reasonable rational assumption that Genesis is fictional - > but it is not /proven/ to be. However, the onus is on believers to > provide evidence that the book is non-fiction. Utter bullshit mate. Evolution is BOTH fact and theory. As evolution is a fact we know humans evolved but according to Genesis Adam had no parents so yes evolution being a fact does prove that Genesis is a fiction. /Flibble |
| Paul <pepstein5@gmail.com>: May 19 03:01PM -0700 I think that std::string array[4]; defines array as an array of 4 empty strings. However, I can't justify this assertion (other than by experimenting). For example, int intArray[4]; does not result in an array of four zeroes even though int() evaluates as 0. So the fact that the std::string default constructor results in an empty string doesn't tell me what happens in the case of an array. Could anyone enlighten me? Thanks, Paul |
| "Alf P. Steinbach" <alf.p.steinbach+usenet@gmail.com>: May 20 12:51AM +0200 On 20.05.2016 00:01, Paul wrote: > fact that the std::string default constructor results in an empty > string doesn't tell me what happens in the case of an array. > Could anyone enlighten me? `int` is a built-in type, which is an example of a Plain Old Data type like in C, which has no specified default initialization. So, assuming your array is a local variable, with default initialization nothing happens. • C++11 §8.5/11: If no initializer is specified for an object, the object is default- initialized; ; if no initialization is performed, an object with automatic or dynamic storage duration has indeterminate value. • C++11 §8.5/6: To default-initialize an object of type T means: — if T is a (possibly cv-qualified) class type (Clause 9), the default constructor for T is called (and the initialization is ill-formed if T has no accessible default constructor); — if T is an array type, each element is default-initialized; — otherwise, no initialization is performed. So, by the second dash of §8.5/6, each element of the `int` array is default-initialized, and by the third dash, since `int` is neither class type nor an array, no initialization is performed. And then by §8.5/11 again (now applied to array item) that leaves each item with an indeterminate value. `std::string` is a class type with a default constructor. Default initialization for an object of this type uses the default constructor, which gives you an empty string. This is the first dash of §8.5/6. × × × Things are different if the array is at namespace scope, because namespace scope variables are zero-initialized before anything else. × × × For a local variable you can get zero initialization by specifying value initialization, which reduces to call to default constructor or zero initialization depending on the type. E.g. like this: int ai[4] = {}; // Very zero'ish. Alternatively you can use a `std::vector` instead of a raw array: vector<int> vi( 4 ); // Also very zero'ish. This relies on the `std::vector` constructor to zero things. In contrast, for historical reasons (as `boost::array` for C++03) the `std::array` class has no defined constructor, and doesn't zero things unless you give an initializer. And so array<int, 4> argh; // Very indeterminate. … just gives indeterminate values. :( One might argue that this proves that the Boost route into standardization is problematic. While something is proven as part of Boost, the language evolves (with C++11 supporting list-initialization), changing the context, so that the Boost thing is now much less than an optimal solution. It's a proven design, but only for the previous incarnation of the language; the world is imprefect! Cheers & hth., - Alf |
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