What would ‘std:;’ do in c++?
std: its a label, usable as a target for goto. As pointed by @Adam Rosenfield in a comment, it is a legal label name. C++03 ยง6.1/1: Labels have their own name space and do not interfere with other identifiers.
std: its a label, usable as a target for goto. As pointed by @Adam Rosenfield in a comment, it is a legal label name. C++03 ยง6.1/1: Labels have their own name space and do not interfere with other identifiers.
There are, indeed, performance issues with std:function that must be taken into account whenever using it. The main strength of std::function, namely, its type-erasure mechanism, does not come for free, and we might (but not necessarily must) pay a price for that. std::function is a template class that wraps callable types. However, it is not … Read more
Let’s go over the conditions as they appear: If const T isn’t const (const doesn’t really apply to function types since functions aren’t objects), and T isn’t a reference (const doesn’t apply to references either for the same reason), it’s a function type. int (or any other non-function-non-reference type) wouldn’t fit in because is_const<const int>::value … Read more
The language specification allows implementations to implement <cmath> by declaring (and defining) the standard functions in global namespace and then bringing them into namespace std by means of using-declarations. It is unspecified whether this approach is used 20.5.1.2 Headers 4 […] In the C++ standard library, however, the declarations (except for names which are defined … Read more
Seeing from your G++ version, you need to update it badly. C++11 has only been available since G++ 4.3. The most recent version is 4.7. In versions pre-G++ 4.7, you’ll have to use -std=c++0x, for more recent versions you can use -std=c++11.
There are multiple answers based on what you are doing with the string. 1) Using the string as an id (will not be modified). Passing it in by const reference is probably the best idea here: (std::string const&) 2) Modifying the string but not wanting the caller to see that change. Passing it in by … Read more
You can use std::find as follows: if (std::find(v.begin(), v.end(), “abc”) != v.end()) { // Element in vector. } To be able to use std::find: include <algorithm>.
There were already examples of “core” language features that returned types defined in the std namespace. typeid returns std::type_info and (stretching a point perhaps) sizeof returns std::size_t. In the former case, you already need to include a standard header in order to use this so-called “core language” feature. Now, for initializer lists it happens that … Read more
Well, just do it like the boost guys did it: template <class T> inline void hash_combine(std::size_t& seed, const T& v) { std::hash<T> hasher; seed ^= hasher(v) + 0x9e3779b9 + (seed<<6) + (seed>>2); }
The reason std::string(0) is valid, is due to 0 being a null pointer constant. So 0 matches the string constructor taking a pointer. Then the code runs afoul of the precondition that one may not pass a null pointer to std::string. Only literal 0 would be interpreted as a null pointer constant, if it was … Read more