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gz-cpp-util 1.3
A c++20 library containing various utilities
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Contains utility for strings. More...
#include <string>#include <vector>#include <unordered_map>#include <map>Go to the source code of this file.
Classes | |
| struct | gz::util::string_hash |
Typedefs | |
Map with string type as key, works with strings, string_view and char* | |
| template<typename T > | |
| using | gz::util::unordered_string_map = std::unordered_map< std::string, T, util::string_hash, std::equal_to<> > |
| A unordered_map where you can use string_views to access elements. More... | |
| template<typename T > | |
| using | gz::util::string_map = std::map< std::string, T, util::string_hash, std::equal_to<> > |
| same as unordered_string_map, but using std::map instead of std::unordered_map | |
Functions | |
| template<SplitStringInVectorImplemented T> | |
| std::vector< T > | gz::util::splitStringInVector (const std::string_view &s, const std::string &separator, bool skipEmptyStrings=false) |
| Split a string at a separator into a vector. More... | |
Contains utility for strings.
| using gz::util::unordered_string_map = typedef std::unordered_map<std::string, T, util::string_hash, std::equal_to<> > |
A unordered_map where you can use string_views to access elements.
To retrieve an element using a string view, you need to do this:
The at() member and [] operator do not work with the string_view.
| std::vector< T > gz::util::splitStringInVector | ( | const std::string_view & | s, |
| const std::string & | separator, | ||
| bool | skipEmptyStrings = false |
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| ) |
Split a string at a separator into a vector.
Splits a string at a separator.
Two templates exist, one where the vector will hold std::string_views into s and one where it will hold std::strings. Behavior: