How to shallow copy char* to std::string?

Question

我想知道是否可以浅层复制char*到std::string 。

Answer 1

If you aim at constructing a std::string object with a char* without copying the data: no, this is impossible. std::string owns its resources, it can't refer to another char* . This is also why the appropriate constructor takes a const char* , not a char* : it doesn't modify the data, but copies it.

In C++17, you have std::string_view which is exactly meant for referring to a string (literal) that it doesn't own. Note that this view isn't intended to modify the data, its hence constructed with const char* again, not char* .

Answer 2

All the std containers (and std::string is basically a container of char s) do own their data and have value semantics. Thus, no you cannot shallow copy a char* to a std::string out-of-the-box. There are however different objects that can be used for this purpose, to make the difference obvious they are called ..._view , eg since C++17 there is std::string_view .

Answer 3

Technically it should be possible to implement string constructor or member function taking a pointer to buffer (and size / capacity) with ownership transfer (similar to unique_ptr ). However it won't be very helpful since string buffers are supposed to be created by corresponding allocator so you'll end up with calling string allocator manually and / or with difficult to track bugs when pointer passed wasn't created by correct allocator. Also this would make small string optimization problematic.