如果 new_size 不大于旧的，C++ 标准是否保证 std::string::resize(new_size) 不会导致分配？

Question

#include <string>
#include <cassert>

int main()
{
    auto s = "hello"s;
    auto p = &s[0];

    s.resize(3);
    assert('h' == *p); // always ok?
}

如果 new_size 不大于旧的，C++ 标准是否保证 std::string::resize(new_size) 不会导致分配？

Answer 1

首先请注意，标准通常设定要求而不是保证。

如果 resize 参数小于其当前 size()，则不需要

std::basic_string::resize 不重新分配字符串。事实上，C++17标准说当前字符串被新字符串替换：

If n <= size(), the function replaces the string designated by *this with a string of length n whose elements are a copy of the initial elements of the original string designated by *this.

使用小字符串优化时，调整为较小的大小可能会导致字符串就地存储字符，而不是在动态分配的缓冲区中。

在C++20标准中resize变成了constexpr上面的写法没有了，见string.capacity:

constexpr void resize(size_type n, charT c);

5 #Effects: Alters the value of *this as follows:

(5.1) If n <= size(), erases the last size() - n elements.

(5.2) If n > size(), appends n - size() copies of c.

Answer 2

它不保证指针的有效性：

References, pointers, and iterators referring to the elements of a basic_string sequence may be invalidated by the following uses of that basic_string object:

Passing as an argument to any standard library function taking a reference to non-const basic_string as an argument.

Calling non-const member functions, except operator[], at, data, front, back, begin, rbegin, end, and rend.

resize 是非 const 成员函数，不在保证不会使引用无效的成员函数列表中。

如果 new_size 不大于旧的，C++ 标准是否保证 std::string::resize(new_size) 不会导致分配？

Does the C++ standard guarantee that std::string::resize(new_size) will not cause allocation if the new_size is not greater than the old one?

c++

string

performance

standards

memory-management