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 字符串的容量大小是否总是15的倍数？ 
 
 
 没有; 关于std::string容量的唯一保证是s.capacity() >= s.size() 。 
 
 一个好的实现可能会使容量呈指数增长，这样每次需要重新分配底层数组时，它的容量就会扩大一倍。 这是std::vector所必需的，这样push_back可以具有分期恒定的时间复杂度，但是对于std::string没有这样的要求。 
 
 另外， std::string实现可以执行小型字符串优化，其中比std::string对象本身存储的字符数小于一定数量的字符，而不是存储在动态分配的数组中。 这很有用，因为很多字符串很短，动态分配可能很昂贵。 通常，如果存储字符串所需的字节数小于将指针存储到动态分配的缓冲区中所需的字节数，则会执行小型字符串优化。 
 
 我不知道你的特定实现是否执行小字符串优化。 

 
 
Is the capacity size of string always a multiple value of 15?
 
 
No; the only guarantee about the capacity of a std::string is that s.capacity() >= s.size().
 
A good implementation will probably grow the capacity exponentially so that it doubles in size each time a reallocation of the underlying array is required. This is required for std::vector so that push_back can have amortized constant time complexity, but there is no such requirement for std::string.
 
In addition, a std::string implementation can perform small string optimizations where strings smaller than some number of characters are stored in the std::string object itself, not in a dynamically allocated array. This is useful because many strings are short and dynamic allocation can be expensive. Usually a small string optimization is performed if the number of bytes required to store the string is smaller than the number of bytes required to store the pointers into a dynamically allocated buffer. 
 
Whether or not your particular implementation performs small string optimizations, I don't know.