Does `&s[0]` Guarantee Contiguous Memory Allocation in a C `std::string`?

Accessing Characters via "&s[0]" in a std::string

In C , the std::string class provides a convenient way to manipulate character sequences. However, the underlying representation of string data may differ depending on implementation, leading to questions about memory safety.

Question:

Does using "&s[0]" to access characters in a std::string guarantee contiguous memory allocation?

Answer:

Under the C 98/03 standard, the memory allocation of a std::string is not explicitly defined, leaving it up to the implementation. However, the C 11 standard mandates contiguous storage for std::string objects.

In practice, most implementations, both before and after the standardization of C 11, have adopted contiguous storage for std::string. This means that using "&s[0]" to access characters is generally considered safe.

Standard Guaranteed Access:

According to the C 11 standard, the operator[] for std::string always guarantees access to a valid character reference. Even in cases where the string is empty, "&s[0]" is defined to point to an object with a charT() value, ensuring that it remains a valid memory reference.

This is further reinforced by the definition of data() for std::string, which specifies that the returned pointer "p" should reference consecutive memory locations that match the elements accessible via "&operator[](i)" within the valid range [0,size()].

Historical Note:

In earlier versions of the C 0x standard, the behavior of "&s[0]" with zero-length strings was undefined. However, pre-standardization C 0x and subsequent standard drafts have clarified this behavior, guaranteeing that "&s[0]" points to a valid memory reference in all cases.

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