C++ Memory Allocation for Struct With String

Question

If I have a string member within a struct that's then stored into an array, how does memory get allocated?

struct garage {
   int ncars;
   int nspaces;
   int nmechanics;
   string name;
}

But for that last member, name, string is basically a typedef of basic_string, so its memory gets allocated when it gets defined, right? For example: garage.name = "Cool Cars"; But if I don't define that member YET, and store the struct in an array:

garage nearby_garages[15];
garage g0, g1, g2;

nearby_garages[0] = g0; nearby_garages[1] = g1; nearby_garages[2] = g2;

garage current;
current = nearby_garage[1];
current.name = "Jack's Garage";

string size can vary depending on the length of the string/data. struct size can vary depending on string size, which means the array size can vary depending on struct size, but then the array would fall apart if it was pre-allocated. The only way I can see this working is if string is a pointer to a memory location not sandwiched within the struct. But I don't think that is what's happening here. Help please?

Eric · Accepted Answer · 2014-11-13 01:40:37Z

Your garage only has references so your array can be allocated on the stack with no problem. Internally however, std::string does new/malloc to create memory for your data.

Your garage then holds a reference to a string which holds a pointer to a chunk of memory containing your data. Nothing breaks here because the garage knows at creation that the string will have a pointer to data so the pointer already has a space for it.

When you include literals such as "Jack's Garage", the compiler creates a special place to hold those strings, they are not allocated in the same memory segment.

Finally, when you call current.name = "Jack's Garage", C++ will determine that it needs a conversion between a const char* to a std::string. Fortunately for all of us, such a conversion exists. Your assignment is then transformed to

current.name = std::string("Jack's Garage");

Then the assignment operator of std::string will copy the value to current.name. New memory will be allocated inside garage to hold that value and (probably) that memcpy will be called at a lower level.

Chris · Accepted Answer · 2014-11-13 03:25:53Z

0

std::string is similar in implementation to an std::vector: Essentially a pointer and size, two pointers (begin and end), or one pointer and the ability to query allocator block sizes.

In some cases, it may also implement SSO (Small String Optimization) where the string structure itself has a small buffer for short strings, and switches to using a pointer for longer strings.

Without SSO, the backing store for characters owned by an std::string is allocated upon construction or assignment with a literal (or with another string, if the implementation isn't COW), or re-allocated during a concatenation.

In your code above, current.name = "Jack's Garage", would be the allocation site (without SSO in this case).

edited Nov 13, 2014 at 3:25

answered Nov 13, 2014 at 1:16

Chris

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1 Comment

Sam Araiza Over a year ago

Sneaky and clever. Thanks!

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