为什么C ++中的数组内容有时未初始化？

Question

I am reading the book of effective c++, one some pages it says that:

if you're in the C part of C++ and initialization would probably incur a runtime cost, it's not guaranteed to take place. If you cross into the non-C parts of C++, things sometimes change. This explains why an array (from the C part of C++) isn't necessarily guaranteed to have its contents initialized, but a vector (from the STL part of C++) is.

I wondered why the array which is from c part of c++ is not guaranteed to be initialized, but a vector is? Is not the array from c part is build-in type? but why the vector(from the STL part of C++) guaranteed to be initialized?

Answer 1

Because that's what the standard says. Very early in the specification of C++, it was decided not to change the way types defined in C worked. So an array ( T[] ) is only initialized if 1) the type T has a constructor, 2) there is an explicit initialization or 3) it has static lifetime. Since std::vector isn't inherited from C, there's no C precedent to be concerned with. And since some of the operations you can do on a vector require copying, it has to be initialized. Copying an uninitialized value is undefined behavior.

Note that in C++11, std::array follows the C rules. (This is done to allow aggregate initialization.)

Answer 2

When you define int a[10]; (non-statically), then the contents of a are not initialized . That's just how C and C++ are designed. You didn't ask for it, you don't get it. (If you wanted an initialized array, you would have said int a[10](); .)

In C++, the class std::vector<int> is designed to always have well-defined semantics, no matter what you do. So when you define std::vector<int> v(10) , you get ten value-initialized (ie zeroed) elements (similar to the second example above).

However, don't-pay-for-what-you-don't-need applies regardless, and the latter is probably poor C++. What you'd normally use is reserve , which, like in C, creates memory but no objects, and objects are created when needed and require no further allocation:

std::vector<int> v;            // no allocation (hopefully)
v.reserve(10);                 // one single allocation

for (int i = 0; i != 10; ++i)
    v.push_back(i);            // no allocations, guaranteed

In a nutshell: C only knows about memory . C++ separates the notion of storage and objects . This finer-grained model goes along with a richer interface. In C, you simply cannot express the concept of "I may need this storage, but I don't have anything to put into it yet."

Answer 3

C++ types like vector are actually classes, and they've got a built-in constructor - a procedure which is automatically called when we are creating a variable of this type, and constructors for STL containers can initialize the variable. And when you are using things which come from C, it is much more simple: all you do when you create a variable is just, broadly speaking, taking some part of memory for your own needs. If there is something in this memory, it will stay there just because creating variable of this type (like a simple array) includes only finding memory for it.

Answer 4

the difference is a C-style array is a language primitive, pretty much like an int or a float, so the rule for those is basically do nothing upon declaration.

int i;

here, i will take any value that was at this position in memory before.

But vector (and other STL containers) are objects, and objects have a constructors, that usually do some work.

std::vector<int> v1(10); //create a vector of 10 ints, not initialized, just like int[10] would do
std::vector<int> v2(10, 0);//create a vector of 10 ints, all initialized to zero.