Question

typedef unsigned int Set[10];
Set set1;

I am correct in assuming that this creates a variable of type Set named set1 with 320 bits of storage space?

Answer 1

The number of bits (and bytes¹) allocated may vary. Though, it is guaranteed to create a sequentiell storage for 10 integers which can be references through set1.

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What does the standard say?

The standard doesn’t force a byte to contain N bits since that will make it harder to write compilers for a platform which doesn’t have N bits in a byte.

[open-std.org – n1147.pdf, 3.6/3]

• A byte is composed of a contiguous sequence of bits, the
  number of which is implementation defined. The least significant bit
  is called the low-order bit; the most significant bit is called the
  high-order bit.

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How would I get the number of bits in type T?

There is a constant (CHAR_BIT) defined that holds the number of bits in a char in <climits>.

Since all types consists of N bytes and a char is guaranteed to yield 1 when doing sizeof(char) we can use this constraint to calculate the bits in any arbitrary type.

#include <climits>

template<typename T>
struct sizeof_in_bits {
  enum {
    value = sizeof(T) * CHAR_BIT
  };
};

std::cerr << "output: " << sizeof_in_bits<unsigned int[10]>::value;

output: 320

^{The above output is, as mentioned, implementation-defined.}

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¹ The standard doesn’t force the size of an int to be M bytes. All it cleary states is that an int is to be able to hold at least -32767 through 32767 (if signed), and 0 to 65535 (if unsigned).