C++ Layout of object types Class types
Example
By "class", we mean a type that was defined using the class or struct keyword (but not enum struct or enum class).
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Even an empty class still occupies at least one byte of storage; it will therefore consist purely of padding. This ensures that if
ppoints to an object of an empty class, thenp + 1is a distinct address and points to a distinct object. However, it is possible for an empty class to have a size of 0 when used as a base class. See empty base optimisation.class Empty_1 {}; // sizeof(Empty_1) == 1 class Empty_2 {}; // sizeof(Empty_2) == 1 class Derived : Empty_1 {}; // sizeof(Derived) == 1 class DoubleDerived : Empty_1, Empty_2 {}; // sizeof(DoubleDerived) == 1 class Holder { Empty_1 e; }; // sizeof(Holder) == 1 class DoubleHolder { Empty_1 e1; Empty_2 e2; }; // sizeof(DoubleHolder) == 2 class DerivedHolder : Empty_1 { Empty_1 e; }; // sizeof(DerivedHolder) == 2 -
The object representation of a class type contains the object representations of the base class and non-static member types. Therefore, for example, in the following class:
struct S { int x; char* y; };there is a consecutive sequence of
sizeof(int)bytes within anSobject, called a subobject, that contain the value ofx, and another subobject withsizeof(char*)bytes that contains the value ofy. The two cannot be interleaved. -
If a class type has members and/or base classes with types
t1, t2,...tN, the size must be at leastsizeof(t1) + sizeof(t2) + ... + sizeof(tN)given the preceding points. However, depending on the alignment requirements of the members and base classes, the compiler may be forced to insert padding between subobjects, or at the beginning or end of the complete object.struct AnInt { int i; }; // sizeof(AnInt) == sizeof(int) // Assuming a typical 32- or 64-bit system, sizeof(AnInt) == 4 (4). struct TwoInts { int i, j; }; // sizeof(TwoInts) >= 2 * sizeof(int) // Assuming a typical 32- or 64-bit system, sizeof(TwoInts) == 8 (4 + 4). struct IntAndChar { int i; char c; }; // sizeof(IntAndChar) >= sizeof(int) + sizeof(char) // Assuming a typical 32- or 64-bit system, sizeof(IntAndChar) == 8 (4 + 1 + padding). struct AnIntDerived : AnInt { long long l; }; // sizeof(AnIntDerived) >= sizeof(AnInt) + sizeof(long long) // Assuming a typical 32- or 64-bit system, sizeof(AnIntDerived) == 16 (4 + padding + 8). -
If padding is inserted in an object due to alignment requirements, the size will be greater than the sum of the sizes of the members and base classes. With
n-byte alignment, size will typically be the smallest multiple ofnwhich is larger than the size of all members & base classes. Each membermemNwill typically be placed at an address which is a multiple ofalignof(memN), andnwill typically be the largestalignofout of all members'alignofs. Due to this, if a member with a smalleralignofis followed by a member with a largeralignof, there is a possibility that the latter member will not be aligned properly if placed immediately after the former. In this case, padding (also known as an alignment member ) will be placed between the two members, such that the latter member can have its desired alignment. Conversely, if a member with a largeralignofis followed by a member with a smalleralignof, no padding will usually be necessary. This process is also known as "packing".
Due to classes typically sharing thealignofof their member with the largestalignof, classes will typically be aligned to thealignofof the largest built-in type they directly or indirectly contain.// Assume sizeof(short) == 2, sizeof(int) == 4, and sizeof(long long) == 8. // Assume 4-byte alignment is specified to the compiler. struct Char { char c; }; // sizeof(Char) == 1 (sizeof(char)) struct Int { int i; }; // sizeof(Int) == 4 (sizeof(int)) struct CharInt { char c; int i; }; // sizeof(CharInt) == 8 (1 (char) + 3 (padding) + 4 (int)) struct ShortIntCharInt { short s; int i; char c; int j; }; // sizeof(ShortIntCharInt) == 16 (2 (short) + 2 (padding) + 4 (int) + 1 (char) + // 3 (padding) + 4 (int)) struct ShortIntCharCharInt { short s; int i; char c; char d; int j; }; // sizeof(ShortIntCharCharInt) == 16 (2 (short) + 2 (padding) + 4 (int) + 1 (char) + // 1 (char) + 2 (padding) + 4 (int)) struct ShortCharShortInt { short s; char c; short t; int i; }; // sizeof(ShortCharShortInt) == 12 (2 (short) + 1 (char) + 1 (padding) + 2 (short) + // 2 (padding) + 4 (int)) struct IntLLInt { int i; long long l; int j; }; // sizeof(IntLLInt) == 16 (4 (int) + 8 (long long) + 4 (int)) // If packing isn't explicitly specified, most compilers will pack this as // 8-byte alignment, such that: // sizeof(IntLLInt) == 24 (4 (int) + 4 (padding) + 8 (long long) + // 4 (int) + 4 (padding)) // Assume sizeof(bool) == 1, sizeof(ShortIntCharInt) == 16, and sizeof(IntLLInt) == 24. // Assume default alignment: alignof(ShortIntCharInt) == 4, alignof(IntLLInt) == 8. struct ShortChar3ArrShortInt { short s; char c3[3]; short t; int i; }; // ShortChar3ArrShortInt has 4-byte alignment: alignof(int) >= alignof(char) && // alignof(int) >= alignof(short) // sizeof(ShortChar3ArrShortInt) == 12 (2 (short) + 3 (char[3]) + 1 (padding) + // 2 (short) + 4 (int)) // Note that t is placed at alignment of 2, not 4. alignof(short) == 2. struct Large_1 { ShortIntCharInt sici; bool b; ShortIntCharInt tjdj; }; // Large_1 has 4-byte alignment. // alignof(ShortIntCharInt) == alignof(int) == 4 // alignof(b) == 1 // Therefore, alignof(Large_1) == 4. // sizeof(Large_1) == 36 (16 (ShortIntCharInt) + 1 (bool) + 3 (padding) + // 16 (ShortIntCharInt)) struct Large_2 { IntLLInt illi; float f; IntLLInt jmmj; }; // Large_2 has 8-byte alignment. // alignof(IntLLInt) == alignof(long long) == 8 // alignof(float) == 4 // Therefore, alignof(Large_2) == 8. // sizeof(Large_2) == 56 (24 (IntLLInt) + 4 (float) + 4 (padding) + 24 (IntLLInt))
C++11
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If strict alignment is forced with
alignas, padding will be used to force the type to meet the specified alignment, even when it would otherwise be smaller. For example, with the definition below,Chars<5>will have three (or possibly more) padding bytes inserted at the end so that its total size is 8. It is not possible for a class with an alignment of 4 to have a size of 5 because it would be impossible to make an array of that class, so the size must be "rounded up" to a multiple of 4 by inserting padding bytes.// This type shall always be aligned to a multiple of 4. Padding shall be inserted as // needed. // Chars<1>..Chars<4> are 4 bytes, Chars<5>..Chars<8> are 8 bytes, etc. template<size_t SZ> struct alignas(4) Chars { char arr[SZ]; }; static_assert(sizeof(Chars<1>) == sizeof(Chars<4>), "Alignment is strict.\n");
- If two non-static members of a class have the same access specifier, then the one that comes later in declaration order is guaranteed to come later in the object representation. But if two non-static members have different access specifiers, their relative order within the object is unspecified.
- It is unspecified what order the base class subobjects appear in within an object, whether they occur consecutively, and whether they appear before, after, or between member subobjects.
