__alignof Operator
The latest version of this topic can be found at __alignof Operator.
C++11 introduces the alignof operator that returns the alignment, in bytes, of the specified type. For maximum portability, you should use the alignof operator instead of the Microsoft-specific __alignof operator.
Microsoft Specific
Returns a value of type size_t that is the alignment requirement of the type.
Syntax
__alignof(
type
)
Remarks
For example:
| Expression | Value |
|---|---|
| __alignof( char ) | 1 |
| __alignof( short ) | 2 |
| __alignof( int ) | 4 |
| __alignof( __int64 ) | 8 |
| __alignof( float ) | 4 |
| __alignof( double ) | 8 |
| __alignof( char* ) | 4 |
The __alignof value is the same as the value for sizeof for basic types. Consider, however, this example:
typedef struct { int a; double b; } S;
// __alignof(S) == 8
In this case, the __alignof value is the alignment requirement of the largest element in the structure.
Similarly, for
typedef __declspec(align(32)) struct { int a; } S;
__alignof(S) is equal to 32.
One use for __alignof would be as a parameter to one of your own memory-allocation routines. For example, given the following defined structure S, you could call a memory-allocation routine named aligned_malloc to allocate memory on a particular alignment boundary.
typedef __declspec(align(32)) struct { int a; double b; } S;
int n = 50; // array size
S* p = (S*)aligned_malloc(n * sizeof(S), __alignof(S));
For more information on modifying alignment, see:
For more information on differences in alignment in code for x86 and x64, see: