__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 compatibility with previous versions, _alignof is a synonym for __alignof unless compiler option /Za (Disable language extensions) is specified.

For more information on modifying alignment, see:

For more information on differences in alignment in code for x86 and x64, see:

END Microsoft Specific

See also

Expressions with Unary Operators
Keywords