alignof operator
The alignof operator returns the alignment in bytes of the specified type as a value of type size_t.
Syntax
alignof( type )
Remarks
For example:
| Expression | Value |
|---|---|
alignof( char ) |
1 |
alignof( short ) |
2 |
alignof( int ) |
4 |
alignof( long long ) |
8 |
alignof( float ) |
4 |
alignof( double ) |
8 |
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:
Microsoft-specific
alignof and __alignof are synonyms in the Microsoft compiler. Before it became part of the standard in C++11, the Microsoft-specific __alignof operator provided this functionality. For maximum portability, you should use the alignof operator instead of the Microsoft-specific __alignof operator.
For compatibility with previous versions, _alignof is a synonym for __alignof unless compiler option /Za (Disable language extensions) is specified.