alignof operator （C++11 起）
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查询的对齐要求的类型
Original:
Queries alignment requirements of a type
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目录 |
[编辑] 语法
alignof( type )
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返回对象类型std::size_t
Original:
Returns an object of type std::size_t.
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[编辑] 解释
返回
type，这是无论是完整的类型，数组类型，或引用类型的实例所需的任何调整（以字节为单位）（2的整数次幂）.Original:
Returns alignment in bytes (an integer power of two) required for any instance of the given
type, which is either complete type, an array type, or a reference type.The text has been machine-translated via Google Translate.
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如果类型是引用类型，操作符返回的对齐方式“引用”类型，如果该类型是数组类型，元素类型的对齐要求，返回.
Original:
If the type is reference type, the operator returns the alignment of referenced type; if the type is array type, alignment requirement of the element type is returned.
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类型char，signed char，unsigned char最弱（最小）对齐实现支持.
Original:
The types char, signed char, and unsigned char have the weakest (smallest) alignment supported by the implementation.
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[编辑] 关键字
[编辑] 为例
#include <iostream> struct Empty {}; struct Foo { int f2; float f1; char c; }; int main() { std::cout << "alignment of empty class: " << alignof(Empty) << '\n' << "alignment of pointer : " << alignof(int*) << '\n' << "alignment of char : " << alignof(char) << '\n' << "alignment of Foo : " << alignof(Foo) << '\n' ; }
Output:
alignment of empty class: 1 alignment of pointer : 8 alignment of char : 1 alignment of Foo : 4
[编辑] 另请参阅
| alignas说明 | 指定的变量的存储空间应保持一致的具体数额（C++11）
Original: specifies that the storage for the variable should be aligned by specific amount （C++11） The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. |
| （C++11） |
获得该类型的对齐要求 Original: obtains the type's alignment requirements The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. （类模板） |
