Question

给定无符号整数，我想知道是否有办法确定是否在单个操作中设置了多个位。已经有一种替代方法可以通过逐位检查来实现这一点（如下所示），但我想知道是否有办法集中检查所有位。

typedef enum Foo_X
{
  Foo_0 = 0x1,
  Foo_1 = 0x2,
  Foo_2 = 0x4,
  Foo_3 = 0x8,
} Foo_X;

bool CheckFoo ( UINT Value, Foo_X Foo_to_Check )
{
    if (Value & Foo_to_Check)
    {
        // Foo_to_Check is present
        return true;
    }
}

void main()
{
    UINT value = GetValueFromSomewhere();
    if (CheckFoo(value, Foo_0) && CheckFoo(value, Foo_3))
        // both Foo_0 and Foo_3 are present
    else
        // not both present
}

使用集合方法的示例也如下所示。有任何想法吗？ TIA！

bool CheckFooTogether ( UINT Value, UINT Foos_to_Check )
{
    // check value against Foos_to_Check collectively
}

void main()
{
    UINT value = GetValueFromSomewhere();
    if (CheckFooTogether (value, Foo_0 | Foo_3))
        // both Foo_0 and Foo_3 are present
    else
        // not both present           
}

Answer 1

（Value＆amp; Foos_to_Check）== Foos_to_Check

Answer 2

(value & (FOO_0 | FOO_3)) == (FOO_0 | FOO_3)

在设置true和FOO_0时返回FOO_3。

Answer 3

您可以使用简单的按位and检查来完成此操作：

bool CheckFooTogether ( UINT Value, UINT Foos_to_Check )
{
    // check value against Foos_to_Check collectively
    if( (Value & Foos_to_Check) == Foos_to_Check)
        return true;
    else
        return false;
}

int main()
{
    UINT value =  Foo_3 | Foo_2;
    if (CheckFooTogether (value, Foo_0 | Foo_3))
        // both Foo_0 and Foo_3 are present
        cout << "both present";
    else
        // not both present    
        cout << "not both present";
}

Example

如何统一检查位是否设置为无符号整数？

3 个答案: