本文整理汇总了C++中UtlHashBag::numberOfBuckets方法的典型用法代码示例。如果您正苦于以下问题：C++ UtlHashBag::numberOfBuckets方法的具体用法？C++ UtlHashBag::numberOfBuckets怎么用？C++ UtlHashBag::numberOfBuckets使用的例子？那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类UtlHashBag的用法示例。
在下文中一共展示了UtlHashBag::numberOfBuckets方法的1个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C++代码示例。
示例1: testRemoveReference
void testRemoveReference()
{
// the following two entries collide if the initial bucket size is 16
UtlInt int1(1);
UtlInt int16(16);
UtlInt int2a(2);
UtlInt int2b(2);
UtlInt int3(3);
UtlHashBag bag;
CPPUNIT_ASSERT( bag.numberOfBuckets() == 16 ); // check assumption of collision
// Load all the test objects
CPPUNIT_ASSERT( bag.insert(&int1) == &int1 );
CPPUNIT_ASSERT( bag.insert(&int16) == &int16 );
CPPUNIT_ASSERT( bag.insert(&int2a) == &int2a );
CPPUNIT_ASSERT( bag.insert(&int2b) == &int2b );
CPPUNIT_ASSERT( bag.insert(&int3) == &int3 );
// Check that everything is there
CPPUNIT_ASSERT( bag.entries() == 5 );
CPPUNIT_ASSERT( bag.contains(&int1) );
CPPUNIT_ASSERT( bag.contains(&int16) );
CPPUNIT_ASSERT( bag.contains(&int2a) ); // cannot test for 2a and 2b independently
CPPUNIT_ASSERT( bag.contains(&int3) );
// Take entry 1 out (might collide w/ 16)
CPPUNIT_ASSERT( bag.removeReference(&int1) == &int1 );
// Check that everything except entry 1 is still there, and that 1 is gone
CPPUNIT_ASSERT( bag.entries() == 4 );
CPPUNIT_ASSERT( ! bag.contains(&int1) );
CPPUNIT_ASSERT( bag.contains(&int16) );
CPPUNIT_ASSERT( bag.contains(&int2a) );// cannot test for 2a and 2b independently
CPPUNIT_ASSERT( bag.contains(&int3) );
// Put entry 1 back in (so that 16 will collide w/ it again)
CPPUNIT_ASSERT( bag.insert(&int1) == &int1 );
// Check that everything is there
CPPUNIT_ASSERT( bag.entries() == 5 );
CPPUNIT_ASSERT( bag.contains(&int1) );
CPPUNIT_ASSERT( bag.contains(&int16) );
CPPUNIT_ASSERT( bag.contains(&int2a) );
CPPUNIT_ASSERT( bag.contains(&int3) );
// Take entry 16 out (might collide w/ 1)
CPPUNIT_ASSERT( bag.removeReference(&int16) == &int16 );
// Check that everything except entry 16 is still there, and that 16 is gone
CPPUNIT_ASSERT( bag.entries() == 4 );
CPPUNIT_ASSERT( bag.contains(&int1) );
CPPUNIT_ASSERT( ! bag.contains(&int16) );
CPPUNIT_ASSERT( bag.contains(&int2a) );// cannot test for 2a and 2b independently
CPPUNIT_ASSERT( bag.contains(&int3) );
// remove 2a (and ensure that you don't get back 2b)
CPPUNIT_ASSERT( bag.removeReference(&int2a) == &int2a );
// Check that everything that should be is still there
CPPUNIT_ASSERT( bag.entries() == 3 );
CPPUNIT_ASSERT( bag.contains(&int1) );
CPPUNIT_ASSERT( ! bag.contains(&int16) );
CPPUNIT_ASSERT( bag.find(&int2a) == &int2b ); // equal values, but now there's only one
CPPUNIT_ASSERT( bag.contains(&int3) );
// remove 3 (no collision for this one)
CPPUNIT_ASSERT( bag.removeReference(&int3) == &int3 );
// Check that everything that should be is still there
CPPUNIT_ASSERT( bag.entries() == 2 );
CPPUNIT_ASSERT( bag.contains(&int1) );
CPPUNIT_ASSERT( ! bag.contains(&int16) );
CPPUNIT_ASSERT( bag.find(&int2a) == &int2b ); // equal values, but now there's only one
CPPUNIT_ASSERT( ! bag.contains(&int3) );
// remove 3 again - should fail this time
CPPUNIT_ASSERT( bag.removeReference(&int3) == NULL );
// Check that everything that should be is still there
CPPUNIT_ASSERT( bag.entries() == 2 );
CPPUNIT_ASSERT( bag.contains(&int1) );
CPPUNIT_ASSERT( ! bag.contains(&int16) );
CPPUNIT_ASSERT( bag.find(&int2a) == &int2b ); // equal values, but now there's only one
CPPUNIT_ASSERT( ! bag.contains(&int3) );
}