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) );
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parent = pathback( ) ; pathpop-back( if( parent->left = = current )
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current return;
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template <class Object, class Compare> ConstSetItr<Object, Compare> h ConstSetItr<Object,Compare>::operator++ ( i assertCanAdvance( ) ; advance i ) ; return *this;
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template <class Object, class Compare> ConstSetItr<Object,Compare> ConstSetItr<Object,Compare>::operator++
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ConstSetItr<Object,Compare>old = *this; + + ( *this ) ; return old;
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Figure 1978 Advancing for the s e t class
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1 template <class Object, class Compare> 2 bool ConstSetItr<Object,Compare>:: 3 operator== ( const ConstSetItr & rhs ) const 4 i 5 return root = = rhsroot & & current = = rhscurrent; 6 1 7 8 template <class Object, class Compare> 9 bool ConstSetItr<Object,Compare>:: 10 operator!= i const ConstSetItr & rhs ) const 11 t 12 return ! ( *this == rhs ) ; 13 1 14 15 template <class Object, class Compare> 16 ConstSetItr<Object,Compare>:: 17 ConstSetItri const set<Object,Compare> & source 18 : root( sourceroot ) , current( NULL ) { 1 19 20 template <class Object, class Compare> 21 void ConstSetItr<Object,Compare>::goLeft( ) 22 ( 23 pathpush-back( current ) ; 24 current = current->left; 25 1 26 27 template <class Object, class Compare> 28 void ConstSetItrcObject,Compare>::goRight( ) 29 ( 30 pathpush-back( current 1 ; 31 current = current->right; 32 1 33 34 template <class Object, class Compare> 35 void ConstSetItr~Object,Compare~::goRoot( ) 36 i 37 current = root: 38 while ( !pathemptyi ) ) 39 pathpop-back( ) ; 40 1 41 42 template <class Object, class Compare> 43 bool ConstSetItr<Object,Compare>::hasLeft( ) const 44 i
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45 46 47 48 49 50 51 52
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template <class Object, class Compare> bool ConstSetItr<Object,Compare>::hasRight() const i return current->right ! = current->right->right;
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Figure 1979 Various one-line routines for the s e t class
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1 template <class kvpair, class Compare> 2 class lessKV
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3 I 4 public: 5 boo1 operatori ) ( const kvpair & lhs, 6 const kvpair & rhs ) const 7 { return less ( lhsfirst, rhsfirst ) ; } 8 Compare less; 9 };
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Figure 1980 A function object for the m a p class that uses the key to compare
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S o far, we have assumed that we can store an entire data structure in the main memory of a computer Suppose, however, that we have more data than can fit in main memory, and, as a result, we must have the data structure reside on disk When that happens, the rules of the game change, because the Big-Oh model is no longer meaningful The problem is that a Big-Oh analysis assumes that all operations are equal However that is not true especially when disk I10 is involved On the one hand, a 500-MIPS machine supposedly executes 500 million instructions per second That is fairly fast, mainly because the speed depends largely on electrical properties On the other hand, a disk is mechanical Its speed depends largely on the time required to spin the disk and move a disk head Many disks spin at 7200 RPM Thus in 1 minute, it makes 7200 revolutions; hence one revolution occurs in 11120 of a second, or 83 ms On average we might expect that we have to spin a disk halfway to find what we are Looking for, but this is compensated by the time to move the disk head, so we get an access time of 83 ms (This estimate is very charitable; 9 to 1 I ms access times are more common) Consequently, we can do approximately 120 disk accesses per second This number of accesses sounds good, until we compare it with the processor speed: We have 500 million instructions versus 120 disk accesses Put another way, one disk access is worth about 4,000,000 instructions Of course, everything here is a rough calculation, but the relative speeds are rather clear: Disk accesses are incredibly expensive Furthermore, processor speeds are increasing at a much faster rate than disk speeds (it is disk sizes that are increasing quite quickly) Thus, we are willing to do lots of calculations just to save a disk access In almost all cases, the number of disk accesses dominates the running time By halving the number of disk accesses, we can halve the running time
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When data are too large fit in the number of disk accesses becomes important A disk access is unbelievably expensive to a typical computer instruction
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