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// revised beginning portion of min() // introduces small inefficiency int minVal = ivec[0]; occurs = 0; int size = ivecsize(); for ( int ix = 0; ix < size; ++ix ) { if ( minVal == ivec[ ix ] ) ++occurs; //
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Because ix is initialized to 0, the first iteration of the loop always finds minVal equal to ivec[0], the value to which we've initialized minVal By initializing ix to 1, we can avoid performing the unnecessary comparison and reassignment of minVal This is an admittedly small improvement, and unfortunately, it introduces yet another bug into our program (maybe we should have just left things as they were!) Do you see what's wrong with our revised program
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// revised beginning portion of min() // unfortunately, it introduces a program bug int minVal = ivec[0]; occurs = 0; int size = ivecsize(); for ( int ix = 1; ix < size; ++ix ) { if ( minVal == ivec[ ix ] ) ++occurs; //
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If ivec[0] turns out to be the minimum value, then occurs is never set to 1! The fix is easy, of course, but only after we first see that it's necessary:
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int minVal = ivec[0]; occurs = 1;
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Unfortunately, this still isn't quite right What happens if the user, accidentally or otherwise, passes in an empty vector Attempting to access the first element of an empty vector is incorrect, and is likely to result in a run-time program error We must guard against this possibility, however improbable One solution is the following: (alternative solutions include returning a bool value indicating whether the function succeeded):
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int min( const vector<int> &ivec, int &occurs ) { int size = ivecsize(); // handle anomaly of empty vector // occurs set to 0 indicates empty vector if ( ! size ) { occurs = 0; return 0; } // ok: vector contains at least one element
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int minVal = ivec[ 0 ]; occurs = 1; for ( int ix = 1; ix < size; ++ix ) { if ( minVal == ivec[ ix ] ) ++occurs; else if ( minVal > ivec[ ix ] ){ minVal = ivec[ ix ]; occurs = 1; } } return minVal; }
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An alternative solution to the problem of an empty vector is to have min() return a bool value indicating success or failure, with the minimum value returned by reference:
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// alternative solution to empty vector problem bool min( const vector< int > &ivec, int &minVal, int &occurs );
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Another design choice is to have min() throw an exception should an empty vector be passed to it (See 11 for a discussion of exception handling) Unfortunately, errors such as these are not at all uncommon As programmers, we are going to make mistakes at some point at times even stupid mistakes It happens The important thing is to accept that mistakes happen and to be on the alert for them, testing and reviewing our code as rigorously as circumstances permit The conditional operator can provide a convenient shorthand notation for a simple if-else test For example, the following min() function template
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template <class valueType> inline const valueType& min( valueType &val1, valueType &val2 ) { if ( val1 < val2 ) return val1; return val2; }
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template <class valueType> inline const valueType& min( valueType &val1, valueType &val2 ) { return ( val1 < val2 ) val1 : val2;
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Long chains of if-else statements, such as the following, can often be difficult to read and prone to error when modifying:
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if ( ch == 'a' || ch == 'A' ) ++aCnt; else if ( ch == 'e' || ch == 'E' ) ++eCnt; else if ( ch == 'i' || ch == 'I' ) ++iCnt; else if ( ch == 'o' || ch == 'O' ) ++oCnt; else if ( ch == 'u' || ch == 'U' ) ++uCnt;
An alternative construct to a chain of if-else statements is the switch statement, provided the values being tested against are constant expressions, such as the character constants tested above The switch statement is the topic of the next section Exercise 53 Correct each of the following:
(a) if ( ival1 != ival2 ) ival1 = ival2 else ival1 = ival2 = 0; (b) if ( ival < minval ) minval = ival; occurs = 1; (c) if ( int ival = get_value()) cout "ival = " ival endl; if ( ! ival ) cout "ival = 0\n"; (d) if ( ival = 0 ) ival = get_value(); (e) if ( ival == 0 ) else ival = 0;
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