/********************************************************************** ** Copyright (C) 2000 Trolltech AS. All rights reserved. ** ** numberh.cpp ** ** This file is part of Qt Linguist. ** ** See the file LICENSE included in the distribution for the usage ** and distribution terms. ** ** The file is provided AS IS with NO WARRANTY OF ANY KIND, ** INCLUDING THE WARRANTY OF DESIGN, MERCHANTABILITY AND FITNESS FOR ** A PARTICULAR PURPOSE. ** **********************************************************************/ #include #include #include #include #include #include typedef TQMap TMM; typedef TQValueList TML; static bool isDigitFriendly( int c ) { return ispunct( c ) || isspace( c ); } static int numberLength( const char *s ) { int i = 0; if ( isdigit(s[0]) ) { do { i++; } while ( isdigit(s[i]) || (isDigitFriendly(s[i]) && (isdigit(s[i + 1]) || (isDigitFriendly(s[i + 1]) && isdigit(s[i + 2])))) ); } return i; } /* Returns a version of 'key' where all numbers have been replaced by zeroes. If there were none, returns "". */ static TQCString zeroKey( const char *key ) { TQCString zeroed( strlen(key) + 1 ); char *z = zeroed.data(); int i = 0, j = 0; int len; bool metSomething = FALSE; while ( key[i] != '\0' ) { len = numberLength( key + i ); if ( len > 0 ) { i += len; z[j++] = '0'; metSomething = TRUE; } else { z[j++] = key[i++]; } } z[j] = '\0'; if ( metSomething ) return zeroed; else return ""; } static TQString translationAttempt( const TQString& oldTranslation, const char *oldSource, const char *newSource ) { int p = zeroKey( oldSource ).contains( '0' ); int oldSourceLen = qstrlen( oldSource ); TQString attempt; TQStringList oldNumbers; TQStringList newNumbers; TQMemArray met( p ); TQMemArray matchedYet( p ); int i, j; int k = 0, ell, best; int m, n; int pass; /* This algorithm is hard to follow, so we'll consider an example all along: oldTranslation is "XeT 3.0", oldSource is "TeX 3.0" and newSource is "XeT 3.1". First, we set up two tables: oldNumbers and newNumbers. In our example, oldNumber[0] is "3.0" and newNumber[0] is "3.1". */ for ( i = 0, j = 0; i < oldSourceLen; i++, j++ ) { m = numberLength( oldSource + i ); n = numberLength( newSource + j ); if ( m > 0 ) { oldNumbers.append( TQCString(oldSource + i, m + 1) ); newNumbers.append( TQCString(newSource + j, n + 1) ); i += m; j += n; met[k] = FALSE; matchedYet[k] = 0; k++; } } /* We now go over the old translation, "XeT 3.0", one letter at a time, looking for numbers found in oldNumbers. Whenever such a number is met, it is replaced with its newNumber equivalent. In our example, the "3.0" of "XeT 3.0" becomes "3.1". */ for ( i = 0; i < (int) oldTranslation.length(); i++ ) { attempt += oldTranslation[i]; for ( k = 0; k < p; k++ ) { if ( oldTranslation[i] == oldNumbers[k][matchedYet[k]] ) matchedYet[k]++; else matchedYet[k] = 0; } /* Let's find out if the last character ended a match. We make two passes over the data. In the first pass, we try to match only numbers that weren't matched yet; if that fails, the second pass does the trick. This is useful in some suspicious cases, flagged below. */ for ( pass = 0; pass < 2; pass++ ) { best = p; // an impossible value for ( k = 0; k < p; k++ ) { if ( (!met[k] || pass > 0) && matchedYet[k] == (int) oldNumbers[k].length() && numberLength(oldTranslation.latin1() + (i + 1) - matchedYet[k]) == matchedYet[k] ) { // the longer the better if ( best == p || matchedYet[k] > matchedYet[best] ) best = k; } } if ( best != p ) { attempt.truncate( attempt.length() - matchedYet[best] ); attempt += newNumbers[best]; met[best] = TRUE; for ( k = 0; k < p; k++ ) matchedYet[k] = 0; break; } } } /* We flag two kinds of suspicious cases. They are identified as such with comments such as "{2000?}" at the end. Example of the first kind: old source text "TeX 3.0" translated as "XeT 2.0" is flagged "TeX 2.0 {3.0?}", no matter what the new text is. */ for ( k = 0; k < p; k++ ) { if ( !met[k] ) attempt += TQString( " {" ) + newNumbers[k] + TQString( "?}" ); } /* Example of the second kind: "1 of 1" translated as "1 af 1", with new source text "1 of 2", generates "1 af 2 {1 or 2?}" because it's not clear which of "1 af 2" and "2 af 1" is right. */ for ( k = 0; k < p; k++ ) { for ( ell = 0; ell < p; ell++ ) { if ( k != ell && oldNumbers[k] == oldNumbers[ell] && newNumbers[k] < newNumbers[ell] ) attempt += TQString( " {" ) + newNumbers[k] + TQString( " or " ) + newNumbers[ell] + TQString( "?}" ); } } return attempt; } /* Augments a MetaTranslator with translations easily derived from similar existing (probably obsolete) translations. For example, if "TeX 3.0" is translated as "XeT 3.0" and "TeX 3.1" has no translation, "XeT 3.1" is added to the translator and is marked Unfinished. */ void applyNumberHeuristic( MetaTranslator *tor, bool verbose ) { TMM translated, untranslated; TMM::Iterator t, u; TML all = tor->messages(); TML::Iterator it; int inserted = 0; for ( it = all.begin(); it != all.end(); ++it ) { if ( (*it).type() == MetaTranslatorMessage::Unfinished ) { if ( (*it).translation().isEmpty() ) untranslated.insert( zeroKey((*it).sourceText()), *it ); } else if ( !(*it).translation().isEmpty() ) { translated.insert( zeroKey((*it).sourceText()), *it ); } } for ( u = untranslated.begin(); u != untranslated.end(); ++u ) { t = translated.find( u.key() ); if ( t != translated.end() && !t.key().isEmpty() && qstrcmp((*t).sourceText(), (*u).sourceText()) != 0 ) { MetaTranslatorMessage m( *u ); m.setTranslation( translationAttempt((*t).translation(), (*t).sourceText(), (*u).sourceText()) ); tor->insert( m ); inserted++; } } if ( verbose && inserted != 0 ) qWarning( " number heuristic provided %d translation%s", inserted, inserted == 1 ? "" : "s" ); }