From 6df43f8b17ff16e8b6124dee2bd47cbd05a89e30 Mon Sep 17 00:00:00 2001 From: Anne-Edgar WILKE Date: Fri, 28 Aug 2015 00:38:22 +0200 Subject: [PATCH] Fix COMPOUNDHYPHENMIN=1 compound hyphenation FIRST BUG --------- Problem In a compound word, the word parts of two characters are never hyphenated. Example To reproduce the bug, just go to the directory hyphen-2.8.8 and do the following : echo "\ UTF-8 LEFTHYPHENMIN 1 RIGHTHYPHENMIN 1 COMPOUNDLEFTHYPHENMIN 1 COMPOUNDRIGHTHYPHENMIN 1 .post1 NEXTLEVEL e1 a1 " > hyphen.pat ./example hyphen.pat <(echo postea) The output is post=ea ; but it should be post=e=a. If you replace postea with posteaque in the command above, you get post=e=a=que, which is correct. Indeed, the component "eaque" is now five characters long, so it is hyphenated. If you replace postea with ea, you get e=a, which is also correct ; this is because ea is not a compound word. Solution In the file hyphen.c, line 966, "if (i - begin > 1)" must be replaced with "if (i - begin > 0)". Indeed, the word part is comprised between begin and i inclusively ; its length is i - begin + 1. So, if you want to hyphenate the words parts of length 2 and above, you have to check that i - begin + 1 >= 2, ie i - begin > 0. SECOND BUG ---------- Problem In a compound word, the word parts are never hyphenated between their second to last and their last character. Example To reproduce the bug, do the following : echo "\ UTF-8 LEFTHYPHENMIN 1 RIGHTHYPHENMIN 1 COMPOUNDLEFTHYPHENMIN 1 COMPOUNDRIGHTHYPHENMIN 1 1que. NEXTLEVEL e1 " > hyphen.pat ./example hyphen.pat <(echo meaque) The output is mea=que ; but it should be me=a=que. Again, if you replace meaque with mea, you get me=a, which is correct, because mea is not a compound word. If you replace meaque with eamque, you get e=am=que, as expected ; it shows that there is no similar bug with the first and the second character of word parts. Solution In the file hyphen.c, line 983, "for (j = 0; j < i - begin - 1; j++)" must be replaced with "for (j = 0; j < i - begin; j++)". Indeed, the word part has length i - begin + 1. So there are i - begin possible places for a hyphen. Thus j must take i - begin different values, ie go from 0 to i - begin - 1. --- hyphen.c | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/hyphen.c b/hyphen.c index 79dc072..4954dbd 100644 --- a/hyphen.c +++ b/hyphen.c @@ -963,7 +963,7 @@ int hnj_hyphen_hyph_(HyphenDict *dict, const char *word, int word_size, for (i = 0; i < word_size; i++) rep2[i] = NULL; for (i = 0; i < word_size; i++) if (hyphens[i]&1 || (begin > 0 && i + 1 == word_size)) { - if (i - begin > 1) { + if (i - begin > 0) { int hyph = 0; prep_word[i + 2] = '\0'; /* non-standard hyphenation at compound boundary (Schiffahrt) */ @@ -980,7 +980,7 @@ int hnj_hyphen_hyph_(HyphenDict *dict, const char *word, int word_size, hnj_hyphen_hyph_(dict, prep_word + begin + 1, i - begin + 1 + hyph, hyphens2, &rep2, &pos2, &cut2, clhmin, crhmin, (begin > 0 ? 0 : lend), (hyphens[i]&1 ? 0 : rend)); - for (j = 0; j < i - begin - 1; j++) { + for (j = 0; j < i - begin; j++) { hyphens[begin + j] = hyphens2[j]; if (rep2[j] && rep && pos && cut) { if (!*rep && !*pos && !*cut) {