diff options
Diffstat (limited to 'Src')
| -rw-r--r-- | Src/hist.c | 106 |
1 files changed, 88 insertions, 18 deletions
diff --git a/Src/hist.c b/Src/hist.c index b9f315280..6cb7e1d31 100644 --- a/Src/hist.c +++ b/Src/hist.c @@ -2869,6 +2869,7 @@ bufferwords(LinkList list, char *buf, int *index) int num = 0, cur = -1, got = 0, ne = noerrs; int owb = wb, owe = we, oadx = addedx, ozp = zleparse, onc = nocomments; int ona = noaliases, ocs = zlemetacs, oll = zlemetall; + int forloop = 0; char *p, *addedspaceptr; if (!list) @@ -2942,25 +2943,84 @@ bufferwords(LinkList list, char *buf, int *index) ctxtlex(); if (tok == ENDINPUT || tok == LEXERR) break; - if (tokstr && *tokstr) { - untokenize((p = dupstring(tokstr))); - if (ingetptr() == addedspaceptr + 1) { - /* - * Whoops, we've read past the space we added, probably - * because we were expecting a terminator but when - * it didn't turn up we shrugged our shoulders thinking - * it might as well be a complete string anyway. - * So remove the space. C.f. below for the case - * where the missing terminator caused a lex error. - * We use the same paranoid test. - */ - int plen = strlen(p); - if (plen && p[plen-1] == ' ' && - (plen == 1 || p[plen-2] != Meta)) - p[plen-1] = '\0'; + if (tok == FOR) { + /* + * The way for (( expr1 ; expr2; expr3 )) is parsed is: + * - a FOR tok + * - a DINPAR with no tokstr + * - two DINPARS with tokstr's expr1, expr2. + * - a DOUTPAR with tokstr expr3. + * + * We'll decrement the variable forloop as we verify + * the various stages. + * + * Don't ask me, ma'am, I'm just the programmer. + */ + forloop = 5; + } else { + switch (forloop) { + case 1: + if (tok != DOUTPAR) + forloop = 0; + break; + + case 2: + case 3: + case 4: + if (tok != DINPAR) + forloop = 0; + break; + + default: + /* nothing to do */ + break; + } + } + if (tokstr) { + switch (tok) { + case ENVARRAY: + p = dyncat(tokstr, "=("); + break; + + case DINPAR: + if (forloop) { + /* See above. */ + p = dyncat(tokstr, ";"); + } else { + /* + * Mathematical expressions analysed as a single + * word. That's correct because it behaves like + * double quotes. Whitespace in the middle is + * similarly retained, so just add the parentheses back. + */ + p = tricat("((", tokstr, "))"); + } + break; + + default: + p = dupstring(tokstr); + break; + } + if (*p) { + untokenize(p); + if (ingetptr() == addedspaceptr + 1) { + /* + * Whoops, we've read past the space we added, probably + * because we were expecting a terminator but when + * it didn't turn up we shrugged our shoulders thinking + * it might as well be a complete string anyway. + * So remove the space. C.f. below for the case + * where the missing terminator caused a lex error. + * We use the same paranoid test. + */ + int plen = strlen(p); + if (plen && p[plen-1] == ' ' && + (plen == 1 || p[plen-2] != Meta)) + p[plen-1] = '\0'; + } + addlinknode(list, p); + num++; } - addlinknode(list, p); - num++; } else if (buf) { if (IS_REDIROP(tok) && tokfd >= 0) { char b[20]; @@ -2973,6 +3033,16 @@ bufferwords(LinkList list, char *buf, int *index) num++; } } + if (forloop) { + if (forloop == 1) { + /* + * Final "))" of for loop to match opening, + * since we've just added the preceding element. + */ + addlinknode(list, dupstring("))")); + } + forloop--; + } if (!got && !zleparse) { got = 1; cur = num - 1; |