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28319: (z) splitting oddities
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Peter Stephenson
parent
22899d3788
commit
8aac69241f
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@ -1,5 +1,8 @@
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2010-10-06 Peter Stephenson <pws@csr.com>
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* 28319: Src/hist.c, Test/D04parameter.ztst: ${(z)...}
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splitting oddities and some tests for consistency.
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* 28285: Doc/Zsh/zle.yo, Src/Zle/zle_hist.c: add
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zle-isearch-update and zle-isearch-exit hooks.
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@ -13700,5 +13703,5 @@
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*****************************************************
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* This is used by the shell to define $ZSH_PATCHLEVEL
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* $Revision: 1.5094 $
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* $Revision: 1.5095 $
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*****************************************************
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106
Src/hist.c
106
Src/hist.c
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@ -2869,6 +2869,7 @@ bufferwords(LinkList list, char *buf, int *index)
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int num = 0, cur = -1, got = 0, ne = noerrs;
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int owb = wb, owe = we, oadx = addedx, ozp = zleparse, onc = nocomments;
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int ona = noaliases, ocs = zlemetacs, oll = zlemetall;
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int forloop = 0;
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char *p, *addedspaceptr;
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if (!list)
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@ -2942,25 +2943,84 @@ bufferwords(LinkList list, char *buf, int *index)
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ctxtlex();
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if (tok == ENDINPUT || tok == LEXERR)
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break;
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if (tokstr && *tokstr) {
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untokenize((p = dupstring(tokstr)));
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if (ingetptr() == addedspaceptr + 1) {
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/*
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* Whoops, we've read past the space we added, probably
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* because we were expecting a terminator but when
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* it didn't turn up we shrugged our shoulders thinking
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* it might as well be a complete string anyway.
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* So remove the space. C.f. below for the case
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* where the missing terminator caused a lex error.
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* We use the same paranoid test.
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*/
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int plen = strlen(p);
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if (plen && p[plen-1] == ' ' &&
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(plen == 1 || p[plen-2] != Meta))
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p[plen-1] = '\0';
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if (tok == FOR) {
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/*
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* The way for (( expr1 ; expr2; expr3 )) is parsed is:
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* - a FOR tok
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* - a DINPAR with no tokstr
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* - two DINPARS with tokstr's expr1, expr2.
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* - a DOUTPAR with tokstr expr3.
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*
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* We'll decrement the variable forloop as we verify
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* the various stages.
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*
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* Don't ask me, ma'am, I'm just the programmer.
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*/
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forloop = 5;
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} else {
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switch (forloop) {
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case 1:
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if (tok != DOUTPAR)
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forloop = 0;
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break;
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case 2:
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case 3:
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case 4:
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if (tok != DINPAR)
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forloop = 0;
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break;
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default:
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/* nothing to do */
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break;
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}
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}
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if (tokstr) {
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switch (tok) {
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case ENVARRAY:
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p = dyncat(tokstr, "=(");
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break;
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case DINPAR:
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if (forloop) {
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/* See above. */
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p = dyncat(tokstr, ";");
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} else {
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/*
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* Mathematical expressions analysed as a single
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* word. That's correct because it behaves like
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* double quotes. Whitespace in the middle is
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* similarly retained, so just add the parentheses back.
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*/
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p = tricat("((", tokstr, "))");
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}
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break;
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default:
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p = dupstring(tokstr);
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break;
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}
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if (*p) {
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untokenize(p);
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if (ingetptr() == addedspaceptr + 1) {
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/*
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* Whoops, we've read past the space we added, probably
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* because we were expecting a terminator but when
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* it didn't turn up we shrugged our shoulders thinking
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* it might as well be a complete string anyway.
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* So remove the space. C.f. below for the case
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* where the missing terminator caused a lex error.
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* We use the same paranoid test.
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*/
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int plen = strlen(p);
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if (plen && p[plen-1] == ' ' &&
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(plen == 1 || p[plen-2] != Meta))
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p[plen-1] = '\0';
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}
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addlinknode(list, p);
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num++;
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}
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addlinknode(list, p);
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num++;
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} else if (buf) {
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if (IS_REDIROP(tok) && tokfd >= 0) {
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char b[20];
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@ -2973,6 +3033,16 @@ bufferwords(LinkList list, char *buf, int *index)
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num++;
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}
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}
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if (forloop) {
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if (forloop == 1) {
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/*
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* Final "))" of for loop to match opening,
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* since we've just added the preceding element.
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*/
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addlinknode(list, dupstring("))"));
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}
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forloop--;
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}
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if (!got && !zleparse) {
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got = 1;
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cur = num - 1;
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@ -389,6 +389,34 @@
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>)
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>ten( more
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strings=(
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'foo=(1 2 3)'
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'(( 3 + 1 == 8 / 2 ))'
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'for (( i = 1 ; i < 10 ; i++ ))'
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)
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for string in $strings; do
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array=(${(z)string})
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for (( i = 1; i <= ${#array}; i++ )); do
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print -r -- "${i}:${array[i]}:"
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done
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done
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0:Some syntactical expressions that are hard to split into words with (z).
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>1:foo=(:
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>2:1:
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>3:2:
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>4:3:
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>5:):
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>1:(( 3 + 1 == 8 / 2 )):
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>1:for:
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>2:((:
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# Leading whitespace is removed, because the word proper hasn't started;
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# trailing whitespace is left because the word is terminated by the
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# semicolon or double parentheses. Bit confusing but sort of consistent.
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>3:i = 1 ;:
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>4:i < 10 ;:
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>5:i++ :
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>6:)):
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psvar=(dog)
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setopt promptsubst
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foo='It shouldn'\''t $(happen) to a %1v.'
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