214 lines
6.3 KiB
C
214 lines
6.3 KiB
C
/*****************************************************************************
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*
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* string.c
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*
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* String builtin macros.
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*
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*****************************************************************************/
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#include "m4.h"
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/*****************************************************************************
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*
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* opSubstr
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*
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* Return the substring of $1 starting from $2 and continuing for
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* $3 characters. If $3 is not supplied, then return the entire
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* remainder of the string.
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*
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* If $2 is out of range, then nothing is returned.
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*
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* If $3 is a negative number, then treat it as zero.
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*
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* The extra ptokNil covers us in the case where $# is 1.
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*
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*****************************************************************************/
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DeclareOp(opSubstr)
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{
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if (ctokArgv) {
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TOK tok;
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ITCH itch = (ITCH)atTraditionalPtok(ptokArgv(2));
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if (itch < ctchSPtok(ptokArgv(1))) {
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CTCH ctch;
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if (ctokArgv >= 3) {
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ctch = atTraditionalPtok(ptokArgv(3));
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if ((int)ctch < 0) {
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ctch = 0;
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}
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} else {
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ctch = ctchMax;
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}
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ctch = min(ctch, ctchSPtok(ptokArgv(1)) - itch);
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Assert(itch + ctch <= ctchSPtok(ptokArgv(1)));
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SetStaticPtokPtchCtch(&tok, ptchPtok(ptokArgv(1)) + itch, ctch);
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PushPtok(&tok);
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}
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} else {
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#ifdef STRICT_M4
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Warn("wrong number of arguments to %P", ptokArgv(0));
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#endif
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}
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}
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/*****************************************************************************
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*
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* opIndex
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*
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* Return the zero-based location of the first occurrence of $2 in $1,
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* or -1 if the substring does not appear. If there are multiple
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* matches, the leftmost one is returned.
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*
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* The extra ptokNil covers us in the case where $# is 1.
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*
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* QUIRK! AT&T returns -1 if $1 and $2 are both null strings.
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* GNU returns 0, which is what I do also.
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*
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*****************************************************************************/
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/* SOMEDAY! -- need minimum and maximum arg count */
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DeclareOp(opIndex)
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{
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if (ctokArgv) {
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/*
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* Note carefully: itch and itchMac need to be ints
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* because itchMac can underflow if you try to search
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* for a big string inside a small string.
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*/
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int itch;
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int itchMac = ctchSPtok(ptokArgv(1)) - ctchSPtok(ptokArgv(2));
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for (itch = 0; itch <= itchMac; itch++) {
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if (fEqPtchPtchCtch(ptchPtok(ptokArgv(1)) + itch,
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ptchPtok(ptokArgv(2)),
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ctchSPtok(ptokArgv(2)))) {
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PushAt(itch);
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return;
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}
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}
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PushAt(-1);
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} else {
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PushQuotedPtok(ptokArgv(0));
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}
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}
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/*****************************************************************************
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*
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* opTranslit
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*
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* For each character in the $1, look for a match in $2. If found,
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* produce the corresponding character from $3. If there is no
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* such character, then produce nothing.
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*
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* Note that the algorithm must be as specified, in order for
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*
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* translit(abc,ab,ba)
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*
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* to result in `bac'.
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*
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* We actually walk $1 backwards so we can push directly instead
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* of having to build a temporary token. But the walking of $2
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* must be in the forward direction, so that `translit(a,aa,bc)'
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* results in `b' and not `c'.
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*
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* ptokNil saves us in the case where $# = 1.
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*
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* QUIRK! If given only one argument, AT&T emits $1 unchanged.
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* GNU emits nothing! AT&T is obviously correct, so I side
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* with them on this one.
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*
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*****************************************************************************/
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DeclareOp(opTranslit)
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{
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if (ctokArgv) {
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ITCH itch = ctchArgv(1);
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while ((int)--itch >= 0) {
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TCH tch = ptchArgv(1)[itch];
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ITCH itch;
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for (itch = 0; itch < ctchArgv(2); itch++) {
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if (ptchArgv(2)[itch] == tch) {
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if (itch < ctchArgv(3)) {
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PushTch(ptchArgv(3)[itch]);
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}
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break;
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}
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}
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if (itch >= ctchArgv(2)) {
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PushTch(tch);
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}
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}
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} else {
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PushQuotedPtok(ptokArgv(0));
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}
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}
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/*****************************************************************************
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*
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* opPatsubst
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*
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* Scan $1 for any occurrences of $2. If found, replace them with $3.
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* If $3 is omitted, then the string is deleted.
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*
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* As a special case, if $2 is the null string, then $3 is inserted
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* at the beginning of the string and between each character of $1.
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*
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* NOTE! This is a GNU extension.
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*
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* NOTE! GNU supports regular expressions for $2. We support only
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* literal strings.
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*
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* NOTE! Scanning is required to be forwards, so we temporarily expand
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* into the Exp hold, then pop it off when we're done.
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*
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* QUIRK! If given only one argument, GNU emits nothing!
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* This is clearly wrong, so I emit $1.
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*
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*****************************************************************************/
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DeclareOp(opPatsubst)
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{
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if (ctokArgv) {
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CTCH ctchSrc = ctchArgv(1);
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PTCH ptchSrc = ptchArgv(1);
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CTCH ctchPat = ctchArgv(2); /* ptokNil saves us here */
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PTCH ptchPat = ptchArgv(2);
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TOK tok;
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OpenExpPtok(&tok);
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while (ctchSrc >= ctchPat) {
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if (fEqPtchPtchCtch(ptchPat, ptchSrc, ctchPat)) {
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if (ctokArgv >= 3) {
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AddExpPtok(ptokArgv(3));
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}
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if (ctchSrc == 0) {
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AddExpTch(*ptchSrc);
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ctchSrc--;
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ptchSrc++;
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} else {
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ctchSrc -= ctchPat;
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ptchSrc += ctchPat;
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}
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} else {
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AddExpTch(*ptchSrc);
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ctchSrc--;
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ptchSrc++;
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}
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}
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/* Flush out what's left of the string */
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while (ctchSrc) {
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AddExpTch(*ptchSrc);
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ctchSrc--;
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ptchSrc++;
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}
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CsopExpDopPdivPtok((DIVOP)PushZPtok, 0, &tok);
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} else {
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PushQuotedPtok(ptokArgv(0));
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}
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}
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