CHANGESTR.REX: Difference between revisions
m (updated some comments in this version of the BIF. -- ~~~~) |
m (condensed assignment of two null variables, changed some comments. -- ~~~~) |
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↓ ↓ ↓ ↓ ↓ ↓ */ |
↓ ↓ ↓ ↓ ↓ ↓ */ |
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changestr: parse arg o,h,n,t,b,p /* |
changestr: parse arg o,h,n,t,b,p,$ f /*T,B,P are optional.*/ |
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$='' /*$: the returned string.*/ |
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/*optional arguments ··· */ |
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t=word(t 999999999 , 1) /*maybe use the default? */ |
t=word(t 999999999 , 1) /*maybe use the default? */ |
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b=word(b 1 , 1) /* " " " " */ |
b=word(b 1 , 1) /* " " " " */ |
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L=length(o) /*length of OLD string. */ |
L=length(o) /*length of OLD string. */ |
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if L==0 & t\=0 then return n || h /*changing a null char? */ |
if L==0 & t\=0 then return n || h /*changing a null char? */ |
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f='' /*first part of H if P>1. */ |
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if p\=1 then do /*if P ¬= 1, adjust F & H.*/ |
if p\=1 then do /*if P ¬= 1, adjust F & H.*/ |
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f=left(h, min(p-1, length(h))) /*keep first part intact. */ |
f=left(h, min(p-1, length(h))) /*keep first part intact. */ |
Revision as of 22:38, 20 February 2014
This version of the changestr BIF has more functionality than the standard BIF.
<lang rexx>/*REXX program emulates the CHANGESTR built-in function for older REXXes*/ /*──── This version has more functionality: limit the number of changes.*/ /*──── start of change occurrence#.*/ /*──── start of change position. */
/*╔══════════════════════════ CHANGESTR function ══════════════════════╗ ╔═╩════════════════════════════════════════════════════════════════════╩═╗ ║ The CHANGESTR function is used to replace some or all occurrences of an║ ║ (old) string in a haystack with a new string. The changed string is ║ ║ returned. If the haystack doesn't contain the old string, the ║ ║ original haystack is returned. If the old string is a null string, ║ ║ then the original string is prefixed with the new string. ║ ║ ║ ║ new string to be used►──────────┐ ┌─────◄limit of # changes (times).║ ║ original string (haystack)►──────┐ │ │ [default: ≈ one billion]║ ║ old string to be replaced►──┐ │ │ │ ┌────◄begin at this occurrence #║ ║ {O, H, and N can be null.} │ │ │ │ │ ┌──◄start position (default=1)║ ╚═╦════════════════════════════╗ │ │ │ │ │ │ ╔═════════════════════════╦═╝
╚════════════════════════════╝ │ │ │ │ │ │ ╚═════════════════════════╝ ↓ ↓ ↓ ↓ ↓ ↓ */
changestr: parse arg o,h,n,t,b,p,$ f /*T,B,P are optional.*/ t=word(t 999999999 , 1) /*maybe use the default? */ b=word(b 1 , 1) /* " " " " */ p=word(p 1 , 1) /* " " " " */ if arg() < 3 then signal syntax /*not enough arguments. */ if arg() > 6 then signal syntax /*too many arguments. */ if \datatype(t,'W') then signal syntax /*4th arg not an integer. */ if \datatype(b,'W') then signal syntax /*5th " " " " */ if \datatype(p,'W') then signal syntax /*5th arg " " " */ if t<0 then signal syntax /*4th arg not non-negative*/ if b<1 then signal syntax /*5th arg not positive. */ if p<1 then signal syntax /*6th " " " */ L=length(o) /*length of OLD string. */ if L==0 & t\=0 then return n || h /*changing a null char? */ if p\=1 then do /*if P ¬= 1, adjust F & H.*/
f=left(h, min(p-1, length(h))) /*keep first part intact. */ h=substr(h,p) /*only use this part of H.*/ end /*now, proceed as usual. */
- =0 /*# of changed occurrences*/
do j=1 while # < t /*keep changing, T times. */ parse var h y (o) _ +(L) h /*parse the haystack ··· */ if _== then return f || $ || y /*no more left, return. */ $=$ || y /*append the residual txt.*/ if j<b then $=$ || o /*append OLD if too soon. */ else do /*met the occurrence test.*/ $=$ || n /*append the NEW string.*/ #=#+1 /*bump occurrence number.*/ end end /*j*/ /*Note: most REXX ··· */ /* CHANGESTR BIFs only ···*/
return f || $ || h /* support three options. */</lang>