Thanks Marpon.
I have looked through all the posts here, and nobody had invoked strtok.
(Everything but it, I think)
IMHO, a strtok of genius on your part!
I need someone to defeat my string splitting algo
Re: I need someone to defeat my string splitting algo
A faster version improved 25% with 1 loop only
and adaptative redim of the array
I think i will win the challenge !
but remember that split function it is not a split function as vb was giving
for me it is a sort of split_any_delimiter with no empty string token
if you want an split function with all the tokens (even empty string) its a different story ( and slower of course)
and adaptative redim of the array
I think i will win the challenge !
but remember that split function it is not a split function as vb was giving
for me it is a sort of split_any_delimiter with no empty string token
if you want an split function with all the tokens (even empty string) its a different story ( and slower of course)
Code: Select all
#include "crt/string.bi" 'needed for strtok function
'using strtok(s1 as zstring ptr, s2 as zstring ptr) as zstring ptr only 1 loop
'you can adapt to expected number of tokens to have less redim adaptation, changing the initial i1 value (set to 10 here)
Function String_strtok2(s_in As String , chars As String , result() As String) As Long
Dim As Long ctr
dim as zstring ptr p, p1, p2
dim as long i1 = 10 'initial value to redim result
dim as string s2 = s_in 'copy because strtok will alter the input zstring
p = strptr(s2) 'get pointers
p2= strptr(chars)
p1 = strtok(p , p2)
while p1 <> NULL
ctr +=1
if ctr = 1 THEN
Redim result(1 To i1) 'redim the array first time
elseif ctr > i1 then
i1 *= 2
redim preserve result(1 To i1) 'redim the array to give more space (preserved existing content)
END IF
result(ctr)= *p1
p1 = strtok(NULL, p2)
WEND
If ctr = 0 Then Return 0
Redim preserve result(1 To ctr) 'redim the array to fit exact need (preserved existing content)
Return Ubound(result)
End Function
'====================================================================================Re: I need someone to defeat my string splitting algo
last optimization: with byval s_in As String to avoid to make a copy before using strtok
remind : the strtok function from the clib is very fast but not thread-safe ,
that function incorporates a static value to store the initial pointer and uses it on the consecutive executions,
so that means if you create a multi thread prog using that function it will surely fail.
In most case not a problem, but if you need an other one : thread-safe, contact me,
i 've done 1 some time ago, not as faster as this one (almost 2 x times slower) but even faster than all the versions i've seen here.
remind : the strtok function from the clib is very fast but not thread-safe ,
that function incorporates a static value to store the initial pointer and uses it on the consecutive executions,
so that means if you create a multi thread prog using that function it will surely fail.
In most case not a problem, but if you need an other one : thread-safe, contact me,
i 've done 1 some time ago, not as faster as this one (almost 2 x times slower) but even faster than all the versions i've seen here.
Code: Select all
'warning not thread-safe using strtok(s1 as zstring ptr, s2 as zstring ptr) as zstring ptr only 1 loop
Function String_strtok4(byval s_in As String , byref chars As String , result() As String) As Long
Dim As Long ctr
dim as zstring ptr p, p1, p2
dim as long i1 = 10 'initial value to redim result
p = strptr(s_in) 'get pointers
p2= strptr(chars)
p1 = strtok(p , p2)
while p1 <> NULL
ctr +=1
if ctr = 1 THEN
Redim result(1 To i1) 'redim the array first time
elseif ctr > i1 then
i1 *= 2
redim preserve result(1 To i1) 'redim the array to give more space
END IF
result(ctr)= *p1
p1 = strtok(NULL, p2)
WEND
If ctr = 0 Then Return 0
'redim the array to fit exact need
if Ubound(result) > ctr THEN Redim preserve result(1 To ctr)
Return Ubound(result)
End Function
'====================================================================================Re: I need someone to defeat my string splitting algo
Marpon
2) call from dll
1) compiling in dllremind : the strtok function from the clib is very fast but not thread-safe ,
that function incorporates a static value to store the initial pointer and uses it on the consecutive executions,
so that means if you create a multi thread prog using that function it will surely fail.
2) call from dll
Re: I need someone to defeat my string splitting algo
here my thread-safe version
not as fast as the strtok one but faster than all others found here
in fact it mimics the strtok_r existing in some c lib distributions ( _r for re-entrant)
notice also
Function String_strtok_r(Byval s_in As String , Byref chars As String , result() As String, Byval size1 As Long = 1) As Long
Byval is needed because we need a copy of the string , the f_strtok_r function will alter the input string (putting 0 to delimit the elements)
and in function f_strtok_r : Byref ps1 As Byte Ptr , is mandatory because the function will return back the modified ps1 pointer
Function f_strtok_r(Byref ps1 As Byte Ptr, Byref ps2 As Byte Ptr, Byval i As Long )As Byte Ptr
and the adaptative Redim , you can change the size1 optional parameter to try optimizing litle more according your split estimation
'with that version you can avoid dll and any work-arround , it will work safely!
not as fast as the strtok one but faster than all others found here
in fact it mimics the strtok_r existing in some c lib distributions ( _r for re-entrant)
notice also
Function String_strtok_r(Byval s_in As String , Byref chars As String , result() As String, Byval size1 As Long = 1) As Long
Byval is needed because we need a copy of the string , the f_strtok_r function will alter the input string (putting 0 to delimit the elements)
and in function f_strtok_r : Byref ps1 As Byte Ptr , is mandatory because the function will return back the modified ps1 pointer
Function f_strtok_r(Byref ps1 As Byte Ptr, Byref ps2 As Byte Ptr, Byval i As Long )As Byte Ptr
and the adaptative Redim , you can change the size1 optional parameter to try optimizing litle more according your split estimation
'with that version you can avoid dll and any work-arround , it will work safely!
Code: Select all
#include "crt/string.bi" 'needed for memchr function
Declare Function f_strtok_r(Byref ps1 As Byte Ptr, Byref ps2 As Byte Ptr, Byval i As Long )As Byte Ptr
'thread-safe function
Function String_strtok_r(Byval s_in As String , Byref chars As String , result() As String, Byval size1 As Long = 1) As Long
Dim ptemp As Zstring Ptr
Dim ctr As Long
Dim As Byte Ptr p = Strptr(s_in) 'get pointers
Dim As Byte Ptr d = Strptr(chars)
Dim As Long ilen = Len(chars)
ptemp = p
While ptemp <> NULL
ptemp = f_strtok_r(p, d, ilen) ' p is sent byref and it will be modified on the f_strtok_r function
if ptemp <> NULL Then
ctr += 1
If ctr = 1 Then
Redim result(1 To size1) 'redim the array
Elseif ctr > size1 then
size1 *= 2
Redim Preserve result(1 To size1) 'redim the array
End If
result(ctr)= *ptemp
End If
Wend
If ctr = 0 Then Return 0
If size1 > ctr Then Redim Preserve result(1 To ctr) 'redim the array
Return ctr
End Function
Function f_strtok_r(Byref ps1 As Byte Ptr, Byref ps2 As Byte Ptr, Byval i As Long )As Byte Ptr
Dim p As Byte Ptr
While memchr(ps2, *ps1, i) <> NULL
ps1 += 1
Wend
If *ps1 = 0 Then
ps1 = NULL
Return NULL
End If
p = ps1
While memchr(ps2, *ps1, i) = NULL
ps1 += 1
if *ps1 = 0 Then Exit While
Wend
If *ps1 <> 0 Then
*ps1 = 0
ps1 += 1
End If
Return p
End FunctionRe: I need someone to defeat my string splitting algo
i've collected from my own archives the different versions of split functions
each of them is optimized for speed
Split : vb6 equivalent function , splits the input string using sub-string as delimiter, consecutive delimiters produce empty elements
Split_noempty : variation of Split, but the empty elements are not collected
Split_any : splits the input string using every char as char delimiter, consecutive char delimiters produce empty elements
Split_tok : variation of Split_any, but the empty elements are not collected, using strtok from c lib , not thread-safe but very fast
Split_tok_r : works as Split_tok , but thread-safe and bit slower
some of them do not need any include, but some need
#include "crt/string.bi" 'needed for strtok/memchr functions
i've put the indication on the source
source code and test comparaison
feel free to test and use...
waiting for comments/remarks or faster solutions
each of them is optimized for speed
Split : vb6 equivalent function , splits the input string using sub-string as delimiter, consecutive delimiters produce empty elements
Split_noempty : variation of Split, but the empty elements are not collected
Split_any : splits the input string using every char as char delimiter, consecutive char delimiters produce empty elements
Split_tok : variation of Split_any, but the empty elements are not collected, using strtok from c lib , not thread-safe but very fast
Split_tok_r : works as Split_tok , but thread-safe and bit slower
some of them do not need any include, but some need
#include "crt/string.bi" 'needed for strtok/memchr functions
i've put the indication on the source
source code and test comparaison
Code: Select all
'====================================================================================
'splits TEXT using DELIMIT as sub-string delimiter,
'1 splitted element for each delimiter found or +1 if delimiter does not finish TEXT
' elements can be empty
' vb6 Split equivalent function
'no include needed and thread-safe
'uses array on the heap, so no problem of capacity with the number of elements < 2147483647
Private Function Split(byref TEXT As String , byref DELIMIT As String , RET() As String) As long
Dim As long DMAX = 0
Dim As long I1
Dim As long I2
Dim As ulong size = 1
Dim As ZString Ptr p
Dim As ZString Ptr p1
Dim As ZString Ptr p2
Dim As ZString Ptr ptemp
Dim As long LDelimit = Len(DELIMIT)
Dim As long LT = Len(TEXT)
Dim As uinteger ptr Posi
Dim As uinteger ptr Posi2
If LT = 0 or LDelimit = 0 Or LDelimit > LT Then
ReDim RET(1 to 1)
RET(1) = TEXT ' copy the full TEXT
Return 1
End If
ptemp = allocate(Len(TEXT) + 1)
If ptemp = 0 Then Print "Error Allocating p": End
*ptemp = TEXT ' copy the full text
p1 = ptemp
p = ptemp
' counts the number of element and stores the position on the string
Do While *p
I2 = 0
If p[0] = DELIMIT[0] Then
p2 = p
If LDelimit > 1 Then ' more than 1 character as delimiter
For I1 = 1 To LDelimit - 1
I2 = 0
If p[1] <> DELIMIT[I1] Then Exit For
p += 1
I2 = 1
Next
Else
I2 = 1
End If
If I2 = 1 Then
DMAX += 1
if DMAX = 1 THEN
Posi = allocate((size + 1) * sizeof(uinteger))
If Posi = 0 Then Print "Error Allocating Posi": End
elseif DMAX > size THEN ' need more space to store the position
size *= 2
Posi2 = reallocate(Posi, (size + 1) * sizeof(uinteger))
If Posi2 = 0 Then Print "Error ReAllocating Posi": End
Posi = Posi2
END IF
Posi[DMAX - 1] = cast(uinteger, p1)
*p2 = 0 ' put null byte to finish the string element
p1 = p2 + LDelimit
End If
End If
p += 1
Loop
if p1 < p THEN
DMAX += 1
if DMAX = 1 THEN
size = 1
Posi = allocate( sizeof(uinteger) )
If Posi = 0 Then Print "Error Allocating Posi": End
end if
Posi[DMAX - 1] = cast(uinteger, p1)
END IF
' dimention the array
ReDim RET(1 To DMAX)
' step through the posi array, setting pointers for each element
For I1 = 1 To DMAX
RET(I1) = *cptr(zstring ptr, Posi[I1 - 1]) ' get element, by retrieving the pointer from Posi array
Next
Deallocate (ptemp)
Deallocate (Posi)
Return DMAX
End Function
'====================================================================================
'====================================================================================
'splits TEXT using DELIMIT as sub-string delimiter, almost as Split,
' but only the non-empty elements will be on the RET array
'no include needed and thread-safe
'uses array on the heap, so no problem of capacity with the number of elements < 2147483647
Private Function Split_noempty(byref TEXT As String , byref DELIMIT As String , RET() As String) As long
Dim As long DMAX = 0
Dim As long I1
Dim As long I2
Dim As ulong size = 1
Dim As ZString Ptr p
Dim As ZString Ptr p1
Dim As ZString Ptr p2
Dim As ZString Ptr ptemp
Dim As long LDelimit = Len(DELIMIT)
Dim As long LT = Len(TEXT)
Dim As uinteger ptr Posi
Dim As uinteger ptr Posi2
If LT = 0 Then Return 0
if size < 1 THEN size = 1
If LDelimit = 0 Or LDelimit > LT Then
ReDim RET(1 to 1)
RET(1) = TEXT ' copy the full TEXT
Return 1
End If
ptemp = allocate(Len(TEXT) + 1)
If ptemp = 0 Then Print "Error Allocating p": End
*ptemp = TEXT ' copy the full text
p1 = ptemp
p = ptemp
' counts the number of element and stores the position on the string
Do While *p
I2 = 0
If p[0] = DELIMIT[0] Then
p2 = p
If LDelimit > 1 Then ' more than 1 character as delimiter
For I1 = 1 To LDelimit - 1
I2 = 0
If p[1] <> DELIMIT[I1] Then Exit For
p += 1
I2 = 1
Next
Else
I2 = 1
End If
If I2 = 1 Then
*p2 = 0 ' put null byte to finish the string element
if *p1 <> "" then
DMAX += 1
if DMAX = 1 THEN
Posi = allocate((size + 1) * sizeof(uinteger))
If Posi = 0 Then Print "Error Allocating Posi": End
elseif DMAX > size THEN ' need more space to store the position
size *= 2
Posi2 = reallocate(Posi, (size + 1) * sizeof(uinteger))
If Posi2 = 0 Then Print "Error ReAllocating Posi": End
Posi = Posi2
END IF
Posi[DMAX - 1] = cast(uinteger, p1)
end if
p1 = p2 + LDelimit
End If
End If
p += 1
Loop
if p1 < p and *p1 <> "" THEN
DMAX += 1
if DMAX = 1 THEN
size = 1
Posi = allocate( sizeof(uinteger) )
If Posi = 0 Then Print "Error Allocating Posi": End
end if
Posi[DMAX - 1] = cast(uinteger, p1)
END IF
' dimention the array
ReDim RET(1 To DMAX)
' step through the posi array, setting pointers for each element
For I1 = 1 To DMAX
RET(I1) = *cptr(zstring ptr, Posi[I1 - 1]) ' get element, by retrieving the pointer from Posi array
Next
Deallocate (ptemp)
Deallocate (Posi)
Return DMAX
End Function
'====================================================================================
'====================================================================================
#include "crt/string.bi" 'needed for memchr function
'splits TEXT using every char of DELIMIT as delimiter,
'1 splitted element for each char delimiter found or +1 if char delimiter does not finish TEXT
' elements can be empty
'uses array on the heap, so no problem of capacity with the number of elements < 2147483647
Private Function Split_any(byref TEXT As String , byref DELIMIT As String , RET() As String) As long
Dim As long DMAX = 0
Dim As long I1
Dim As long I2
Dim As long L1 = len(TEXT)
Dim As long L2 = Len(DELIMIT)
Dim As ZString Ptr ptemp
Dim As byte Ptr p
Dim As byte Ptr p1
Dim As byte Ptr p2
Dim As ulong size = 1
Dim As uinteger ptr Posi
Dim As uinteger ptr Posi2
Dim as zstring ptr pdelim = strptr(DELIMIT)
If L2 = 0 or L1 = 0 Then
ReDim RET(1 to 1)
RET(1) = TEXT ' copy the full TEXT
Return 1
End If
ptemp = allocate(L1 + 1)
*ptemp = TEXT ' copy the full TEXT
p1 = ptemp
p = ptemp
' counts the number of element and stores the position on the string
Do While p[0]
I2 = 0
p2 = pdelim
if l2 = 1 THEN
if p2[0] = p[0] then I2 = 1
else
if memchr(pdelim, p[0], L2)<> 0 THEN I2 = 1
END IF
If I2 = 1 Then
DMAX += 1
if DMAX = 1 THEN
Posi = allocate( (size + 1) * sizeof(uinteger) )
If Posi = 0 Then Print "Error Allocating Posi": End
elseif DMAX > size THEN ' need more space to store the position
size *= 2
Posi2 = reallocate( Posi, (size + 1) * sizeof(uinteger))
If Posi2 = 0 Then Print "Error ReAllocating Posi": End
Posi = Posi2
END IF
Posi[DMAX - 1] = cast(uinteger, p1)
*p = 0 ' put null byte to finish the string element
p1 = p + 1
End If
p += 1
Loop
if p1 < p THEN
DMAX += 1
if DMAX = 1 THEN
size = 1
Posi = allocate( sizeof(uinteger) )
If Posi = 0 Then Print "Error Allocating Posi": End
end if
Posi[DMAX - 1] = cast(uinteger, p1)
END IF
' dimention the array
ReDim RET(1 To DMAX)
For I1 = 1 To DMAX
RET(I1) = *cptr(zstring ptr, Posi[I1 - 1]) ' get element, by retrieving the pointer from Posi array
Next
Deallocate (ptemp)
Deallocate (Posi)
Return DMAX
End Function
'====================================================================================
#include "crt/string.bi" 'needed for strtok function
'splits TEXT using every char of DELIMIT as delimiter,
'works as Split_any but only the non-empty elements will be on the RET array
'
'warning not thread-safe using strtok(s1 as zstring ptr, s2 as zstring ptr) as zstring ptr only 1 loop
Private Function Split_tok(byref TEXT As String , byref DELIMIT As String , RET() As String) As Long
Dim As Long ctr
dim as zstring ptr p, p1, p2
dim as ulong i1 = 1 'initial value to redim RET
if(len(TEXT) = 0 or len(DELIMIT) = 0) then
Redim RET(1 To 1)
RET(1) = TEXT
return 1
end if
dim as string s1 = TEXT 'copy TEXT to avoid alteration of the input
p = strptr(s1) 'get pointers
p2= strptr(DELIMIT)
p1 = strtok(p , p2)
while p1 <> 0
ctr +=1
if ctr = 1 THEN
Redim RET(1 To i1) 'redim the array first time
elseif ctr > i1 then
i1 *= 2
redim preserve RET(1 To i1) 'redim the array to give more space
END IF
RET(ctr)= *p1
p1 = strtok(0, p2)
WEND
If ctr = 0 Then Return 0
'redim the array to fit exact need
if i1 > ctr THEN Redim preserve RET(1 To ctr)
Return ctr
End Function
'====================================================================================
'====================================================================================
#include "crt/string.bi" 'needed for memchr function
'helper function for String_tok_r
Private Function h_tok_r(byref ps1 as byte ptr, byref ps2 as byte ptr, byval i as long )as byte ptr
dim p as byte ptr
if i > 1 THEN
while memchr(ps2, *ps1 , i) <> 0
ps1 += 1
wend
else
while *ps1 = *ps2
ps1 += 1
wend
END IF
if *ps1 = 0 THEN
ps1 = 0
return 0
END IF
p = ps1
if i > 1 THEN
while memchr(ps2, *ps1 , i) = 0
ps1 += 1
if *ps1 = 0 then exit while
wend
else
while *ps1 <> *ps2
ps1 += 1
if *ps1 = 0 then exit while
wend
end if
if *ps1 <> 0 THEN
*ps1 = 0
ps1 += 1
end if
return p
end function
'------------------------------------------------------------------------------------
'#include "crt/string.bi" 'needed for memchr function
'thread-safe alternative to Split_tok
'splits TEXT using every char of DELIMIT as delimiter,
'works as Split_any but only the non-empty elements will be on the RET array
Private Function Split_tok_r(byref TEXT As String , byref DELIMIT As String , RET() As String) As Long
Dim ptemp As Zstring Ptr
Dim ctr As Long
Dim size1 As uLong
dim as long ilen = len(DELIMIT)
if(len(TEXT) = 0 or ilen = 0) then
Redim RET(1 To 1)
RET(1) = TEXT
return 1
end if
dim as string s1 = TEXT 'copy TEXT to avoid alteration of the input
dim As byte Ptr p = strptr(s1) 'get pointers
dim As byte Ptr d = strptr(DELIMIT)
ptemp = p
while ptemp <> 0
ptemp = h_tok_r(p, d, ilen) ' p is sent byref and it will be modified on the f_strtok_r function
if ptemp <> 0 THEN
ctr += 1
if ctr = 1 THEN
size1 = 1
Redim RET(1 To size1) 'redim the array
elseif ctr > size1 then
size1 *= 2
redim preserve RET(1 To size1) 'redim the array
END IF
RET(ctr)= *ptemp
end if
WEND
If ctr = 0 Then Return 0
if size1 > ctr then Redim preserve RET(1 To ctr) 'redim the array
Return ctr
end function
'====================================================================================
'====================================================================================
'tests
'====================================================================================
'####################################################################################
dim as string s1 = "fortests fforforfor verifications how it is f"
dim as string s2 = "for"
'####################################################################################
dim as long icount
dim as long x
dim as string a()
icount = Split(s1 , s2 , a())
'====================================================================================
print "Split"
print "initial = >" & s1 & "<" & chr(10) & "delim = >"& s2 & "<" & chr(10)
If LBound( a ) > UBound( a ) Then
Print "array is empty"
Else
if ubound(a) <> - 1 then
print "icount = " & icount & " lbound = " & lbound(a) & " ubound = " & ubound(a)
for x = lbound(a) to icount
print " x= " & x , ">"& a(x)&"<"
NEXT
erase(a)
end if
End If
Print : print
icount = Split_noempty(s1 , s2 , a())
'====================================================================================
print "Split_noempty"
print "initial = >" & s1 & "<" & chr(10) & "delim = >"& s2 & "<" & chr(10)
If LBound( a ) > UBound( a ) Then
Print "array is empty"
Else
if ubound(a) <> - 1 then
print "icount = " & icount & " lbound = " & lbound(a) & " ubound = " & ubound(a)
for x = lbound(a) to icount
print " x= " & x , ">"& a(x)&"<"
NEXT
erase(a)
end if
End If
Print : print
icount = Split_any(s1 , s2 , a())
'====================================================================================
print "Split_any"
print "initial = >" & s1 & "<" & chr(10) & "delim = >"& s2 & "<" & chr(10)
If LBound( a ) > UBound( a ) Then
Print "array is empty"
Else
if ubound(a) <> - 1 then
print "icount = " & icount & " lbound = " & lbound(a) & " ubound = " & ubound(a)
for x = lbound(a) to icount
print " x= " & x , ">"& a(x)&"<"
NEXT
erase(a)
end if
End If
Print : print
icount = Split_tok(s1 , s2 , a())
'====================================================================================
print "Split_tok"
print "initial = >" & s1 & "<" & chr(10) & "delim = >"& s2 & "<" & chr(10)
If LBound( a ) > UBound( a ) Then
Print "array is empty"
Else
if ubound(a) <> - 1 then
print "icount = " & icount & " lbound = " & lbound(a) & " ubound = " & ubound(a)
for x = lbound(a) to icount
print " x= " & x , ">"& a(x)&"<"
NEXT
erase(a)
end if
End If
Print : print
icount = Split_tok_r(s1 , s2 , a())
'====================================================================================
print "Split_tok_r"
print "initial = >" & s1 & "<" & chr(10) & "delim = >"& s2 & "<" & chr(10)
If LBound( a ) > UBound( a ) Then
Print "array is empty"
Else
if ubound(a) <> - 1 then
print "icount = " & icount & " lbound = " & lbound(a) & " ubound = " & ubound(a)
for x = lbound(a) to icount
print " x= " & x , ">"& a(x)&"<"
NEXT
erase(a)
end if
End If
Print : print
sleepwaiting for comments/remarks or faster solutions
-
Tourist Trap
- Posts: 2958
- Joined: Jun 02, 2015 16:24
Re: I need someone to defeat my string splitting algo
Thanks, that looks like the ultimate splitter collection.marpon wrote:i've collected from my own archives the different versions of split functions
each of them is optimized for speed
feel free to test and use...
waiting for comments/remarks or faster solutions
Re: I need someone to defeat my string splitting algo
I use V1ctors split and it works for me well
I tested yours on long string and it crashes
I tested yours on long string and it crashes
Code: Select all
dim as string txt
dim as integer i
open "data1.txt" for input as #1
line input #1, txt
close #1
'test v1ctor
'split_v1ctor(txt, ":", a())
'for i = 0 to ubound(a)-1
'? a (i)
'next i
split(txt, ":", a())
for i = 0 to ubound(a)-1
? a (i)
next i
Code: Select all
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0 D:3172.753000000001 6684.203 0 3166.104000000001 6698.984 0 D:3166.104000000001 6698.984 0 3159.191 6715.000999999999 0 D:3159.191 6715.000999999999 0 3152.152 6731.912 0 D:3152.152 6731.912 0 3145.125 6749.375 0 D:3145.125 6749.375 0 3138.248 6767.048 0 D:3138.248 6767.048 0 3131.659 6784.589 0 D:3131.659 6784.589 0 3125.496 6801.656 0 D:3125.496 6801.656 0 3119.897 6817.907 0 D:3119.897 6817.907 0 3115 6833 0 D:3115 6833 0 3109.004000000001 6852.787000000001 0 D:3109.004000000001 6852.787000000001 0 3103.512000000001 6869.736000000002 0 D:3103.512000000001 6869.736000000002 0 3098.368 6884.129 0 D:3098.368 6884.129 0 3093.416 6896.248000000001 0 D:3093.416 6896.248000000001 0 3088.5 6906.375 0 D:3088.5 6906.375 0 3083.464 6914.792 0 D:3083.464 6914.792 0 3078.152 6921.781 0 D:3078.152 6921.781 0 3072.408 6927.624000000001 0 D:3072.408 6927.624000000001 0 3066.076 6932.603 0 D:3066.076 6932.603 0 3059 6937 0 D:3059 6937 0 3054.300000000001 6939.779000000001 0 D:3054.300000000001 6939.779000000001 0 3049.840000000001 6942.672000000001 0 D:3049.840000000001 6942.672000000001 0 3045.68 6945.612999999999 0 D:3045.68 6945.612999999999 0 3041.880000000001 6948.536000000001 0 D:3041.880000000001 6948.536000000001 0 3038.5 6951.375 0 D:3038.5 6951.375 0 3035.6 6954.063999999999 0 D:3035.6 6954.063999999999 0 3033.24 6956.536999999999 0 D:3033.24 6956.536999999999 0 3031.48 6958.728 0 D:3031.48 6958.728 0 3030.38 6960.571 0 D:3030.38 6960.571 0 3030 6962 0 D:3030 6962 0 3029.503000000001 6963.196000000001 0 D:3029.503000000001 6963.196000000001 0 3028.064000000001 6964.368000000001 0 D:3028.064000000001 6964.368000000001 0 3025.761 6965.492 0 D:3025.761 6965.492 0 3022.672 6966.544000000001 0 D:3022.672 6966.544000000001 0 3018.875 6967.5 0 D:3018.875 6967.5 0 3014.448 6968.335999999999 0 D:3014.448 6968.335999999999 0 3009.469 6969.027999999999 0 D:3009.469 6969.027999999999 0 3004.016 6969.552 0 D:3004.016 6969.552 0 2998.167 6969.884 0 D:2998.167 6969.884 0 2992 6970 0 D:2992 6970 0 2982.572000000001 6969.854000000001 0 D:2982.572000000001 6969.854000000001 0 2974.536000000001 6969.312000000002 0 D:2974.536000000001 6969.312000000002 0 2967.664 6968.218 0 D:2967.664 6968.218 0 2961.728 6966.416000000001 0 D:2961.728 6966.416000000001 0 2956.5 6963.75 0 D:2956.5 6963.75 0 2951.752 6960.063999999999 0 D:2951.752 6960.063999999999 0 2947.256 6955.201999999999 0 D:2947.256 6955.201999999999 0 2942.784 6949.008000000001 0 D:2942.784 6949.008000000001 0 2938.108 6941.326 0 D:2938.108 6941.326 0 2933 6932 0 D:Re: I need someone to defeat my string splitting algo
@ike
you probably made a mistake , my split function returns an array with lbound = 1 , that means the first element is a(1) not a(0)
according the code you let, i suppose the split_v1ctor is returning and array with lbound = 0
please check with following code
you will see it works without any problem
if you want to have lbound = 0 change all the redim's in my split function...
but at the end, if you are happy with split_v1ctor , why to change, mine only tries to be the fastest, i did not compare with split_v1ctor
could you post the code?
you probably made a mistake , my split function returns an array with lbound = 1 , that means the first element is a(1) not a(0)
according the code you let, i suppose the split_v1ctor is returning and array with lbound = 0
please check with following code
Code: Select all
dim as string txt
dim as integer i
open "data1.txt" for input as #1
line input #1, txt
close #1
print "len = " & len(txt)
dim as string a()
i = Split(txt, ":", a())
print "nb elements = " & i
print "lbound = " & lbound(a)
print "ubound = " & ubound(a)
? " press key to continue"
sleep
for i = lbound(a) to ubound(a)
? a (i)
next i
? " press key to finish"
sleepif you want to have lbound = 0 change all the redim's in my split function...
but at the end, if you are happy with split_v1ctor , why to change, mine only tries to be the fastest, i did not compare with split_v1ctor
could you post the code?