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2

This function was working in the production system and I'm not aware of changing anything (MikTeX did some package updates which may have had an effect).

A very much stripped down MWE is

documentclassarticle

directlua
 function underscore(s) 
 s = tostring(s+2080) 
 tex.print(s) 
 end 


DeclareRobustCommandBinl[1]directluaunderscore([[#1]])% 

begindocument

Use MBinl1 or MBinl2

enddocument 

I used to get M_1 and M_2 with proper subscripts, but now the correct unicode characters U+2081 and U+2082 are being generated but have acquired a trailing ".0" as can be seen in the output. N.B. I've omitted prepending "char"" to s before passing the string back.

Output :

Use M2081.0 or M2082.0

luatex

share|improve this question

edited 8 hours ago

DLyons

asked 8 hours ago

DLyonsDLyons

1956

add a comment | 

2

This function was working in the production system and I'm not aware of changing anything (MikTeX did some package updates which may have had an effect).

A very much stripped down MWE is

documentclassarticle

directlua
 function underscore(s) 
 s = tostring(s+2080) 
 tex.print(s) 
 end 


DeclareRobustCommandBinl[1]directluaunderscore([[#1]])% 

begindocument

Use MBinl1 or MBinl2

enddocument 

I used to get M_1 and M_2 with proper subscripts, but now the correct unicode characters U+2081 and U+2082 are being generated but have acquired a trailing ".0" as can be seen in the output. N.B. I've omitted prepending "char"" to s before passing the string back.

Output :

Use M2081.0 or M2082.0

luatex

share|improve this question

edited 8 hours ago

DLyons

asked 8 hours ago

DLyonsDLyons

1956

add a comment | 

This function was working in the production system and I'm not aware of changing anything (MikTeX did some package updates which may have had an effect).

A very much stripped down MWE is

documentclassarticle

directlua
 function underscore(s) 
 s = tostring(s+2080) 
 tex.print(s) 
 end 


DeclareRobustCommandBinl[1]directluaunderscore([[#1]])% 

begindocument

Use MBinl1 or MBinl2

enddocument 

I used to get M_1 and M_2 with proper subscripts, but now the correct unicode characters U+2081 and U+2082 are being generated but have acquired a trailing ".0" as can be seen in the output. N.B. I've omitted prepending "char"" to s before passing the string back.

Output :

Use M2081.0 or M2082.0

luatex

share|improve this question

edited 8 hours ago

DLyons

asked 8 hours ago

DLyonsDLyons

1956

documentclassarticle

directlua
 function underscore(s) 
 s = tostring(s+2080) 
 tex.print(s) 
 end 


DeclareRobustCommandBinl[1]directluaunderscore([[#1]])% 

begindocument

Use MBinl1 or MBinl2

enddocument 

Use M2081.0 or M2082.0

share|improve this question

edited 8 hours ago

DLyons

asked 8 hours ago

DLyonsDLyons

1956

share|improve this question

edited 8 hours ago

DLyons

asked 8 hours ago

DLyonsDLyons

1956

asked 8 hours ago

DLyonsDLyons

1956

asked 8 hours ago

DLyonsDLyons

1956

2 Answers
2

active

oldest

votes

4

That's due to the change from lua 5.2 to lua 5.3.

You can force the number to be an integer e.g. with math.floor:

documentclassarticle

directlua
 function underscore(s)
 s = tostring(math.floor(s+2080))
 tex.print(s)
 end


DeclareRobustCommandBinl[1]directluaunderscore([[#1]])%

begindocument

Use MBinl1 or MBinl2

enddocument 

share|improve this answer

answered 8 hours ago

Ulrike FischerUlrike Fischer

205k9313705

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks @Ulrike Fischer. That fixed it.
  
  – DLyons
  8 hours ago
  

  
  
  
  

add a comment | 

2

s = tostring(math.floor(s+2080)) 

share|improve this answer

answered 8 hours ago

Red-CloudRed-Cloud

3,803215

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks Red-Cloud. Millisecs behind Ulrike.
  
  – DLyons
  8 hours ago
  

  
  
  
  

add a comment | 

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4

That's due to the change from lua 5.2 to lua 5.3.

You can force the number to be an integer e.g. with math.floor:

documentclassarticle

directlua
 function underscore(s)
 s = tostring(math.floor(s+2080))
 tex.print(s)
 end


DeclareRobustCommandBinl[1]directluaunderscore([[#1]])%

begindocument

Use MBinl1 or MBinl2

enddocument 

share|improve this answer

answered 8 hours ago

Ulrike FischerUlrike Fischer

205k9313705

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks @Ulrike Fischer. That fixed it.
  
  – DLyons
  8 hours ago
  

  
  
  
  

add a comment | 

4

That's due to the change from lua 5.2 to lua 5.3.

You can force the number to be an integer e.g. with math.floor:

documentclassarticle

directlua
 function underscore(s)
 s = tostring(math.floor(s+2080))
 tex.print(s)
 end


DeclareRobustCommandBinl[1]directluaunderscore([[#1]])%

begindocument

Use MBinl1 or MBinl2

enddocument 

share|improve this answer

answered 8 hours ago

Ulrike FischerUlrike Fischer

205k9313705

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks @Ulrike Fischer. That fixed it.
  
  – DLyons
  8 hours ago
  

  
  
  
  

add a comment | 

That's due to the change from lua 5.2 to lua 5.3.

You can force the number to be an integer e.g. with math.floor:

documentclassarticle

directlua
 function underscore(s)
 s = tostring(math.floor(s+2080))
 tex.print(s)
 end


DeclareRobustCommandBinl[1]directluaunderscore([[#1]])%

begindocument

Use MBinl1 or MBinl2

enddocument 

share|improve this answer

answered 8 hours ago

Ulrike FischerUlrike Fischer

205k9313705

documentclassarticle

directlua
 function underscore(s)
 s = tostring(math.floor(s+2080))
 tex.print(s)
 end


DeclareRobustCommandBinl[1]directluaunderscore([[#1]])%

begindocument

Use MBinl1 or MBinl2

enddocument 

share|improve this answer

answered 8 hours ago

Ulrike FischerUlrike Fischer

205k9313705

answered 8 hours ago

Ulrike FischerUlrike Fischer

205k9313705

answered 8 hours ago

Ulrike FischerUlrike Fischer

205k9313705

2

s = tostring(math.floor(s+2080)) 

share|improve this answer

answered 8 hours ago

Red-CloudRed-Cloud

3,803215

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks Red-Cloud. Millisecs behind Ulrike.
  
  – DLyons
  8 hours ago
  

  
  
  
  

add a comment | 

2

s = tostring(math.floor(s+2080)) 

share|improve this answer

answered 8 hours ago

Red-CloudRed-Cloud

3,803215

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks Red-Cloud. Millisecs behind Ulrike.
  
  – DLyons
  8 hours ago
  

  
  
  
  

add a comment | 

s = tostring(math.floor(s+2080)) 

share|improve this answer

answered 8 hours ago

Red-CloudRed-Cloud

3,803215

s = tostring(math.floor(s+2080)) 

share|improve this answer

answered 8 hours ago

Red-CloudRed-Cloud

3,803215

answered 8 hours ago

Red-CloudRed-Cloud

3,803215

answered 8 hours ago

Red-CloudRed-Cloud

3,803215