Subtleties when using def and ifnum conditionals

Clash Royale CLAN TAG#URR8PPP

up vote
2
down vote

favorite

1

I am having some difficultly comparing two numbers passed as arguments using ifnum. Consider the command

defcomparenum[#1,#2]

 defx#1
 defy#2
 ifnumx=y
 TRUE
 else
 FALSE
 fi

comparenum[1,1] gives TRUE as I would expect. However, if I replace the above with

ifnumx=y
 1
else
 0
fi

The same call comparenum[1,1] produces 0 (replacing y with y, or indeed #2 seems to fix this). What is the reason for this behaviour? I would like to use the second construction so that I can nest the statement inside another conditional as in the top answer to How to form “if … or … then” conditionals in TeX?

tex-core conditionals

share|improve this question

edited 16 mins ago

asked 1 hour ago

Pippip19

132

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• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You may want to look into etoolbox and its ifnumequal test. Together with ifboolexpr you can easily nest stuff. But you can also look into expl3 and its vast array of tests.
  – moewe
  53 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your first code produces TRUE but does not produce FALSE if you call comparenum[1,2].
  – Sigur
  52 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sigur I just realised the same. defx#1 defy#1 could explain that ;-) But I'm wondering why the def's are needed at all... ifnum#1=#2 should do more or less the same.
  – moewe
  51 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you want to be able to use 1 and 0 as "return" values you should try ifnumx=yrelax
  – moewe
  50 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @moewe, ohhhhh, I missed that!!! Good!
  – Sigur
  50 mins ago
  

  
  
  
  

 | 
show 3 more comments

up vote
2
down vote

favorite

1

I am having some difficultly comparing two numbers passed as arguments using ifnum. Consider the command

defcomparenum[#1,#2]

 defx#1
 defy#2
 ifnumx=y
 TRUE
 else
 FALSE
 fi

comparenum[1,1] gives TRUE as I would expect. However, if I replace the above with

ifnumx=y
 1
else
 0
fi

The same call comparenum[1,1] produces 0 (replacing y with y, or indeed #2 seems to fix this). What is the reason for this behaviour? I would like to use the second construction so that I can nest the statement inside another conditional as in the top answer to How to form “if … or … then” conditionals in TeX?

tex-core conditionals

share|improve this question

edited 16 mins ago

asked 1 hour ago

Pippip19

132

New contributor

Pippip19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You may want to look into etoolbox and its ifnumequal test. Together with ifboolexpr you can easily nest stuff. But you can also look into expl3 and its vast array of tests.
  – moewe
  53 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your first code produces TRUE but does not produce FALSE if you call comparenum[1,2].
  – Sigur
  52 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sigur I just realised the same. defx#1 defy#1 could explain that ;-) But I'm wondering why the def's are needed at all... ifnum#1=#2 should do more or less the same.
  – moewe
  51 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you want to be able to use 1 and 0 as "return" values you should try ifnumx=yrelax
  – moewe
  50 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @moewe, ohhhhh, I missed that!!! Good!
  – Sigur
  50 mins ago
  

  
  
  
  

 | 
show 3 more comments

up vote
2
down vote

favorite

1

up vote
2
down vote

favorite

1

1

I am having some difficultly comparing two numbers passed as arguments using ifnum. Consider the command

defcomparenum[#1,#2]

 defx#1
 defy#2
 ifnumx=y
 TRUE
 else
 FALSE
 fi

comparenum[1,1] gives TRUE as I would expect. However, if I replace the above with

ifnumx=y
 1
else
 0
fi

The same call comparenum[1,1] produces 0 (replacing y with y, or indeed #2 seems to fix this). What is the reason for this behaviour? I would like to use the second construction so that I can nest the statement inside another conditional as in the top answer to How to form “if … or … then” conditionals in TeX?

tex-core conditionals

share|improve this question

edited 16 mins ago

asked 1 hour ago

Pippip19

132

New contributor

Pippip19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

I am having some difficultly comparing two numbers passed as arguments using ifnum. Consider the command

defcomparenum[#1,#2]

 defx#1
 defy#2
 ifnumx=y
 TRUE
 else
 FALSE
 fi

comparenum[1,1] gives TRUE as I would expect. However, if I replace the above with

ifnumx=y
 1
else
 0
fi

The same call comparenum[1,1] produces 0 (replacing y with y, or indeed #2 seems to fix this). What is the reason for this behaviour? I would like to use the second construction so that I can nest the statement inside another conditional as in the top answer to How to form “if … or … then” conditionals in TeX?

tex-core conditionals

tex-core conditionals

share|improve this question

edited 16 mins ago

asked 1 hour ago

Pippip19

132

New contributor

Pippip19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 16 mins ago

asked 1 hour ago

Pippip19

132

New contributor

Pippip19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

share|improve this question

edited 16 mins ago

edited 16 mins ago

edited 16 mins ago

asked 1 hour ago

Pippip19

132

New contributor

Pippip19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 1 hour ago

Pippip19

132

asked 1 hour ago

Pippip19

132

132

New contributor

Pippip19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

Pippip19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Pippip19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You may want to look into etoolbox and its ifnumequal test. Together with ifboolexpr you can easily nest stuff. But you can also look into expl3 and its vast array of tests.
  – moewe
  53 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your first code produces TRUE but does not produce FALSE if you call comparenum[1,2].
  – Sigur
  52 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sigur I just realised the same. defx#1 defy#1 could explain that ;-) But I'm wondering why the def's are needed at all... ifnum#1=#2 should do more or less the same.
  – moewe
  51 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you want to be able to use 1 and 0 as "return" values you should try ifnumx=yrelax
  – moewe
  50 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @moewe, ohhhhh, I missed that!!! Good!
  – Sigur
  50 mins ago
  

  
  
  
  

 | 
show 3 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You may want to look into etoolbox and its ifnumequal test. Together with ifboolexpr you can easily nest stuff. But you can also look into expl3 and its vast array of tests.
  – moewe
  53 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your first code produces TRUE but does not produce FALSE if you call comparenum[1,2].
  – Sigur
  52 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sigur I just realised the same. defx#1 defy#1 could explain that ;-) But I'm wondering why the def's are needed at all... ifnum#1=#2 should do more or less the same.
  – moewe
  51 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you want to be able to use 1 and 0 as "return" values you should try ifnumx=yrelax
  – moewe
  50 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @moewe, ohhhhh, I missed that!!! Good!
  – Sigur
  50 mins ago
  

  
  
  
  

1

1

You may want to look into etoolbox and its ifnumequal test. Together with ifboolexpr you can easily nest stuff. But you can also look into expl3 and its vast array of tests.
– moewe
53 mins ago

You may want to look into etoolbox and its ifnumequal test. Together with ifboolexpr you can easily nest stuff. But you can also look into expl3 and its vast array of tests.
– moewe
53 mins ago

Your first code produces TRUE but does not produce FALSE if you call comparenum[1,2].
– Sigur
52 mins ago

Your first code produces TRUE but does not produce FALSE if you call comparenum[1,2].
– Sigur
52 mins ago

@Sigur I just realised the same. defx#1 defy#1 could explain that ;-) But I'm wondering why the def's are needed at all... ifnum#1=#2 should do more or less the same.
– moewe
51 mins ago

@Sigur I just realised the same. defx#1 defy#1 could explain that ;-) But I'm wondering why the def's are needed at all... ifnum#1=#2 should do more or less the same.
– moewe
51 mins ago

If you want to be able to use 1 and 0 as "return" values you should try ifnumx=yrelax
– moewe
50 mins ago

If you want to be able to use 1 and 0 as "return" values you should try ifnumx=yrelax
– moewe
50 mins ago

@moewe, ohhhhh, I missed that!!! Good!
– Sigur
50 mins ago

@moewe, ohhhhh, I missed that!!! Good!
– Sigur
50 mins ago

 | 
show 3 more comments

1 Answer
1

active

oldest

votes

up vote
2
down vote



accepted

Your second code is equivalent to

defcomparenum[#1,#2] defx#1 defy#1 ifnumx=y1 else0 fi

(because spaces and end-of-lines are ignored after a control word). Now you should be able to see the main problem with the code. When TeX evaluates the conditional, it needs two numbers and does full expansion until finding tokens that cannot be interpreted as digits.

So it expands x and = stops the search for digits while also starting the lookup for the next number; y is expanded and 1 follows. So the call comparenum[1,1] translates into

ifnum1=11 else0 fi

which of course returns false.

You solve the issue with

defcomparenum[#1,#2]%
 defx#1%
 defy#1%
 ifnumx=yrelax
 1% 
 else
 0%
 fi

The relax token stops the lookup for digits.

On the other hand, this construct is not expandable. You can make an expandable version with

defcomparenum[#1,#2]%
 ifnum#1=number#2spacespace
 1%
 else
 0%
 fi

The first space expands to a space token that stops number from looking up for more digits and is then ignored by rule; the second space stops the search for digit related to ifnum and is again ignored.

An e-TeX version would be

defcomparenum[#1,#2]%
 ifnum#1=numexpr#2relax
 1%
 else
 0%
 fi

Please, note the % that protect the end-of-lines avoiding that they make spaces in the output.

share|improve this answer

answered 19 mins ago

egreg

691k8518383086

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1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Your second code is equivalent to

defcomparenum[#1,#2] defx#1 defy#1 ifnumx=y1 else0 fi

(because spaces and end-of-lines are ignored after a control word). Now you should be able to see the main problem with the code. When TeX evaluates the conditional, it needs two numbers and does full expansion until finding tokens that cannot be interpreted as digits.

So it expands x and = stops the search for digits while also starting the lookup for the next number; y is expanded and 1 follows. So the call comparenum[1,1] translates into

ifnum1=11 else0 fi

which of course returns false.

You solve the issue with

defcomparenum[#1,#2]%
 defx#1%
 defy#1%
 ifnumx=yrelax
 1% 
 else
 0%
 fi

The relax token stops the lookup for digits.

On the other hand, this construct is not expandable. You can make an expandable version with

defcomparenum[#1,#2]%
 ifnum#1=number#2spacespace
 1%
 else
 0%
 fi

The first space expands to a space token that stops number from looking up for more digits and is then ignored by rule; the second space stops the search for digit related to ifnum and is again ignored.

An e-TeX version would be

defcomparenum[#1,#2]%
 ifnum#1=numexpr#2relax
 1%
 else
 0%
 fi

Please, note the % that protect the end-of-lines avoiding that they make spaces in the output.

share|improve this answer

answered 19 mins ago

egreg

691k8518383086

add a comment | 

up vote
2
down vote



accepted

Your second code is equivalent to

defcomparenum[#1,#2] defx#1 defy#1 ifnumx=y1 else0 fi

(because spaces and end-of-lines are ignored after a control word). Now you should be able to see the main problem with the code. When TeX evaluates the conditional, it needs two numbers and does full expansion until finding tokens that cannot be interpreted as digits.

So it expands x and = stops the search for digits while also starting the lookup for the next number; y is expanded and 1 follows. So the call comparenum[1,1] translates into

ifnum1=11 else0 fi

which of course returns false.

You solve the issue with

defcomparenum[#1,#2]%
 defx#1%
 defy#1%
 ifnumx=yrelax
 1% 
 else
 0%
 fi

The relax token stops the lookup for digits.

On the other hand, this construct is not expandable. You can make an expandable version with

defcomparenum[#1,#2]%
 ifnum#1=number#2spacespace
 1%
 else
 0%
 fi

The first space expands to a space token that stops number from looking up for more digits and is then ignored by rule; the second space stops the search for digit related to ifnum and is again ignored.

An e-TeX version would be

defcomparenum[#1,#2]%
 ifnum#1=numexpr#2relax
 1%
 else
 0%
 fi

Please, note the % that protect the end-of-lines avoiding that they make spaces in the output.

share|improve this answer

answered 19 mins ago

egreg

691k8518383086

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Your second code is equivalent to

defcomparenum[#1,#2] defx#1 defy#1 ifnumx=y1 else0 fi

(because spaces and end-of-lines are ignored after a control word). Now you should be able to see the main problem with the code. When TeX evaluates the conditional, it needs two numbers and does full expansion until finding tokens that cannot be interpreted as digits.

So it expands x and = stops the search for digits while also starting the lookup for the next number; y is expanded and 1 follows. So the call comparenum[1,1] translates into

ifnum1=11 else0 fi

which of course returns false.

You solve the issue with

defcomparenum[#1,#2]%
 defx#1%
 defy#1%
 ifnumx=yrelax
 1% 
 else
 0%
 fi

The relax token stops the lookup for digits.

On the other hand, this construct is not expandable. You can make an expandable version with

defcomparenum[#1,#2]%
 ifnum#1=number#2spacespace
 1%
 else
 0%
 fi

The first space expands to a space token that stops number from looking up for more digits and is then ignored by rule; the second space stops the search for digit related to ifnum and is again ignored.

An e-TeX version would be

defcomparenum[#1,#2]%
 ifnum#1=numexpr#2relax
 1%
 else
 0%
 fi

Please, note the % that protect the end-of-lines avoiding that they make spaces in the output.

share|improve this answer

answered 19 mins ago

egreg

691k8518383086

Your second code is equivalent to

defcomparenum[#1,#2] defx#1 defy#1 ifnumx=y1 else0 fi

(because spaces and end-of-lines are ignored after a control word). Now you should be able to see the main problem with the code. When TeX evaluates the conditional, it needs two numbers and does full expansion until finding tokens that cannot be interpreted as digits.

So it expands x and = stops the search for digits while also starting the lookup for the next number; y is expanded and 1 follows. So the call comparenum[1,1] translates into

ifnum1=11 else0 fi

which of course returns false.

You solve the issue with

defcomparenum[#1,#2]%
 defx#1%
 defy#1%
 ifnumx=yrelax
 1% 
 else
 0%
 fi

The relax token stops the lookup for digits.

On the other hand, this construct is not expandable. You can make an expandable version with

defcomparenum[#1,#2]%
 ifnum#1=number#2spacespace
 1%
 else
 0%
 fi

The first space expands to a space token that stops number from looking up for more digits and is then ignored by rule; the second space stops the search for digit related to ifnum and is again ignored.

An e-TeX version would be

defcomparenum[#1,#2]%
 ifnum#1=numexpr#2relax
 1%
 else
 0%
 fi

Please, note the % that protect the end-of-lines avoiding that they make spaces in the output.

share|improve this answer

answered 19 mins ago

egreg

691k8518383086

share|improve this answer

share|improve this answer

answered 19 mins ago

egreg

691k8518383086

answered 19 mins ago

egreg

691k8518383086

answered 19 mins ago

egreg

691k8518383086

691k8518383086

add a comment | 

add a comment | 

Pippip19 is a new contributor. Be nice, and check out our Code of Conduct.

 

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