What are the values for parameters of an edef at the time of definition?

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Suppose we have the following short document

documentclassarticle

deffoo#1if#1XXelse not Xfi

edefbar#1if#1XXelse not Xfi

begindocument
foo: X = fooX, Y = fooY

bar: X = barX, Y = barY
enddocument

which results in

foo: X = X, Y = not X

bar: X = not X, Y = not X

The first case is quite simple to understand, expansion of foo happens at the time the macro is called, #1 is replaced by X or Y, and the if is executed with either of these two characters.

What is not so clear is the result of bar. I understand that the expansion of if here happens at the time the macro is defined and thus the concrete value of #1 not yet available; TeX leaves "holes" for the parameters in the expanded replacement text to be filled in later when the macro is actually called. But to properly expand the if here, some value for #1 is needed.

The parameter doesn't seem to be just left empty, otherwise bar shouldn't give not X, because if XX would always be true. They also don't seem to be relaxed, as

edefbaz#1ifx#1relax yeselse nofi

always gives no.

Then what values are used for the parameters at the time of definition?

tex-core expansion

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asked 1 hour ago

siracusa

3,6021926

add a comment | 

up vote
4
down vote

favorite

1

Suppose we have the following short document

documentclassarticle

deffoo#1if#1XXelse not Xfi

edefbar#1if#1XXelse not Xfi

begindocument
foo: X = fooX, Y = fooY

bar: X = barX, Y = barY
enddocument

which results in

foo: X = X, Y = not X

bar: X = not X, Y = not X

The first case is quite simple to understand, expansion of foo happens at the time the macro is called, #1 is replaced by X or Y, and the if is executed with either of these two characters.

What is not so clear is the result of bar. I understand that the expansion of if here happens at the time the macro is defined and thus the concrete value of #1 not yet available; TeX leaves "holes" for the parameters in the expanded replacement text to be filled in later when the macro is actually called. But to properly expand the if here, some value for #1 is needed.

The parameter doesn't seem to be just left empty, otherwise bar shouldn't give not X, because if XX would always be true. They also don't seem to be relaxed, as

edefbaz#1ifx#1relax yeselse nofi

always gives no.

Then what values are used for the parameters at the time of definition?

tex-core expansion

share|improve this question

asked 1 hour ago

siracusa

3,6021926

add a comment | 

up vote
4
down vote

favorite

1

up vote
4
down vote

favorite

1

1

Suppose we have the following short document

documentclassarticle

deffoo#1if#1XXelse not Xfi

edefbar#1if#1XXelse not Xfi

begindocument
foo: X = fooX, Y = fooY

bar: X = barX, Y = barY
enddocument

which results in

foo: X = X, Y = not X

bar: X = not X, Y = not X

The first case is quite simple to understand, expansion of foo happens at the time the macro is called, #1 is replaced by X or Y, and the if is executed with either of these two characters.

What is not so clear is the result of bar. I understand that the expansion of if here happens at the time the macro is defined and thus the concrete value of #1 not yet available; TeX leaves "holes" for the parameters in the expanded replacement text to be filled in later when the macro is actually called. But to properly expand the if here, some value for #1 is needed.

The parameter doesn't seem to be just left empty, otherwise bar shouldn't give not X, because if XX would always be true. They also don't seem to be relaxed, as

edefbaz#1ifx#1relax yeselse nofi

always gives no.

Then what values are used for the parameters at the time of definition?

tex-core expansion

share|improve this question

asked 1 hour ago

siracusa

3,6021926

Suppose we have the following short document

documentclassarticle

deffoo#1if#1XXelse not Xfi

edefbar#1if#1XXelse not Xfi

begindocument
foo: X = fooX, Y = fooY

bar: X = barX, Y = barY
enddocument

which results in

foo: X = X, Y = not X

bar: X = not X, Y = not X

The first case is quite simple to understand, expansion of foo happens at the time the macro is called, #1 is replaced by X or Y, and the if is executed with either of these two characters.

What is not so clear is the result of bar. I understand that the expansion of if here happens at the time the macro is defined and thus the concrete value of #1 not yet available; TeX leaves "holes" for the parameters in the expanded replacement text to be filled in later when the macro is actually called. But to properly expand the if here, some value for #1 is needed.

The parameter doesn't seem to be just left empty, otherwise bar shouldn't give not X, because if XX would always be true. They also don't seem to be relaxed, as

edefbaz#1ifx#1relax yeselse nofi

always gives no.

Then what values are used for the parameters at the time of definition?

tex-core expansion

tex-core expansion

share|improve this question

asked 1 hour ago

siracusa

3,6021926

share|improve this question

asked 1 hour ago

siracusa

3,6021926

share|improve this question

share|improve this question

asked 1 hour ago

siracusa

3,6021926

asked 1 hour ago

siracusa

3,6021926

asked 1 hour ago

siracusa

3,6021926

3,6021926

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
3
down vote

When TeX does edef<macro><parameter text>₁<replacement text>₂ it sets aside the <macro> the replacement text and ₁, then does full expansion to <replacement text> until finding the matching ₂. Finally it does

def<macro><parameter text><full expansion of the replacement text>

The full expansion of if#1XXelse not Xfi is not X, because # and 1 are different character tokens (by character code).

Thus your edef becomes the same as

defbar#1not X

As another example,

edefrightbracestringstring}

is perfectly legal, even if it appears to have unbalanced braces, because when the expansion is being done, the first } has already been tokenized as }₁₂ when encountered. This proves that TeX doesn't first absorb the whole <replacement text>, but performs expansion just like in normal circumstances, with the difference that the matching }₂ stops the process and resumes the assignment.

share|improve this answer

answered 30 mins ago

egreg

685k8418273077

add a comment | 

up vote
2
down vote

I would say that it compares # with 1 hance always goes for the false branch. (Compares second test where it compares # with # and chooses true branch)

edefbar#1if#1#1 (TRUE)else #1 (FALSE)fi

showbar


edefbar#1if##1 (TRUE)else (FALSE)fi

showbar

bye

Gives

> bar=macro:
#1->#1 (FALSE).
l.3 showbar

> bar=macro:
#1->1 (TRUE).
l.8 showbar

share|improve this answer

answered 1 hour ago

jfbu

42.3k63136

• 
  
  
  

  
  

  
  
  
  
  
  

  
  try also edefbar#1if###1 (TRUE)else (FALSE)fi, bar X.
  – jfbu
  58 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, if one tries (with pdftex) edefbar#1unlessif#1XXelse not Xfi, showbar will output that bar is a macro doing #1->XX.
  – egreg
  46 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @egreg good test case indeed.
  – jfbu
  40 mins ago
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote

When TeX does edef<macro><parameter text>₁<replacement text>₂ it sets aside the <macro> the replacement text and ₁, then does full expansion to <replacement text> until finding the matching ₂. Finally it does

def<macro><parameter text><full expansion of the replacement text>

The full expansion of if#1XXelse not Xfi is not X, because # and 1 are different character tokens (by character code).

Thus your edef becomes the same as

defbar#1not X

As another example,

edefrightbracestringstring}

is perfectly legal, even if it appears to have unbalanced braces, because when the expansion is being done, the first } has already been tokenized as }₁₂ when encountered. This proves that TeX doesn't first absorb the whole <replacement text>, but performs expansion just like in normal circumstances, with the difference that the matching }₂ stops the process and resumes the assignment.

share|improve this answer

answered 30 mins ago

egreg

685k8418273077

add a comment | 

up vote
3
down vote

When TeX does edef<macro><parameter text>₁<replacement text>₂ it sets aside the <macro> the replacement text and ₁, then does full expansion to <replacement text> until finding the matching ₂. Finally it does

def<macro><parameter text><full expansion of the replacement text>

The full expansion of if#1XXelse not Xfi is not X, because # and 1 are different character tokens (by character code).

Thus your edef becomes the same as

defbar#1not X

As another example,

edefrightbracestringstring}

is perfectly legal, even if it appears to have unbalanced braces, because when the expansion is being done, the first } has already been tokenized as }₁₂ when encountered. This proves that TeX doesn't first absorb the whole <replacement text>, but performs expansion just like in normal circumstances, with the difference that the matching }₂ stops the process and resumes the assignment.

share|improve this answer

answered 30 mins ago

egreg

685k8418273077

add a comment | 

up vote
3
down vote

up vote
3
down vote

When TeX does edef<macro><parameter text>₁<replacement text>₂ it sets aside the <macro> the replacement text and ₁, then does full expansion to <replacement text> until finding the matching ₂. Finally it does

def<macro><parameter text><full expansion of the replacement text>

The full expansion of if#1XXelse not Xfi is not X, because # and 1 are different character tokens (by character code).

Thus your edef becomes the same as

defbar#1not X

As another example,

edefrightbracestringstring}

is perfectly legal, even if it appears to have unbalanced braces, because when the expansion is being done, the first } has already been tokenized as }₁₂ when encountered. This proves that TeX doesn't first absorb the whole <replacement text>, but performs expansion just like in normal circumstances, with the difference that the matching }₂ stops the process and resumes the assignment.

share|improve this answer

answered 30 mins ago

egreg

685k8418273077

When TeX does edef<macro><parameter text>₁<replacement text>₂ it sets aside the <macro> the replacement text and ₁, then does full expansion to <replacement text> until finding the matching ₂. Finally it does

def<macro><parameter text><full expansion of the replacement text>

The full expansion of if#1XXelse not Xfi is not X, because # and 1 are different character tokens (by character code).

Thus your edef becomes the same as

defbar#1not X

As another example,

edefrightbracestringstring}

is perfectly legal, even if it appears to have unbalanced braces, because when the expansion is being done, the first } has already been tokenized as }₁₂ when encountered. This proves that TeX doesn't first absorb the whole <replacement text>, but performs expansion just like in normal circumstances, with the difference that the matching }₂ stops the process and resumes the assignment.

share|improve this answer

answered 30 mins ago

egreg

685k8418273077

share|improve this answer

share|improve this answer

answered 30 mins ago

egreg

685k8418273077

answered 30 mins ago

egreg

685k8418273077

answered 30 mins ago

egreg

685k8418273077

685k8418273077

add a comment | 

add a comment | 

up vote
2
down vote

I would say that it compares # with 1 hance always goes for the false branch. (Compares second test where it compares # with # and chooses true branch)

edefbar#1if#1#1 (TRUE)else #1 (FALSE)fi

showbar


edefbar#1if##1 (TRUE)else (FALSE)fi

showbar

bye

Gives

> bar=macro:
#1->#1 (FALSE).
l.3 showbar

> bar=macro:
#1->1 (TRUE).
l.8 showbar

share|improve this answer

answered 1 hour ago

jfbu

42.3k63136

• 
  
  
  

  
  

  
  
  
  
  
  

  
  try also edefbar#1if###1 (TRUE)else (FALSE)fi, bar X.
  – jfbu
  58 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, if one tries (with pdftex) edefbar#1unlessif#1XXelse not Xfi, showbar will output that bar is a macro doing #1->XX.
  – egreg
  46 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @egreg good test case indeed.
  – jfbu
  40 mins ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

I would say that it compares # with 1 hance always goes for the false branch. (Compares second test where it compares # with # and chooses true branch)

edefbar#1if#1#1 (TRUE)else #1 (FALSE)fi

showbar


edefbar#1if##1 (TRUE)else (FALSE)fi

showbar

bye

Gives

> bar=macro:
#1->#1 (FALSE).
l.3 showbar

> bar=macro:
#1->1 (TRUE).
l.8 showbar

share|improve this answer

answered 1 hour ago

jfbu

42.3k63136

• 
  
  
  

  
  

  
  
  
  
  
  

  
  try also edefbar#1if###1 (TRUE)else (FALSE)fi, bar X.
  – jfbu
  58 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, if one tries (with pdftex) edefbar#1unlessif#1XXelse not Xfi, showbar will output that bar is a macro doing #1->XX.
  – egreg
  46 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @egreg good test case indeed.
  – jfbu
  40 mins ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

I would say that it compares # with 1 hance always goes for the false branch. (Compares second test where it compares # with # and chooses true branch)

edefbar#1if#1#1 (TRUE)else #1 (FALSE)fi

showbar


edefbar#1if##1 (TRUE)else (FALSE)fi

showbar

bye

Gives

> bar=macro:
#1->#1 (FALSE).
l.3 showbar

> bar=macro:
#1->1 (TRUE).
l.8 showbar

share|improve this answer

answered 1 hour ago

jfbu

42.3k63136

I would say that it compares # with 1 hance always goes for the false branch. (Compares second test where it compares # with # and chooses true branch)

edefbar#1if#1#1 (TRUE)else #1 (FALSE)fi

showbar


edefbar#1if##1 (TRUE)else (FALSE)fi

showbar

bye

Gives

> bar=macro:
#1->#1 (FALSE).
l.3 showbar

> bar=macro:
#1->1 (TRUE).
l.8 showbar

share|improve this answer

answered 1 hour ago

jfbu

42.3k63136

share|improve this answer

share|improve this answer

answered 1 hour ago

jfbu

42.3k63136

answered 1 hour ago

jfbu

42.3k63136

answered 1 hour ago

jfbu

42.3k63136

42.3k63136

• 
  
  
  

  
  

  
  
  
  
  
  

  
  try also edefbar#1if###1 (TRUE)else (FALSE)fi, bar X.
  – jfbu
  58 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, if one tries (with pdftex) edefbar#1unlessif#1XXelse not Xfi, showbar will output that bar is a macro doing #1->XX.
  – egreg
  46 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @egreg good test case indeed.
  – jfbu
  40 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  try also edefbar#1if###1 (TRUE)else (FALSE)fi, bar X.
  – jfbu
  58 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, if one tries (with pdftex) edefbar#1unlessif#1XXelse not Xfi, showbar will output that bar is a macro doing #1->XX.
  – egreg
  46 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @egreg good test case indeed.
  – jfbu
  40 mins ago
  

  
  
  
  

try also edefbar#1if###1 (TRUE)else (FALSE)fi, bar X.
– jfbu
58 mins ago

try also edefbar#1if###1 (TRUE)else (FALSE)fi, bar X.
– jfbu
58 mins ago

Yes, if one tries (with pdftex) edefbar#1unlessif#1XXelse not Xfi, showbar will output that bar is a macro doing #1->XX.
– egreg
46 mins ago

Yes, if one tries (with pdftex) edefbar#1unlessif#1XXelse not Xfi, showbar will output that bar is a macro doing #1->XX.
– egreg
46 mins ago

@egreg good test case indeed.
– jfbu
40 mins ago

@egreg good test case indeed.
– jfbu
40 mins ago

add a comment | 

 

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