A curious dependency between newcommand and curly braces
An example is a bit artificial, but the point is that I do not understand why
{} influence the outcome of newcommand in the way described below.
Consider the following definition:
newcommandprintit[1]{#1}
whose outcome is obvious. However, a small alteration:
newcommand{printit[1]}{#1}
and e.g. printit{2} yields:
I have no idea why. Could you please shed some light on what is going on here?
macros braces grouping
asked 3 hours ago
Mad HatterMad Hatter
580419
add a comment |
An example is a bit artificial, but the point is that I do not understand why
{} influence the outcome of newcommand in the way described below.
Consider the following definition:
newcommandprintit[1]{#1}
whose outcome is obvious. However, a small alteration:
newcommand{printit[1]}{#1}
and e.g. printit{2} yields:
I have no idea why. Could you please shed some light on what is going on here?
macros braces grouping
asked 3 hours ago
Mad HatterMad Hatter
580419
1
defnew@command#1{% @testopt{@newcommand#1}0}from latex.ltx, i.e. LaTeX tests whether there is an optional argument in the first argument ofnewcommand, i.e.printit[1]and returns0, the{2}in its usage is not regarded as an argument. You could useprintitwithout{}and it will still print0, use it asprintit[2]and it will print2. In fact you have defined a macro with optional arguments in a non-obvious way
– Christian Hupfer
2 hours ago
The correct syntax isnewcommandprintit[1]{#1}ornewcommand{printit}[1]{#1}(The number of parameters to be expected is given in square brackets, outside of the{}). In case people come across this and are confused by the above syntax.
– Peter Grill
2 hours ago
@PeterGrill Yes, I know the correct syntax. But I was curious why grouping printit[1] results in the (rather unexpected) outcome.
– Mad Hatter
40 mins ago
@MadHatter: Yep, I figured that. My comment was intended for others who may come across this and be confused.
– Peter Grill
3 mins ago
add a comment |
An example is a bit artificial, but the point is that I do not understand why
{} influence the outcome of newcommand in the way described below.
Consider the following definition:
newcommandprintit[1]{#1}
whose outcome is obvious. However, a small alteration:
newcommand{printit[1]}{#1}
and e.g. printit{2} yields:
I have no idea why. Could you please shed some light on what is going on here?
macros braces grouping
asked 3 hours ago
Mad HatterMad Hatter
580419
An example is a bit artificial, but the point is that I do not understand why
{} influence the outcome of newcommand in the way described below.
Consider the following definition:
newcommandprintit[1]{#1}
whose outcome is obvious. However, a small alteration:
newcommand{printit[1]}{#1}
and e.g. printit{2} yields:
I have no idea why. Could you please shed some light on what is going on here?
macros braces grouping
macros braces grouping
asked 3 hours ago
Mad HatterMad Hatter
580419
asked 3 hours ago
Mad HatterMad Hatter
580419
asked 3 hours ago
Mad HatterMad Hatter
580419
asked 3 hours ago
Mad HatterMad Hatter
580419
asked 3 hours ago
Mad HatterMad Hatter
580419
580419
1
defnew@command#1{% @testopt{@newcommand#1}0}from latex.ltx, i.e. LaTeX tests whether there is an optional argument in the first argument ofnewcommand, i.e.printit[1]and returns0, the{2}in its usage is not regarded as an argument. You could useprintitwithout{}and it will still print0, use it asprintit[2]and it will print2. In fact you have defined a macro with optional arguments in a non-obvious way
– Christian Hupfer
2 hours ago
The correct syntax isnewcommandprintit[1]{#1}ornewcommand{printit}[1]{#1}(The number of parameters to be expected is given in square brackets, outside of the{}). In case people come across this and are confused by the above syntax.
– Peter Grill
2 hours ago
@PeterGrill Yes, I know the correct syntax. But I was curious why grouping printit[1] results in the (rather unexpected) outcome.
– Mad Hatter
40 mins ago
@MadHatter: Yep, I figured that. My comment was intended for others who may come across this and be confused.
– Peter Grill
3 mins ago
add a comment |
1
defnew@command#1{% @testopt{@newcommand#1}0}from latex.ltx, i.e. LaTeX tests whether there is an optional argument in the first argument ofnewcommand, i.e.printit[1]and returns0, the{2}in its usage is not regarded as an argument. You could useprintitwithout{}and it will still print0, use it asprintit[2]and it will print2. In fact you have defined a macro with optional arguments in a non-obvious way
– Christian Hupfer
2 hours ago
The correct syntax isnewcommandprintit[1]{#1}ornewcommand{printit}[1]{#1}(The number of parameters to be expected is given in square brackets, outside of the{}). In case people come across this and are confused by the above syntax.
– Peter Grill
2 hours ago
@PeterGrill Yes, I know the correct syntax. But I was curious why grouping printit[1] results in the (rather unexpected) outcome.
– Mad Hatter
40 mins ago
@MadHatter: Yep, I figured that. My comment was intended for others who may come across this and be confused.
– Peter Grill
3 mins ago
1
1
defnew@command#1{% @testopt{@newcommand#1}0} from latex.ltx, i.e. LaTeX tests whether there is an optional argument in the first argument of newcommand, i.e. printit[1] and returns 0, the {2} in its usage is not regarded as an argument. You could use printit without {} and it will still print 0, use it as printit[2] and it will print 2. In fact you have defined a macro with optional arguments in a non-obvious way– Christian Hupfer
2 hours ago
defnew@command#1{% @testopt{@newcommand#1}0} from latex.ltx, i.e. LaTeX tests whether there is an optional argument in the first argument of newcommand, i.e. printit[1] and returns 0, the {2} in its usage is not regarded as an argument. You could use printit without {} and it will still print 0, use it as printit[2] and it will print 2. In fact you have defined a macro with optional arguments in a non-obvious way– Christian Hupfer
2 hours ago
The correct syntax is
newcommandprintit[1]{#1} or newcommand{printit}[1]{#1} (The number of parameters to be expected is given in square brackets, outside of the {}). In case people come across this and are confused by the above syntax.– Peter Grill
2 hours ago
The correct syntax is
newcommandprintit[1]{#1} or newcommand{printit}[1]{#1} (The number of parameters to be expected is given in square brackets, outside of the {}). In case people come across this and are confused by the above syntax.– Peter Grill
2 hours ago
@PeterGrill Yes, I know the correct syntax. But I was curious why grouping printit[1] results in the (rather unexpected) outcome.
– Mad Hatter
40 mins ago
@PeterGrill Yes, I know the correct syntax. But I was curious why grouping printit[1] results in the (rather unexpected) outcome.
– Mad Hatter
40 mins ago
@MadHatter: Yep, I figured that. My comment was intended for others who may come across this and be confused.
– Peter Grill
3 mins ago
@MadHatter: Yep, I figured that. My comment was intended for others who may come across this and be confused.
– Peter Grill
3 mins ago
add a comment |
1 Answer
1
active
oldest
votes
If LaTeX was being written now,
newcommand{printit[1]}{#1}
would give an error message that printit[1] was not a single token.
The correct syntax is
newcommand{printit}[1]{#1}
or
newcommandprintit[1]{#1}
LaTeX does nothing special to allow both forms, it is just on the general TeX macro syntax rules that braces can be omitted if a macro argument is a single token.
If you pass two (or in this case, four) tokens to newcommand then anything that happens is essentially just accidental behaviour, as the system could not really afford the space or time required to catch "unlikely" errors.
As to what happens,
documentclass{article}
begin{document}
newcommand{printit[1]}{#1}
showprintit
expandaftershowcsnamestringprintitendcsname
end{document}
produces the terminal log
> printit=macro:
->@protected@testopt printit \printit {0}.
l.7 showprintit
?
> \printit=long macro:
[#1]->#1.
The command defines printit to have an optional first argument, although the default value of 0 being supplied was intended as the default value for newcommand itself.
it is what makes newcommandfoo the same as newcommandfoo[0] and define a command with no arguments.
The inner part of the command handling the optional argument is \printit which just drops the square brackets.
So
printit{2}
is
printit[0]{2}
which is
0{2}
which typesets as 02
answered 1 hour ago
David CarlisleDavid Carlisle
485k4111191862
Thanks for a thorough explanation!
– Mad Hatter
38 mins ago
add a comment |
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1 Answer
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1 Answer
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oldest
votes
If LaTeX was being written now,
newcommand{printit[1]}{#1}
would give an error message that printit[1] was not a single token.
The correct syntax is
newcommand{printit}[1]{#1}
or
newcommandprintit[1]{#1}
LaTeX does nothing special to allow both forms, it is just on the general TeX macro syntax rules that braces can be omitted if a macro argument is a single token.
If you pass two (or in this case, four) tokens to newcommand then anything that happens is essentially just accidental behaviour, as the system could not really afford the space or time required to catch "unlikely" errors.
As to what happens,
documentclass{article}
begin{document}
newcommand{printit[1]}{#1}
showprintit
expandaftershowcsnamestringprintitendcsname
end{document}
produces the terminal log
> printit=macro:
->@protected@testopt printit \printit {0}.
l.7 showprintit
?
> \printit=long macro:
[#1]->#1.
The command defines printit to have an optional first argument, although the default value of 0 being supplied was intended as the default value for newcommand itself.
it is what makes newcommandfoo the same as newcommandfoo[0] and define a command with no arguments.
The inner part of the command handling the optional argument is \printit which just drops the square brackets.
So
printit{2}
is
printit[0]{2}
which is
0{2}
which typesets as 02
answered 1 hour ago
David CarlisleDavid Carlisle
485k4111191862
Thanks for a thorough explanation!
– Mad Hatter
38 mins ago
add a comment |
If LaTeX was being written now,
newcommand{printit[1]}{#1}
would give an error message that printit[1] was not a single token.
The correct syntax is
newcommand{printit}[1]{#1}
or
newcommandprintit[1]{#1}
LaTeX does nothing special to allow both forms, it is just on the general TeX macro syntax rules that braces can be omitted if a macro argument is a single token.
If you pass two (or in this case, four) tokens to newcommand then anything that happens is essentially just accidental behaviour, as the system could not really afford the space or time required to catch "unlikely" errors.
As to what happens,
documentclass{article}
begin{document}
newcommand{printit[1]}{#1}
showprintit
expandaftershowcsnamestringprintitendcsname
end{document}
produces the terminal log
> printit=macro:
->@protected@testopt printit \printit {0}.
l.7 showprintit
?
> \printit=long macro:
[#1]->#1.
The command defines printit to have an optional first argument, although the default value of 0 being supplied was intended as the default value for newcommand itself.
it is what makes newcommandfoo the same as newcommandfoo[0] and define a command with no arguments.
The inner part of the command handling the optional argument is \printit which just drops the square brackets.
So
printit{2}
is
printit[0]{2}
which is
0{2}
which typesets as 02
answered 1 hour ago
David CarlisleDavid Carlisle
485k4111191862
Thanks for a thorough explanation!
– Mad Hatter
38 mins ago
add a comment |
If LaTeX was being written now,
newcommand{printit[1]}{#1}
would give an error message that printit[1] was not a single token.
The correct syntax is
newcommand{printit}[1]{#1}
or
newcommandprintit[1]{#1}
LaTeX does nothing special to allow both forms, it is just on the general TeX macro syntax rules that braces can be omitted if a macro argument is a single token.
If you pass two (or in this case, four) tokens to newcommand then anything that happens is essentially just accidental behaviour, as the system could not really afford the space or time required to catch "unlikely" errors.
As to what happens,
documentclass{article}
begin{document}
newcommand{printit[1]}{#1}
showprintit
expandaftershowcsnamestringprintitendcsname
end{document}
produces the terminal log
> printit=macro:
->@protected@testopt printit \printit {0}.
l.7 showprintit
?
> \printit=long macro:
[#1]->#1.
The command defines printit to have an optional first argument, although the default value of 0 being supplied was intended as the default value for newcommand itself.
it is what makes newcommandfoo the same as newcommandfoo[0] and define a command with no arguments.
The inner part of the command handling the optional argument is \printit which just drops the square brackets.
So
printit{2}
is
printit[0]{2}
which is
0{2}
which typesets as 02
answered 1 hour ago
David CarlisleDavid Carlisle
485k4111191862
If LaTeX was being written now,
newcommand{printit[1]}{#1}
would give an error message that printit[1] was not a single token.
The correct syntax is
newcommand{printit}[1]{#1}
or
newcommandprintit[1]{#1}
LaTeX does nothing special to allow both forms, it is just on the general TeX macro syntax rules that braces can be omitted if a macro argument is a single token.
If you pass two (or in this case, four) tokens to newcommand then anything that happens is essentially just accidental behaviour, as the system could not really afford the space or time required to catch "unlikely" errors.
As to what happens,
documentclass{article}
begin{document}
newcommand{printit[1]}{#1}
showprintit
expandaftershowcsnamestringprintitendcsname
end{document}
produces the terminal log
> printit=macro:
->@protected@testopt printit \printit {0}.
l.7 showprintit
?
> \printit=long macro:
[#1]->#1.
The command defines printit to have an optional first argument, although the default value of 0 being supplied was intended as the default value for newcommand itself.
it is what makes newcommandfoo the same as newcommandfoo[0] and define a command with no arguments.
The inner part of the command handling the optional argument is \printit which just drops the square brackets.
So
printit{2}
is
printit[0]{2}
which is
0{2}
which typesets as 02
answered 1 hour ago
David CarlisleDavid Carlisle
485k4111191862
edited 35 mins ago
answered 1 hour ago
David CarlisleDavid Carlisle
485k4111191862
answered 1 hour ago
David CarlisleDavid Carlisle
485k4111191862
answered 1 hour ago
David CarlisleDavid Carlisle
485k4111191862
485k4111191862
Thanks for a thorough explanation!
– Mad Hatter
38 mins ago
add a comment |
Thanks for a thorough explanation!
– Mad Hatter
38 mins ago
Thanks for a thorough explanation!
– Mad Hatter
38 mins ago
Thanks for a thorough explanation!
– Mad Hatter
38 mins ago
add a comment |
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1
defnew@command#1{% @testopt{@newcommand#1}0}from latex.ltx, i.e. LaTeX tests whether there is an optional argument in the first argument ofnewcommand, i.e.printit[1]and returns0, the{2}in its usage is not regarded as an argument. You could useprintitwithout{}and it will still print0, use it asprintit[2]and it will print2. In fact you have defined a macro with optional arguments in a non-obvious way– Christian Hupfer
2 hours ago
The correct syntax is
newcommandprintit[1]{#1}ornewcommand{printit}[1]{#1}(The number of parameters to be expected is given in square brackets, outside of the{}). In case people come across this and are confused by the above syntax.– Peter Grill
2 hours ago
@PeterGrill Yes, I know the correct syntax. But I was curious why grouping printit[1] results in the (rather unexpected) outcome.
– Mad Hatter
40 mins ago
@MadHatter: Yep, I figured that. My comment was intended for others who may come across this and be confused.
– Peter Grill
3 mins ago