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Where does nom's “$i” macro argument come from?
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar Manara
Data science time! April 2019 and salary with experience
The Ask Question Wizard is Live!Casting pthread's start_routine argument with a macroOverloading Macro on Number of ArgumentsHow to implement a macro system into my LISPUnderstanding DEFER and OBSTRUCT macrosMacro call vs macro definition environment in JuliaStrange definitions of TRUE and FALSE macrosFixing “no rules expected the token” macro errorWhy does avr-gcc tell me about missing argument while expanding my [fancy] macro?How to use a type (ty) within a macro to construct a struct instance in Rust?PHP Internals: How does TSRMLS_FETCH Work?
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I'm trying to understand how Rust macro captures work and am looking at the nom parser library.
Location nom/src/bytes.rs declares macro tag! which captures with ($i:expr, $tag: expr).
However throughout the places where tag! is used it is used with only one argument, i.e. tag!("+").
Example: nom-lua52/op.rs. I've read tutorials about macros however this construct seems to be special.
It seems like the input $i is somehow implicit?
macros rust nom
edited Mar 22 at 16:22
loganfsmyth
107k21229188
asked Jul 16 '18 at 18:08
Konrad EiseleKonrad Eisele
1,058816
add a comment |
I'm trying to understand how Rust macro captures work and am looking at the nom parser library.
Location nom/src/bytes.rs declares macro tag! which captures with ($i:expr, $tag: expr).
However throughout the places where tag! is used it is used with only one argument, i.e. tag!("+").
Example: nom-lua52/op.rs. I've read tutorials about macros however this construct seems to be special.
It seems like the input $i is somehow implicit?
macros rust nom
edited Mar 22 at 16:22
loganfsmyth
107k21229188
asked Jul 16 '18 at 18:08
Konrad EiseleKonrad Eisele
1,058816
add a comment |
I'm trying to understand how Rust macro captures work and am looking at the nom parser library.
Location nom/src/bytes.rs declares macro tag! which captures with ($i:expr, $tag: expr).
However throughout the places where tag! is used it is used with only one argument, i.e. tag!("+").
Example: nom-lua52/op.rs. I've read tutorials about macros however this construct seems to be special.
It seems like the input $i is somehow implicit?
macros rust nom
edited Mar 22 at 16:22
loganfsmyth
107k21229188
asked Jul 16 '18 at 18:08
Konrad EiseleKonrad Eisele
1,058816
I'm trying to understand how Rust macro captures work and am looking at the nom parser library.
Location nom/src/bytes.rs declares macro tag! which captures with ($i:expr, $tag: expr).
However throughout the places where tag! is used it is used with only one argument, i.e. tag!("+").
Example: nom-lua52/op.rs. I've read tutorials about macros however this construct seems to be special.
It seems like the input $i is somehow implicit?
macros rust nom
macros rust nom
edited Mar 22 at 16:22
loganfsmyth
107k21229188
asked Jul 16 '18 at 18:08
Konrad EiseleKonrad Eisele
1,058816
edited Mar 22 at 16:22
loganfsmyth
107k21229188
asked Jul 16 '18 at 18:08
Konrad EiseleKonrad Eisele
1,058816
edited Mar 22 at 16:22
loganfsmyth
107k21229188
edited Mar 22 at 16:22
loganfsmyth
107k21229188
edited Mar 22 at 16:22
loganfsmyth
107k21229188
107k21229188
asked Jul 16 '18 at 18:08
Konrad EiseleKonrad Eisele
1,058816
asked Jul 16 '18 at 18:08
Konrad EiseleKonrad Eisele
1,058816
asked Jul 16 '18 at 18:08
Konrad EiseleKonrad Eisele
1,058816
1,058816
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
The trick is that the tag!() macro call is inside the value! macro, which will match against tag! and pass it an additional parameter. This is described in the file implementing tag!.
answered Jul 16 '18 at 18:24
starbluestarblue
45.7k1178138
1
Thanks, Now I see the trick:$submac!($i, $($args)*).
– Konrad Eisele
Jul 16 '18 at 18:28
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
votes
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oldest
votes
The trick is that the tag!() macro call is inside the value! macro, which will match against tag! and pass it an additional parameter. This is described in the file implementing tag!.
answered Jul 16 '18 at 18:24
starbluestarblue
45.7k1178138
1
Thanks, Now I see the trick:$submac!($i, $($args)*).
– Konrad Eisele
Jul 16 '18 at 18:28
add a comment |
The trick is that the tag!() macro call is inside the value! macro, which will match against tag! and pass it an additional parameter. This is described in the file implementing tag!.
answered Jul 16 '18 at 18:24
starbluestarblue
45.7k1178138
1
Thanks, Now I see the trick:$submac!($i, $($args)*).
– Konrad Eisele
Jul 16 '18 at 18:28
add a comment |
The trick is that the tag!() macro call is inside the value! macro, which will match against tag! and pass it an additional parameter. This is described in the file implementing tag!.
answered Jul 16 '18 at 18:24
starbluestarblue
45.7k1178138
The trick is that the tag!() macro call is inside the value! macro, which will match against tag! and pass it an additional parameter. This is described in the file implementing tag!.
answered Jul 16 '18 at 18:24
starbluestarblue
45.7k1178138
answered Jul 16 '18 at 18:24
starbluestarblue
45.7k1178138
answered Jul 16 '18 at 18:24
starbluestarblue
45.7k1178138
answered Jul 16 '18 at 18:24
starbluestarblue
45.7k1178138
45.7k1178138
1
Thanks, Now I see the trick:$submac!($i, $($args)*).
– Konrad Eisele
Jul 16 '18 at 18:28
add a comment |
1
Thanks, Now I see the trick:$submac!($i, $($args)*).
– Konrad Eisele
Jul 16 '18 at 18:28
1
1
Thanks, Now I see the trick:
$submac!($i, $($args)*).– Konrad Eisele
Jul 16 '18 at 18:28
Thanks, Now I see the trick:
$submac!($i, $($args)*).– Konrad Eisele
Jul 16 '18 at 18:28
add a comment |
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