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Why boldmath fails in a tikz node?
LaTeX equivalent of ConTeXt buffersRotate a node but not its content: the case of the ellipse decorationHow to define the default vertical distance between nodes?Numerical conditional within tikz keys?TikZ/ERD: node (=Entity) label on the insideTikZ: Drawing an arc from an intersection to an intersectionDrawing rectilinear curves in Tikz, aka an Etch-a-Sketch drawingLine up nested tikz enviroments or how to get rid of themHow to draw a square and its diagonals with arrows?Parallel arrows between nodes of varying size
I think that probably boldmath is deprecated, although I can't find any "official" reference, but anyway:
documentclass[border=10pt]standalone
usepackagetikz
begindocument
begintikzpicture
path (0,0) node[draw]boldmath (+) (1,1);
endtikzpicture
enddocument
Fails to compile with a strange:
! Package tikz Error: Giving up on this path. Did you forget a semicolon?.
See the tikz package documentation for explanation.
Type H <return> for immediate help.
...
l.6 path (0,0) node[draw]boldmath (+)
(1,1);
?
If I double the braces in the node, as
path (0,0) node[draw]boldmath (+) (1,1);
it works ok:
Is this expected behavior?
tikz-pgf boldmath
asked Apr 26 at 16:29
RmanoRmano
8,48121649
add a comment |
I think that probably boldmath is deprecated, although I can't find any "official" reference, but anyway:
documentclass[border=10pt]standalone
usepackagetikz
begindocument
begintikzpicture
path (0,0) node[draw]boldmath (+) (1,1);
endtikzpicture
enddocument
Fails to compile with a strange:
! Package tikz Error: Giving up on this path. Did you forget a semicolon?.
See the tikz package documentation for explanation.
Type H <return> for immediate help.
...
l.6 path (0,0) node[draw]boldmath (+)
(1,1);
?
If I double the braces in the node, as
path (0,0) node[draw]boldmath (+) (1,1);
it works ok:
Is this expected behavior?
tikz-pgf boldmath
asked Apr 26 at 16:29
RmanoRmano
8,48121649
5
boldmathis not deprecated.
– David Carlisle
Apr 26 at 17:03
4
I seem to recall that tikz only allows a certain number of expansions while looking for the;and the full definition ofboldmathis rather complicated so perhaps it is too much....
– David Carlisle
Apr 26 at 17:07
Ah, ok, I thought it was deprecated for this joshua.smcvt.edu/latex2e/… but I see this is not official...
– Rmano
Apr 26 at 20:28
Not only unofficial it is completely wrong.
– David Carlisle
Apr 26 at 20:36
add a comment |
I think that probably boldmath is deprecated, although I can't find any "official" reference, but anyway:
documentclass[border=10pt]standalone
usepackagetikz
begindocument
begintikzpicture
path (0,0) node[draw]boldmath (+) (1,1);
endtikzpicture
enddocument
Fails to compile with a strange:
! Package tikz Error: Giving up on this path. Did you forget a semicolon?.
See the tikz package documentation for explanation.
Type H <return> for immediate help.
...
l.6 path (0,0) node[draw]boldmath (+)
(1,1);
?
If I double the braces in the node, as
path (0,0) node[draw]boldmath (+) (1,1);
it works ok:
Is this expected behavior?
tikz-pgf boldmath
asked Apr 26 at 16:29
RmanoRmano
8,48121649
I think that probably boldmath is deprecated, although I can't find any "official" reference, but anyway:
documentclass[border=10pt]standalone
usepackagetikz
begindocument
begintikzpicture
path (0,0) node[draw]boldmath (+) (1,1);
endtikzpicture
enddocument
Fails to compile with a strange:
! Package tikz Error: Giving up on this path. Did you forget a semicolon?.
See the tikz package documentation for explanation.
Type H <return> for immediate help.
...
l.6 path (0,0) node[draw]boldmath (+)
(1,1);
?
If I double the braces in the node, as
path (0,0) node[draw]boldmath (+) (1,1);
it works ok:
Is this expected behavior?
tikz-pgf boldmath
tikz-pgf boldmath
asked Apr 26 at 16:29
RmanoRmano
8,48121649
asked Apr 26 at 16:29
RmanoRmano
8,48121649
asked Apr 26 at 16:29
RmanoRmano
8,48121649
asked Apr 26 at 16:29
RmanoRmano
8,48121649
asked Apr 26 at 16:29
RmanoRmano
8,48121649
8,48121649
5
boldmathis not deprecated.
– David Carlisle
Apr 26 at 17:03
4
I seem to recall that tikz only allows a certain number of expansions while looking for the;and the full definition ofboldmathis rather complicated so perhaps it is too much....
– David Carlisle
Apr 26 at 17:07
Ah, ok, I thought it was deprecated for this joshua.smcvt.edu/latex2e/… but I see this is not official...
– Rmano
Apr 26 at 20:28
Not only unofficial it is completely wrong.
– David Carlisle
Apr 26 at 20:36
add a comment |
5
boldmathis not deprecated.
– David Carlisle
Apr 26 at 17:03
4
I seem to recall that tikz only allows a certain number of expansions while looking for the;and the full definition ofboldmathis rather complicated so perhaps it is too much....
– David Carlisle
Apr 26 at 17:07
Ah, ok, I thought it was deprecated for this joshua.smcvt.edu/latex2e/… but I see this is not official...
– Rmano
Apr 26 at 20:28
Not only unofficial it is completely wrong.
– David Carlisle
Apr 26 at 20:36
5
5
boldmath is not deprecated.– David Carlisle
Apr 26 at 17:03
boldmath is not deprecated.– David Carlisle
Apr 26 at 17:03
4
4
I seem to recall that tikz only allows a certain number of expansions while looking for the
; and the full definition of boldmath is rather complicated so perhaps it is too much....– David Carlisle
Apr 26 at 17:07
I seem to recall that tikz only allows a certain number of expansions while looking for the
; and the full definition of boldmath is rather complicated so perhaps it is too much....– David Carlisle
Apr 26 at 17:07
Ah, ok, I thought it was deprecated for this joshua.smcvt.edu/latex2e/… but I see this is not official...
– Rmano
Apr 26 at 20:28
Ah, ok, I thought it was deprecated for this joshua.smcvt.edu/latex2e/… but I see this is not official...
– Rmano
Apr 26 at 20:28
Not only unofficial it is completely wrong.
– David Carlisle
Apr 26 at 20:36
Not only unofficial it is completely wrong.
– David Carlisle
Apr 26 at 20:36
add a comment |
1 Answer
1
active
oldest
votes
TikZ has the key font (as well as node font for such things) with which it works.
documentclass[border=10pt]standalone
usepackagetikz
begindocument
begintikzpicture
path (1,0) node[draw,font=boldmath] (+);
path (2,0) node[draw,node font=boldmath] (+);
path (3,0) node[draw] (+);
endtikzpicture
enddocument
answered Apr 26 at 18:17
marmotmarmot
124k6160303
add a comment |
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1 Answer
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1 Answer
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oldest
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active
oldest
votes
TikZ has the key font (as well as node font for such things) with which it works.
documentclass[border=10pt]standalone
usepackagetikz
begindocument
begintikzpicture
path (1,0) node[draw,font=boldmath] (+);
path (2,0) node[draw,node font=boldmath] (+);
path (3,0) node[draw] (+);
endtikzpicture
enddocument
answered Apr 26 at 18:17
marmotmarmot
124k6160303
add a comment |
TikZ has the key font (as well as node font for such things) with which it works.
documentclass[border=10pt]standalone
usepackagetikz
begindocument
begintikzpicture
path (1,0) node[draw,font=boldmath] (+);
path (2,0) node[draw,node font=boldmath] (+);
path (3,0) node[draw] (+);
endtikzpicture
enddocument
answered Apr 26 at 18:17
marmotmarmot
124k6160303
add a comment |
TikZ has the key font (as well as node font for such things) with which it works.
documentclass[border=10pt]standalone
usepackagetikz
begindocument
begintikzpicture
path (1,0) node[draw,font=boldmath] (+);
path (2,0) node[draw,node font=boldmath] (+);
path (3,0) node[draw] (+);
endtikzpicture
enddocument
answered Apr 26 at 18:17
marmotmarmot
124k6160303
TikZ has the key font (as well as node font for such things) with which it works.
documentclass[border=10pt]standalone
usepackagetikz
begindocument
begintikzpicture
path (1,0) node[draw,font=boldmath] (+);
path (2,0) node[draw,node font=boldmath] (+);
path (3,0) node[draw] (+);
endtikzpicture
enddocument
answered Apr 26 at 18:17
marmotmarmot
124k6160303
answered Apr 26 at 18:17
marmotmarmot
124k6160303
answered Apr 26 at 18:17
marmotmarmot
124k6160303
answered Apr 26 at 18:17
marmotmarmot
124k6160303
124k6160303
add a comment |
add a comment |
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5
boldmathis not deprecated.– David Carlisle
Apr 26 at 17:03
4
I seem to recall that tikz only allows a certain number of expansions while looking for the
;and the full definition ofboldmathis rather complicated so perhaps it is too much....– David Carlisle
Apr 26 at 17:07
Ah, ok, I thought it was deprecated for this joshua.smcvt.edu/latex2e/… but I see this is not official...
– Rmano
Apr 26 at 20:28
Not only unofficial it is completely wrong.
– David Carlisle
Apr 26 at 20:36