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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













7















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:



bold plus in a box



Is this expected behavior?










share|improve this question

















  • 5





    boldmath is not deprecated.

    – David Carlisle
    3 hours ago






  • 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
    3 hours ago












  • Ah, ok, I thought it was deprecated for this joshua.smcvt.edu/latex2e/… but I see this is not official...

    – Rmano
    29 mins ago











  • Not only unofficial it is completely wrong.

    – David Carlisle
    22 mins ago















7















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:



bold plus in a box



Is this expected behavior?










share|improve this question

















  • 5





    boldmath is not deprecated.

    – David Carlisle
    3 hours ago






  • 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
    3 hours ago












  • Ah, ok, I thought it was deprecated for this joshua.smcvt.edu/latex2e/… but I see this is not official...

    – Rmano
    29 mins ago











  • Not only unofficial it is completely wrong.

    – David Carlisle
    22 mins ago













7












7








7


0






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:



bold plus in a box



Is this expected behavior?










share|improve this question














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:



bold plus in a box



Is this expected behavior?







tikz-pgf boldmath






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 4 hours ago









RmanoRmano

8,34121649




8,34121649







  • 5





    boldmath is not deprecated.

    – David Carlisle
    3 hours ago






  • 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
    3 hours ago












  • Ah, ok, I thought it was deprecated for this joshua.smcvt.edu/latex2e/… but I see this is not official...

    – Rmano
    29 mins ago











  • Not only unofficial it is completely wrong.

    – David Carlisle
    22 mins ago












  • 5





    boldmath is not deprecated.

    – David Carlisle
    3 hours ago






  • 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
    3 hours ago












  • Ah, ok, I thought it was deprecated for this joshua.smcvt.edu/latex2e/… but I see this is not official...

    – Rmano
    29 mins ago











  • Not only unofficial it is completely wrong.

    – David Carlisle
    22 mins ago







5




5





boldmath is not deprecated.

– David Carlisle
3 hours ago





boldmath is not deprecated.

– David Carlisle
3 hours ago




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
3 hours ago






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
3 hours ago














Ah, ok, I thought it was deprecated for this joshua.smcvt.edu/latex2e/… but I see this is not official...

– Rmano
29 mins ago





Ah, ok, I thought it was deprecated for this joshua.smcvt.edu/latex2e/… but I see this is not official...

– Rmano
29 mins ago













Not only unofficial it is completely wrong.

– David Carlisle
22 mins ago





Not only unofficial it is completely wrong.

– David Carlisle
22 mins ago










1 Answer
1






active

oldest

votes


















4














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


enter image description here






share|improve this answer























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    1 Answer
    1






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    active

    oldest

    votes






    active

    oldest

    votes









    4














    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


    enter image description here






    share|improve this answer



























      4














      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


      enter image description here






      share|improve this answer

























        4












        4








        4







        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


        enter image description here






        share|improve this answer













        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


        enter image description here







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 2 hours ago









        marmotmarmot

        121k6159297




        121k6159297



























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