% This is file texdimens.tex, part of texdimens package, which % is distributed under the LPPL 1.3c. Copyright (c) 2021 Jean-François Burnol % 2021/11/04 v0.99d \edef\texdimensendinput{\endlinechar\the\endlinechar% \catcode`\noexpand _=\the\catcode`\_% \catcode`\noexpand @=\the\catcode`\@\relax\noexpand\endinput}% \endlinechar13\relax% \catcode`\_=11 \catcode`\@=11 % only for using \p@ of Plain. Check exists? % % Mathematics ("down" and "up" macros) % =========== % % In the entire discussion here, "uu" stands for some core unit, % or some unit corresponding to a dimension > 1pt. For the case % of a unit corresponding to a dimension < 1pt, i.e. to % \texdimenwithunit macro added at 0.99, refer to the % comments of issue #2 on the tracker site. % % Is T sp attainable from unit "uu"?. % If not, what is largest dimension < Tsp which is attainable? % Here we suppose T>0. % % phi>1, psi=1/phi, psi<1. % % U(N,phi)=trunc(N phi) is the strictly increasing sequence, % indexed by non-negative integers, of attainable dimensions. % (in sp unit) % % U(N)<= T < U(N+1) iff N = ceil((T+1)psi) - 1 % U(M)< T <= U(M+1) iff M = ceil(T psi) - 1 % % Stumbling block % --------------- % % The stumbling block is that computing "ceil((T+1)psi) - 1" without % overflow is not obvious: yes \numexpr/\dimexpr allow so-called % "scaling operations" but only in the "rounding up" variant. % % If we attempt computing the ceil(x) function via round(x+0.5), % for example with psi=100/7227 which corresponds to the unit "in", % this necessitates evaluating: % % round((((T+1)*200)+7227)/14454) % % But as far as I can tell currently, for this we need to be able % to evaluate without overflow (T+1)*200+7227 and this limits to % T's which are (roughly) such that 100 T is less than \maxdimen. % % A work-around % ------------- % % The rest of the discussion is about an algorithm providing an % alternative route to N, using \numexpr/\dimexpr/TeX facilities, % and working with (almost, as we will see) the full range of allowed % T's, 0 < T <= \maxdimen. (that the algorithm works for T=0 is to be % checked manually after the main discussion). % % Let's return to the U(N)<= T < U(N+1) and U(M)< T <= U(M+1) equations. % % Either (recall in all of this T > 0): % % case1: M = N, i.e. T is not attainable, M=N < T psi < (T+1) psi <= N+1 % case2: M = N - 1, i.e. T is attained, T psi <= N < (T+1) psi, T = trunc(N phi) % % Let X = round(T psi). And let Y = trunc(X phi). We will explain later % how X and Y can be computed using \numexpr/\dimexpr/TeX. % % case1: X can be N or N+1. It will be N+1 iff Y > T. % case2: X can be N or N-1. It will be N iff trunc((X+1)phi)>T. % % This is not convenient: if Y < T it could be that we are in case 2 % but to decide we must check if trunc((X+1) phi) = T or not, so % this means a second computation. % % If psi < 0.5 % ------------ % % The situation then simplifies: % % case1: X can be N or N+1. It will be N+1 iff Y = trunc(X phi) > T. % case2: X is necessarily N. % % Thus: % a) compute X = round(T psi) % b) compute Y = trunc(X phi) and test if Y > T. If true, we % were in case 1, replace X by X - 1, else we were either % in case 1 or case 2, and we leave X as it is. % We have thus found N. % % The operation Y = trunc(X phi) can be achieved this way: % i) use \the\dimexpr to convert X sp into D pt, % ii) use \the\numexpr\dimexpr to convert "D uu" into sp. % These steps give Y. % % This way we find the maximal dimension at most T sp exactly % representable in "uu" unit. % % The computations of X and Y can be done independently of sign of T. % But the final test has to be changed to Y < T if T < 0 and then % one must replace X by X+1. So we must filter out the sign of the input. % % If the goal is only to find a decimal D such that "D uu" is % exactly T sp in the case this is possible, then things are simpler % because from X = round(T psi) we get D such as X sp is same as D pt % and "D uu" will work. % We don't have to take sign into account for this computation. % But if T sp was not attainable we don't know if this X will give % a D such that D uu < T sp or D uu > T sp. % % If psi > 0.5 % ------------ % % For example unit "bp" has phi=803/800. % % It is then not true that if T sp is attainable, the X = round(T psi) % will always work. % % But it is true that R = round((T + 0.5) psi) will always work. % Here we must use -0.5 if T < 0, though. % % This R=round((T+0.5) psi) can always be computed via \numexpr because 2T+1 % will not trigger arithmetic overflow. % % So this gives an approach to find a D such that "D uu" is exactly % T sp when this is possible. % % If Tsp (positive) is not attainable, this R however can produce % either N or N+1. % % But we can decide what happened by computing Z = trunc(R phi). % If and only if Z > T this means R was N+1. % % It is slightly less costly to compute X = round(T psi) than % R = round((T + 0.5) psi), % but if we then realize that trunc(X phi) < T we do not yet know % if trunc((X+1) phi) = T or is > T. So we proceed via R, not X, % to not have to make a second computation if a dimension comparison % test goes awry. % % To recapitulate: we have our algorithm for all units to find out % maximal dimension exactly attainable in "uu" unit and at most equal % to (positive) T sp. % % Unfortunately the check that Y (in case psi < 0.5) or Z (in case psi > % 0.5) verifies or not Y > T may trigger a Dimension too large error if % T sp was near non-attainable \maxdimen. It turns out this sad % situation happens only for the units `dd`, `nc`, and `in`, and T sp % very close to \maxdimen (like for all units apart from `pt`, `bp`, % `nd`, the \maxdimen is not attainable, and by bad luck for `dd`, `nc`, % and `in`, the X will correspond to a decimal D such that Duu>\maxdimen % is the nearest virtually attaible dimensions from above not from % below; see the README.md for the tabulation of the maximal usable inputs). % % Regarding the \texdimen macros, and units with phi > 2, I % hesitated using either the round((T+0.5)psi) or round(T psi), but for % Tsp = \maxdimen, both formulas turned out to give the same result for % all such units, so I chose for these \texdimen macros and the % units with phi>2 to use the simpler round(T psi) which does not need % to check the sign of T. % % For the "up" and "down" macros, we again use the round(T psi), but do % have to check the sign anyhow. We could also have used the % round((T+0.5)psi) which requires a sign check too, but it costs a bit % more. It would have allowed though to share the same codebase for all % units, here we have to prepare some slightly different shared macros % for the first batch bp, nd, dd and the second batch mm, pc, nc, cc, % cm, in. % % Implementation % ============== % \def\texdimenfirstofone#1{#1}% {\catcode`p 12\catcode`t 12 \csname expandafter\endcsname\gdef\csname texdimenstrippt\endcsname#1pt{#1}}% % % down macros: % for units with phi < 2: \def\texdimendown_A#1{\if-#1\texdimendown_neg\fi\texdimendown_B#1}% \def\texdimendown_B#1;#2;{\expandafter\texdimendown_c\the\numexpr(2*#1+1)#2;#1;}% % for units with phi > 2: \def\texdimendown_a#1{\if-#1\texdimendown_neg\fi\texdimendown_b#1}% \def\texdimendown_b#1;#2;{\expandafter\texdimendown_c\the\numexpr#1#2;#1;}% % shared macros: \def\texdimendown_c#1;{\expandafter\texdimendown_d\the\dimexpr#1sp;#1;}% {\catcode`P 12\catcode`T 12\lowercase{\gdef\texdimendown_d#1PT};#2;#3;#4;% {\ifdim#1#4>#3sp \texdimendown_e{#2}\fi\texdimenfirstofone{#1}}% }% % this #2 will be \fi \def\texdimendown_e#1#2#3#4{#2\expandafter\texdimenstrippt\the\dimexpr\numexpr#1-1sp\relax}% % negative branch: % The problem here is that if input very small, output can be 0.0, and we % do not want -0.0 as output. % So let's do this somewhat brutally and non-efficiently. % Anyhow, negative inputs are not our priority. % #1 is \fi here and #2 is \texdimendown_b or _B: \def\texdimendown_neg#1#2-#3;#4;#5;{#1\expandafter\texdimenstrippt\the\dimexpr-#2#3;#4;#5;pt\relax}% % % up macros: \def\texdimenup_A#1{\if-#1\texdimenup_neg\fi\texdimenup_B#1}% \def\texdimenup_B#1;#2;{\expandafter\texdimenup_c\the\numexpr(2*#1+1)#2;#1;}% \def\texdimenup_a#1{\if-#1\texdimenup_neg\fi\texdimenup_b#1}% \def\texdimenup_b#1;#2;{\expandafter\texdimenup_c\the\numexpr#1#2;#1;}% \def\texdimenup_c#1;{\expandafter\texdimenup_d\the\dimexpr#1sp;#1;}% {\catcode`P 12\catcode`T 12\lowercase{\gdef\texdimenup_d#1PT};#2;#3;#4;% {\ifdim#1#4<#3sp \texdimenup_e{#2}\fi\texdimenfirstofone{#1}}% }% % this #2 will be \fi \def\texdimenup_e#1#2#3#4{#2\expandafter\texdimenstrippt\the\dimexpr\numexpr#1+1sp\relax}% % negative branch: % Here we can me more expeditive than for the "down" macros. % But this breaks f-expandability. % #1 will be \fi and #2 is \texdimenup_b or _B: \def\texdimenup_neg#1#2-{#1-#2}% % % pt % \def\texdimenpt#1{\expandafter\texdimenstrippt\the\dimexpr#1\relax}% % % bp 7227/7200 = 803/800 % \def\texdimenbp#1{\expandafter\texdimenbp_\the\numexpr\dimexpr#1;}% \def\texdimenbp_#1#2;{% \expandafter\texdimenstrippt\the\dimexpr\numexpr(2*#1#2+\if-#1-\fi1)*400/803sp\relax }% % \texdimenbpdown: maximal dim exactly expressible in bp and at most equal to input \def\texdimenbpdown#1{\expandafter\texdimendown_A\the\numexpr\dimexpr#1;*400/803;bp;}% % \texdimenbpup: minimal dim exactly expressible in bp and at least equal to input \def\texdimenbpup#1{\expandafter\texdimenup_A\the\numexpr\dimexpr#1;*400/803;bp;}% % % nd 685/642 % \def\texdimennd#1{\expandafter\texdimennd_\the\numexpr\dimexpr#1;}% \def\texdimennd_#1#2;{% \expandafter\texdimenstrippt\the\dimexpr\numexpr(2*#1#2+\if-#1-\fi1)*321/685sp\relax }% % \texdimennddown: maximal dim exactly expressible in nd and at most equal to input \def\texdimennddown#1{\expandafter\texdimendown_A\the\numexpr\dimexpr#1;*321/685;nd;}% % \texdimenndup: minimal dim exactly expressible in nd and at least equal to input \def\texdimenndup#1{\expandafter\texdimenup_A\the\numexpr\dimexpr#1;*321/685;nd;}% % % dd 1238/1157 % \def\texdimendd#1{\expandafter\texdimendd_\the\numexpr\dimexpr#1;}% \def\texdimendd_#1#2;{% \expandafter\texdimenstrippt\the\dimexpr\numexpr(2*#1#2+\if-#1-\fi1)*1157/2476sp\relax }% % \texdimendddown: maximal dim exactly expressible in dd and at most equal to input \def\texdimendddown#1{\expandafter\texdimendown_A\the\numexpr\dimexpr#1;*1157/2476;dd;}% % \texdimenddup: minimal dim exactly expressible in dd and at least equal to input \def\texdimenddup#1{\expandafter\texdimenup_A\the\numexpr\dimexpr#1;*1157/2476;dd;}% % % mm 7227/2540 phi now >2, use from here on the simpler approach % \def\texdimenmm#1{\expandafter\texdimenstrippt\the\dimexpr(#1)*2540/7227\relax}% % \texdimenmmdown: maximal dim exactly expressible in mm and at most equal to input \def\texdimenmmdown#1{\expandafter\texdimendown_a\the\numexpr\dimexpr#1;*2540/7227;mm;}% % \texdimenmmup: minimal dim exactly expressible in mm and at least equal to input \def\texdimenmmup#1{\expandafter\texdimenup_a\the\numexpr\dimexpr#1;*2540/7227;mm;}% % % pc 12/1 % \def\texdimenpc#1{\expandafter\texdimenstrippt\the\dimexpr(#1)/12\relax}% % \texdimenpcdown: maximal dim exactly expressible in pc and at most equal to input \def\texdimenpcdown#1{\expandafter\texdimendown_a\the\numexpr\dimexpr#1;/12;pc;}% % \texdimenpcup: minimal dim exactly expressible in pc and at least equal to input \def\texdimenpcup#1{\expandafter\texdimenup_a\the\numexpr\dimexpr#1;/12;pc;}% % % nc 1370/107 % \def\texdimennc#1{\expandafter\texdimenstrippt\the\dimexpr(#1)*107/1370\relax}% % \texdimenncdown: maximal dim exactly expressible in nc and at most equal to input \def\texdimenncdown#1{\expandafter\texdimendown_a\the\numexpr\dimexpr#1;*107/1370;nc;}% % \texdimenncup: minimal dim exactly expressible in nc and at least equal to input \def\texdimenncup#1{\expandafter\texdimenup_a\the\numexpr\dimexpr#1;*107/1370;nc;}% % % cc 14856/1157 % \def\texdimencc#1{\expandafter\texdimenstrippt\the\dimexpr(#1)*1157/14856\relax}% % \texdimenccdown: maximal dim exactly expressible in cc and at most equal to input \def\texdimenccdown#1{\expandafter\texdimendown_a\the\numexpr\dimexpr#1;*1157/14856;cc;}% % \texdimenccup: minimal dim exactly expressible in cc and at least equal to input \def\texdimenccup#1{\expandafter\texdimenup_a\the\numexpr\dimexpr#1;*1157/14856;cc;}% % % cm 7227/254 % \def\texdimencm#1{\expandafter\texdimenstrippt\the\dimexpr(#1)*254/7227\relax}% % \texdimencmdown: maximal dim exactly expressible in cm and at most equal to input \def\texdimencmdown#1{\expandafter\texdimendown_a\the\numexpr\dimexpr#1;*254/7227;cm;}% % \texdimencmup: minimal dim exactly expressible in cm and at least equal to input \def\texdimencmup#1{\expandafter\texdimenup_a\the\numexpr\dimexpr#1;*254/7227;cm;}% % % in 7227/100 % \def\texdimenin#1{\expandafter\texdimenstrippt\the\dimexpr(#1)*100/7227\relax}% % \texdimenindown: maximal dim exactly expressible in in and at most equal to input \def\texdimenindown#1{\expandafter\texdimendown_a\the\numexpr\dimexpr#1;*100/7227;in;}% % \texdimeninup: minimal dim exactly expressible in in and at least equal to input \def\texdimeninup#1{\expandafter\texdimenup_a\the\numexpr\dimexpr#1;*100/7227;in;}% % both in and cm % Mathematics ("both" macros) % =========== % % Let a and b be two non-negative integers such that U = floor(a 7227/100) = % floor(b 7227/254). It can be proven that a=50k, b=127k for some integer k. % The proof is left to reader. So U = floor(7227 k /2) for some k. % % Let's now find the largest such U <= T. So U = floor(k 7227/2)<= T which is % equivalent (as k is integer) to k 7227/2 <= T + 1/2, i.e. % % kmax = floor((2T+1)/7227) % % If we used for x>0 the formula floor(x)=round(x-1/2)= we would end % up basically with some 4T hence overflow problems even in \numexpr. % Here I used <.> to denote rounding in the sense of \numexpr. It is not % 1-periodical due to how negative inputs are handled, but here x-1/2>-1/2. % % The following lemma holds: let T be a non-negative integer then % % floor((2T+1)/7227) = <(2T - 3612)/7227> % % So we can compute this k, hence get a=50k, b=127k, all within \numexpr and % avoiding overflow. % % Implementation % ============== % % Regarding the output in pt or sp, we seem to need floor(k 7227/2). % The computation of floor(k 7227/2) as <(7227 k - 1)/2> would require to % check if k==0 so we do it rather as <(7227 k + 1)/2> - 1. No overflow % can arise as k = 297147 for \maxdimen, and then 7227 k = 2**31 - 2279 and % there is ample room for 7227k+1 using \numexpr. % % But this step, as well as initial step to get kmax will require to separate % handling of negative input from positive one. % % Alternative % ----------- % % For non-negative T we can compute U = ((T+1)/7227)*7227. If U <= T keep it, % else if U > T, replace it by U - 3614. This is alternative road to the maximal % floor(k 7227/2) at most equal to T. % % There is some slight under-efficiency to share macros across the 3 end targets % as I added one layer of parentheses. \def\texdimenbothincm#1{\expandafter\texdimenstrippt\the\dimexpr \expandafter\texdimenboth_a\the\numexpr\dimexpr#1;127);}% \def\texdimenbothcmin#1{\expandafter\texdimenstrippt\the\dimexpr \expandafter\texdimenboth_a\the\numexpr\dimexpr#1;50);}% \def\texdimenbothincmpt#1{\expandafter\texdimenstrippt\the\dimexpr \expandafter\texdimenboth_a\the\numexpr\dimexpr#1;7227+1)/2-1;}% \let\texdimenbothcminpt\texdimenbothincmpt \def\texdimenboth_a#1{\if-#1\texdimenboth_neg\fi\texdimenboth_b#1}% % The opening parenthesis ( is closed in #2, it was added to share "pt" output % with the two others \def\texdimenboth_b#1;#2;{\numexpr(((2*#1-3612)/7227)*#2sp\relax}% % negative branch. This is expanded in a \dimexpr so we can insert the - % in front of the \numexpr. % #1 is \fi here and #2 is \texdimenboth_b \def\texdimenboth_neg#1#2-#3;#4;{#1-\numexpr(((2*#3-3612)/7227)*#4sp\relax}% % % \texdimenbothincmsp is done separately as I found no easy way to share % its macros with the others; alternative would have been to make it the % core, and derive the others from it, (\texdimencm{\texdimenbothincmsp{...}sp}) % but then they would be less efficient than their current versions. % (it is a bit ironical to worry about not creating too many macros % in such a small package, by the way) \def\texdimenbothincmsp#1{\the\numexpr\expandafter\texdimenbothsp_a\the\numexpr\dimexpr#1;}% \def\texdimenbothsp_a#1{\if-#1\texdimenbothsp_neg\fi\texdimenbothsp_b#1}% \def\texdimenbothsp_b#1;{(((2*#1-3612)/7227)*7227+1)/2-1\relax}% % #1 is \fi % we need to regrab here or to add a \numexpr..\relax layer to % \texdimenbothsp_b (parentheses could do but using 0-(...) syntax) % finally doing the job of \texdimenbothsp_b directly \def\texdimenbothsp_neg#1#2-#3;{#1-\numexpr(((2*#3-3612)/7227)*7227+1)/2-1\relax\relax}% % \let\texdimenbothcminsp\texdimenbothincmsp % (\texdimenwithunit, added at 0.99) % Mathematics % =========== % % As explained in the README.md, the ex and em units are % handled by TeX as if multiplying by a conversion factor f/65536 % (here f sp = 1ex resp. = 1em). % In particular, for any decimal D, input "D em" is handled the exact % same way as input "D\dimexpr 1em\relax"; this is not % the case for the core units except for pt and pc (and sp), whose % conversion factors are the sole ones with a power of 2 denominator % (respectively 1, 1, and 65536). The further difference is that % for the core units apart from sp, the conversion factor is >1. % % We assume for this discussion T is non-negative. % If f/65536 > 1, the analysis is as above : some dimensions T sp % are not attainable as D uu, but the formula % N=round((2T+1)*32768/f) % will give a suitable decimal D via \the\dimexpr N sp\relax. % (if T=0, we get N=0 as 32768/f<0.5) % This D will let TeX convert D uu into T sp, if the dimension % is attainable else it will be a closest match % either from above or below (not necessarily nearest overall). % % If f/65536=1, attention that above formula would give N=1 for % T=0 (was bug #4). % % If f/65536<1, all dimensions Tsp are attainable as D uu. Indeed % D uu is parsed by TeX via N=round(D*65536), then T=trunc(N*phi), % with phi=f/65536. Starting from T we need to find an N such that % T/phi <= N< (T+1)/phi. We first consider v=(T+0.5)/phi. As its % distance to the extremities is 0.5/phi>0.5,(*) its rounding M % to an integer verifies automatically T/phi < M < (T+1)/phi, so % is a candidate. The TeX core conversion of Msp to a Dpt with % D a decimal of at most 5 fractional digits will provide a D % such that indeed M=round(D*65536). % % (*) attention that this fails if phi=1, we definitely can not % use the N=round((2T+1)*32768/f) formula for f=65536, it gives % N=T+1 in place of N=T... It is funny the formula works for all % f apart from f=65536... % % Attention that if T=0, the M=round(0.5/phi) will be at least 1, % so the produced decimal D will not be 0.0, although it will % be true that D \dimexpr f sp\relax produces the zero dimension, % the above analysis being correct even for T=0. % % Implementation % ============== % % \texdimenwithunit{dim1}{dim2}. dim2>0 assumed. % We first get f from dim2 and branch according to whether f>65536, % f=65536 or f<65536. % We will also need to check the sign of T (dim1=T sp). % We then compute in both branches round((2T+1)*32768/f), but % in an indirect way in the f<65536 branch to avoid overflow. % We will need also to intercept T=0 to avoid producing a non % pleasing positive D in that case (still correct for f<65536; % this is issue #3). % % In the f<65536 branch we first do the Euclidean division % 2T+1 = k*2*f + R with 0<=R<2f. The k is obtained as round((2T+1-f)/(2f)) % which can be computed in a numexpr (and never gives -1 even for T=0) % Then (2T+1)*32768/f=65536*k + R*32768/f % Then R*32768/f<=65536-32768/f<65536-32768/65536=65536-0.5 % Hence the numexpr evaluation gives an integer B<65536. % % N.B.: if f>=65536, we still have R*32768/f<65536 as R<2f % so the only difference is that the B could be here 65536 % % From \the\dimexpr Bsp, we get a decimal E < 1, so E=0.ddddd % (or less digits) and the looked for D will be the contatenation % k.ddddd with k as obtained earlier. This procedure has no possible % arithmetic overflow. % % #2 is assumed positive. The case #2=1pt=65536sp is special, and % must be filtered out (issues #4, #6). % pre-multiplication of #1 by 2 here \def\texdimenwithunit#1#2{\expandafter\texdimenwithunit_ \the\numexpr\dimexpr#2\expandafter;\the\numexpr2*\dimexpr#1;}% \let\texdimenorthat\texdimenfirstofone \def\texdimendothis#1#2\texdimenorthat#3{\fi#1}% \def\texdimenwithunit_#1;#2{% \ifnum#1=\p@\texdimendothis\texdimenwithunit_p@\fi \ifnum#1>\p@\texdimendothis\texdimenwithunit_A\fi \texdimenorthat\texdimenwithunit_B#2#1;% }% % we premultiplied the first argument by 2... must be undone now! % and we must avoid overflow. Very underoptimal, but user is not % supposed to do something as silly as actually using this unit=1pt case! \def\texdimenwithunit_p@#1#2;#3;{\expandafter\texdimenstrippt\the\dimexpr\numexpr#1#3/2sp\relax}% % unit>1pt, handle this as for bp. Attention it would be wrong for unit=1pt! \def\texdimenwithunit_A#1#2;#3;{\expandafter\texdimenstrippt \the\dimexpr\numexpr(#1#3+\if-#1-\fi1)*32768/#2sp\relax }% % unit<1pt. Attention it would be wrong for unit=1pt! % if dim1<0, simply negate result for dim1>0 as it can not possibly be 0.0 % Indeed (2T+1)*32768/f will be at least 3*32768/f so its rounding at least 2. % Add a check for dim1=0 to fix issue #3 \def\texdimenzerominusfork #10-#2#3\krof {#2}% \def\texdimenwithunit_B#1{\texdimenzerominusfork #1-\texdimenwithunit_Bzero 0#1\texdimenwithunit_Bneg 0-\texdimenwithunit_Ba \krof#1}% \def\texdimenwithunit_Ba#1#2;#3;{\expandafter\texdimenwithunit_Bb\the\numexpr#1#3+1;#2;}% \def\texdimenwithunit_Bb#1;#2;{\expandafter\texdimenwithunit_Bc\the\numexpr(#1-#2)/(2*#2);#1;#2;}% % I was hesitating between leaving k in the stream (breaking f-expandability) % and then remove the "0" and trailing "pt" from 0.ddddd pt, but opted % finally for doing \the\numexpr..+0.ddddd which is f-expandable and allows % recycling strippt here with no need of another utility. % % This means (see the nota bene above) that we could apply this procedure % also for f>65536, because at worst we will get a \the\numexprk+1.0, which % gives the correct result. I tested and found about 39% longer execution time % if dim2>1pt does same calculations as for dim2<1pt, and at the same time % dim2<1pt was improved about 11% from skipping the conditional \def\texdimenwithunit_Bc#1;#2;#3;{\the\numexpr#1+\expandafter\texdimenstrippt \the\dimexpr\numexpr(#2-#1*2*#3)*32768/#3sp\relax}% % Here, definitely not caring about f-expandability. Or efficiency. \def\texdimenwithunit_Bneg-{-\texdimenwithunit_Ba{}}% \def\texdimenwithunit_Bzero#1;#2;{0.0}% \texdimensendinput