如果你曾经用高级编程语言编写过程序，那么你一定熟悉函数的概念，即返回值的子程序。例如，如果你用 C 语言编写程序，那么你可以编写一个函数来计算整数参数的绝对值
int abs(int a) {
return a > 0 ? a: -a;
}
你可以通过以下方式调用此函数，并将其结果赋给一个变量，例如：
int v = abs(a)`
此外，您可以直接使用返回的值，而无需将其分配给变量，例如，
abs(abs(a)-abs(b))
使用 TeX 编程最接近这一点的是什么？挑战在于最小化调用者和被调用者之间的耦合。当然，由于 TeX 主要涉及字符和文本标记，因此更自然的例子是以下函数调用链：
int ispalindrome(const char *s) {
return strcmp(s, strrev(s)) == 0;
}
答案1
从计算机科学的角度来看，宏是函数。这里的问题是 TeX 没有提供许多数学和字符串处理指令。但假设我们有这些函数，您的abs函数可以直接转换为宏定义（TeXish 或 LaTeXish）：
\def\abs#1{%
\ifgreater{#1}{0}{#1}{\neg{#1}}}
（事实上，TeX 有\ifnum：
\def\abs#1{%
\ifnum#1>0\relax#1\else\neg{#1}\fi}
）
并且可以调用：
\abs{42}
并嵌套：
\abs{\minus{\abs{a}}{\abs{b}}}
答案2
编辑（三年后）回答这个问题导致了在 LaTeX3 中对浮点计算的完全重新实现。
TeX 中有两种类型的（复杂）宏：一类（“可扩展”宏），其纯粹通过扩展来工作，将其参数转换为其他东西，而不需要存储中间结果或以其他方式影响 TeX 机器的状态，另一类（“不可扩展”宏），其会改变状态。
TeX 本身只能使用不可扩展的整数进行计算，通过\advance、\multiply和\divide，本质上+=、*=和//=。这些赋值是不可扩展的。除了简单地执行2+3*4+5或之外，还\add{\add{2}{\mul{3}{4}}}{5}必须执行等同于
x = 2
y = 3
y *= 4
x += y
x += 5
使用中间变量，这是不切实际的。eTeX 原语\numexpr让我们可以评估涉及整数、、、、、、的可扩展表达式。(正如Joseph所描述的，它让我们可以定义类似函数的宏)+-*/
\def\add#1#2{\the\numexpr(#1)*(#2)\relax}
\def\abs#1{\the\numexpr\ifnum\numexpr#1<0-\fi(#1)\relax}
\add{\abs{-3}}{4} % = 7
许多软件包（pgfmath、fp、...）提供不可扩展的宏来计算浮点数，实用性各不相同。自去年以来，l3fp可扩展地评估浮点表达式。一个典型的例子：
\fp_eval:n { abs(-3)+4*ln(5.6e7) }
出于历史目的，我在下面留下了 l3fp 原型的代码。
编辑（2011）我曾编写了一个解析步骤，以便更加方便用户使用，但它有错误，所以我删除了它。我还彻底重写了代码几次，并实现了减法和除法。代码：
\documentclass{article}
\usepackage{expl3}
\ExplSyntaxOn
% ====
% Floating point numbers which we'll call ``<ZBfp>'', of the form
% <sign> {<int>}{<4d>}{<4d>}{<4d>}T{<int expr>}
% where
% <sign> is "+" or "-" (mandatory)
% <int> is at most 4 digits, and never equal to zero (except cases below)
% <4d> represents 4 digits exactly
% <int expr> is something that "\number\numexpr...\relax" is happy with.
%
% Exceptions:
% - The <sign> can also be "X" for undefined.
% X{0000}{0000}{0000}{0000}T{0}
% This is converted to a sign of "+" if we use it further in the calculation.
% +{0000}{0000}{0000}{0000}T{0}
%
% Example:
% + {0234}{2345}{3456}{4567}T{0} means 234.234534564567
% - {1111}{2222}{3333}{4444}T{2} means -111122223333.4444
%
% ==== Helpers
\cs_new:Npn \ZBfp_brace_last:nw #1 #2 #{#1{#2}}
\cs_new:Npn \ZBfp_brace_last:nnnnnw #1 #2 #3 #4 #5 #6 # { #1 #2 #3 #4 #5 {#6} }
\cs_new:Npn \use_Bi_iiE:nn #1 #2 { {#1 #2} }
\cs_new:Npn \use_iv_delimit_by_q_stop:nnnnw #1#2#3#4#5\q_stop {#4}
\tl_const:Nn \c_ZBfp_braced_plus_tl { {+} }
\tl_const:Nn \c_ZBfp_braced_minus_tl { {-} }
% Constants
\tl_const:Nn \c_zero_ZBfp {+{0000}{0000}{0000}{0000}T{0}}
\tl_const:Nn \c_undefined_ZBfp {X{0000}{0000}{0000}{0000}T{0}}
% ==== Adding mantissas.
%
% Used as
% \ZBfp_add_mantissa:nnnnnnnn {<4d>}{<4d>}{<4d>}{<4d>}
% {<4d>}{<4d>}{<4d>}{<4d>} {<sign>} T {<exp>}
% to get
% <sign> {<4d>}{<4d>}{<4d>}{<4d>} T{<exp>}
%
% We split numbers efficiently by pre-carrying.
% Note that all the extra "-1", "+9999", "10000" sum up to zero.
% Also, it would be more efficient to replace those constants by registers?
\cs_new:Npn \ZBfp_add_mantissa:nnnnnnnn #1#2#3#4 #5#6#7#8{
\exp_after:wN \ZBfp_add_mantissa_aux:N
\tex_the:D \etex_numexpr:D -1 + \exp_after:wN \ZBfp_brace_last:nw
\tex_the:D \etex_numexpr:D #1 + #5 + 9999 + \exp_after:wN \ZBfp_brace_last:nw
\tex_the:D \etex_numexpr:D #2 + #6 + 9999 + \exp_after:wN \ZBfp_brace_last:nw
\tex_the:D \etex_numexpr:D #3 + #7 + 9999 + \exp_after:wN \ZBfp_brace_last:nw
\tex_the:D \etex_numexpr:D #4 + #8 + 10000
}
% Important: that "#8" above is followed by a brace group, \marg{sign}.
% We are left with
% "\ZBfp_add_mantissa_aux:N" <1d>{4d}{4d}{4d}{4d}{<sign>}T{<exp>}
% where <1d> is either "0" or "1", the overall carry.
\cs_new:Npn \ZBfp_add_mantissa_aux:N #1 {
\if_meaning:w 0 #1
\exp_after:wN \ZBfp_add_mantissa_aux_z:nnnnnnw
\else:
\exp_after:wN \ZBfp_add_mantissa_aux_i:nnnnnnw
\fi:
#1
}
% If the first digit is "0", we discard it and keep four brace groups.
% Otherwise, we keep it in a brace group, together with three more.
% In both cases, we place the sign and the exponent at the right place.
% Note that the exponent is not evaluated.
\cs_new:Npn \ZBfp_add_mantissa_aux_z:nnnnnnw #1 #2 #3 #4 #5 #6 T #7 {
#6 {#2}{#3}{#4}{#5} T{#7} }
\cs_new:Npn \ZBfp_add_mantissa_aux_i:nnnnnnw #1 #2 #3 #4 #5 #6 T #7 {
#6 {000 #1}{#2}{#3}{#4} T{#7+1} }
% ===== Subtracting mantissa
% Used as
% \ZBfp_sub_back_mantissa:nnnnnnnn <A> <B> {<sign>} T {<exp>}
% where <A> and <B> have the form {<4d>}{<4d>}{<4d>}{<4d>}. This
% calculates $<sign>*( - <A> + <B> )$ (hence the name "back") and
% f-expands to the result in the form
% <sign> {<4d>}{<4d>}{<4d>}{<4d>} T{<exp>}
%
% Again, pre-carrying. We check after the calculation if the
% subtraction should have been done in the other direction.
% Thus, doing subtraction as - <big mantissa> + <small mantissa>
% should be twice slower than the other way around, but that's
% the price to pay to get a more optimized subtraction
% - <small> + <big>, used more frequently in the ``full''
% subtraction and in sines.
%
% We use bigger shifts that we had for addition, because we want
% to ensure that every intermediate numexpr we use has five digits
% for "\ZBfp_brace_last:nw" to work.
\cs_new:Npn \ZBfp_sub_back_mantissa:nnnnnnnn #1#2#3#4 #5#6#7#8 {
\exp_after:wN \ZBfp_sub_back_mantissa_aux:w
\tex_the:D \etex_numexpr:D #5 - #1 - 2 + \exp_after:wN \ZBfp_brace_last:nw
\tex_the:D \etex_numexpr:D #6 - #2 + 19998 + \exp_after:wN \ZBfp_brace_last:nw
\tex_the:D \etex_numexpr:D #7 - #3 + 19998 + \exp_after:wN \ZBfp_brace_last:nw
\tex_the:D \etex_numexpr:D #8 - #4 + 20000
}
% Again, following "#8" is a brace group, \marg{sign}.
% We are left with
% "\ZBfp_sub_back_mantissa_aux:w" <w>{4d}{4d}{4d}{<sign>}T{<exp>},
% where <w> is any integer between -9999 and +9999. Here we need to take
% care of the fact that e.g. -11{2222}{2222}{2222} means
% "-11 + .222222222222", not "-11.222222222222"! Convince yourself
% that if <w> is zero, then we are in the ``positive'' case.
\cs_new:Npn \ZBfp_sub_back_mantissa_aux:w #1 #{
\if_num:w #1 < 0 \exp_stop_f:
\exp_after:wN \ZBfp_sub_back_mantissa_neg:nnnnn
\else:
\exp_after:wN \ZBfp_sub_back_mantissa_pos:nnnn
\fi:
{#1}
}
% \ZBfp_sub_back_mantissa_neg:nnnnn {-<4d>}{<4d>}{<4d>}{<4d>} {<sign>} T {<exp>}
% If the result of the subtraction was negative, then we need
% to correct it and correct the sign.
%
% > "#1", "#2", "#3", "#4" form the body of the number
% > "#5" is the <sign>
% > "#6" would be the exponent expression
%
% Here, we know that "#1" is negative. In fact, our number is
% equal to <sign>"(#1 + .#2#3#4)". We calculate "|#1|-.#2#3#4",
% which is positive, and place the correct sign (namely, $-$<sign>)
% in a brace group after it, then pass the whole thing to
% "\ZBfp_add_mantissa_pos:nnnnnw".
%
% Once more, pre-carry, to make sure everything has 5 digits when passed
% to "\ZBfp_brace_last:nw".
%
% On the first piece below, "-2 - #1 + ...". From the level below
% (i.e. the "..."), we get "1", or sometimes "2", and we know that
% "10000<#1<0", so the whole thing is "0<=...<10000", at most 4 digits.
%
\cs_new:Npn \ZBfp_sub_back_mantissa_neg:nnnnn #1 #2 #3 #4 #5 {
\exp_after:wN \ZBfp_brace_last:nw
\exp_after:wN \ZBfp_sub_back_mantissa_pos:nnnn
\tex_the:D \etex_numexpr:D -2 - #1 + \exp_after:wN \ZBfp_brace_last:nw
\tex_the:D \etex_numexpr:D 19998 - #2 + \exp_after:wN \ZBfp_brace_last:nw
\tex_the:D \etex_numexpr:D 19998 - #3 + \exp_after:wN \ZBfp_brace_last:nw
\tex_the:D \etex_numexpr:D 20000 - #4
\if:w -#5
\exp_after:wN \c_ZBfp_braced_plus_tl
\else:
\exp_after:wN \c_ZBfp_braced_minus_tl
\fi:
}
% Note that the final line gets fully expanded to "{+}" or "{-}"
% when \TeX\ is trying to finish the number "#4". Specifically, the sign
% we get is the opposite of "#5". Any faster test (braces are important)?
% "\ZBfp_sub_back_mantissa_pos:nnnn" {<4d>}{<4d>}{<4d>}{<4d>} {<sign>} T {<exp>}
% Once we have taken care of the signs, we need to check whether
% the first, second, etc. pieces is/are zero, and shift exponents
% accordingly.
% > "#1", "#2", "#3", "#4" form the body of the number,
% > "#5" is the sign
% > "#7" is the exponent
\cs_new:Npn \ZBfp_sub_back_mantissa_pos:nnnn #1 #2 #3 #4 {
\cs:w ZBfp_sub_maux_
\if_num:w #1 = 0 \exp_stop_f: i
\if_num:w #2 = 0 \exp_stop_f: i
\if_num:w #3 = 0 \exp_stop_f: i
\if_num:w #4 = 0 \exp_stop_f: i
\fi:
\fi:
\fi:
\fi:
:w
\cs_end:
{#1}
{#2}
{#3}
{#4}
}
% Just reformat digits in the right way.
\cs_new:Npx\ZBfp_sub_maux_iiii:w #1#2#3#4 #5T#6 {\c_zero_ZBfp}
\cs_new:Npn\ZBfp_sub_maux_iii:w#1#2#3#4 #5T#6 {#5 {#4}{0000}{0000}{0000}T{#6-3}}
\cs_new:Npn\ZBfp_sub_maux_ii:w #1#2#3#4 #5T#6 {#5 {#3} {#4} {0000}{0000}T{#6-2}}
\cs_new:Npn\ZBfp_sub_maux_i:w #1#2#3#4 #5T#6 {#5 {#2} {#3} {#4} {0000}T{#6-1}}
\cs_new:Npn\ZBfp_sub_maux_:w #1#2#3#4 #5T#6 {#5 {#1} {#2} {#3} {#4} T {#6} }
% ======= Full addition and subtraction.
% Syntax:
% "\ZBfp_add:nn" \marg{ZBfp expr.} \marg{ZBfp expr.}
% "\ZBfp_sub:nn" \marg{ZBfp expr.} \marg{ZBfp expr.}
% where <ZBfp> are floating point expressions, which will be
% fully f-expanded before the calculation.
% Plan:
% > expand the left <ZBfp expr.>
% > swap, and expand the right <ZBfp expr.>
% > check signs
% > shift digits according to exponent
\cs_new:Npn \ZBfp_add:nn #1 {
\exp_after:wN \ZBfp_add_aux:NwNn
\tex_romannumeral:D -`\0 #1 \q_stop +
}
\cs_new:Npn \ZBfp_sub:nn #1 {
\exp_after:wN \ZBfp_add_aux:NwNn
\tex_romannumeral:D -`\0 #1 \q_stop -
}
% We now have
% "\ZBfp_add_aux:NwNn" <sign A> <A> T {<eA>} \q_stop <operation> {<expr B>}
% Here <sign A> <A> T {<eA>} is our first floating point number:
% <A> is the body, {<4d>}{<4d>}{<4d>}{<4d>}, and <eA> is the exponent.
% Also, <operation> is "+" or "-".
% We then move <expr B> to the front and f-expand it.
\cs_new:Npn \ZBfp_add_aux:NwNn #1 #2 \q_stop #3 #4 {
\exp_after:wN \ZBfp_add_signs:NNN \exp_after:wN #1 \exp_after:wN #3
\tex_romannumeral:D -`\0 #4 #2
}
% We now have
% "\ZBfp_add_aux:NwNn" <sign A> <operation> <sign B> <B>T{<eB>} <A>T{<eA>}
% If the product <sign A><operation><sign B> is +,
% use "\ZBfp_add_exponent:w", else "\ZBfp_sub_exponent:w".
\cs_new:Npn \ZBfp_add_signs:NNN #1 #2 #3 {
\cs:w ZBfp_
\if:w #1 \if:w #2 #3 + \else: - \fi: add \else: sub \fi:
_exponent:w \cs_end:
#1
}
% We now have "\ZBfp_..._exponent:w" <sign A> <B>T{<eB>} <A>T{<eA>}
% Take care of exponents. If they are the same, do nothing.
% If they differ, shift in the relevant direction using
% "\ZBfp_decimate_do". We first check the equality case
% because we will arrange to be in that case for our calculations
% of sine and cosine (which we want to optimize).
\cs_new:Npn \ZBfp_add_exponent:w #1 #2 T #3 #4 T #5 {
\if_num:w #3 > #5 \exp_stop_f:
\exp_after:wN \use_i:nn
\else:
\exp_after:wN \use_ii:nn
\fi:
{
\exp_after:wN \ZBfp_decimate_do:nNnnnn \exp_after:wN {
\tex_the:D \etex_numexpr:D #3 - #5 }
\ZBfp_add_mantissa:nnnnnnnn #4 #2 {#1} T {#3}
}
{
\exp_after:wN \ZBfp_decimate_do:nNnnnn \exp_after:wN {
\tex_the:D \etex_numexpr:D #5 - #3 }
\ZBfp_add_mantissa:nnnnnnnn #2 #4 {#1} T {#5}
}
}
% "\ZBfp_decimate_do:nNnnnn {<shift>} <func> {<4d>}{<4d>}{<4d>}{<4d>}
% will shift the blocks of digits by <shift> blocks to the right,
% adding "{0000}" blocks at the start when shifting, then place
% <func> in front of 4 blocks of 4 digits. Requires $<shift> \geq 0$.
% When shifting results in a zero number, instead of performing the
% computation, we simply remove <func> and the 4 blocks of digits,
% presumably leaving only the other operand on the input stream.
% In other words, if the difference in exponent is ${}>3$, the small
% number is to small ($10000^{-4}$), and we throw it away.
\cs_new:Npn \ZBfp_decimate_do:nNnnnn #1 {
\if_case:w #1 \exp_stop_f:
% zero shift: do nothing.
\or: \exp_after:wN \ZBfp_decimate_i:Nnnnn
\or: \exp_after:wN \ZBfp_decimate_ii:Nnnnn
\or: \exp_after:wN \ZBfp_decimate_iii:Nnnnn
\else: \exp_after:wN \use_none:nnnnn
\fi:
}
% "\ZBfp_decimate_1:w" <func> {<4d>} {<4d>} {<4d>} {<4d>}
\cs_new:Npn \ZBfp_decimate_i:Nnnnn #1 #2#3#4#5 {#1 {0000} {#2} {#3} {#4}}
\cs_new:Npn \ZBfp_decimate_ii:Nnnnn #1 #2#3#4#5 {#1 {0000}{0000} {#2} {#3}}
\cs_new:Npn \ZBfp_decimate_iii:Nnnnn #1 #2#3#4#5 {#1 {0000}{0000}{0000}{#2}}
% "\ZBfp_sub_exponent:w" <sign A> <B>T{<eB>} <A>T{<eA>}
% For subtraction, signs are tricky. Given the above, we want to
% calculate $<sign A>*(<A>*10000^{<eA>} - <B>*10000^{<eB>})$.
% \begin{itemize}
% \item
% If $ <eB> \leq <eA> $, then we will ``decimate'' <B> (possibly
% with a zero <shift> when there is equality), and essentially
% calculate "sub_back" <B'> <A> <sign A> T {<eA>}, where <B'> is
% the result of decimating. That gives
% $ <sign A> * (- <B> + <A>) * 10000^{<eA>} $.
% \item
% Otherwise, we need to change the sign. Unfortunately, that sign
% has to be sent quite deep (just before the "T" of the exponent).
% Maybe that could be improved, but right now, we first calculate
% the correct sign and place it at the right spot using
% "\ZBfp_sub_exponent_aux:Nw". Some will complain that the
% conditional is not terminated when placing the sign, but no worry,
% it gets terminated since we just continue executing things after
% placing the sign.
%
% Then decimate <A>. And calculate "sub_back" <A'> <B> {- <sign A>} T {<eB>},
% which is equal to $ - <sign A> *(-<A>+<B>) * 10000^{<eB>}$.
% \end{itemize}
\cs_new:Npn \ZBfp_sub_exponent:w #1 #2 T #3 #4 T #5 {
\if_num:w #3 > #5 \exp_stop_f:
\exp_after:wN \use_i:nn
\else:
\exp_after:wN \use_ii:nn
\fi:
{
\exp_after:wN \ZBfp_sub_exponent_aux:Nw \if:w -#1 + \else: - \fi:
\exp_after:wN \ZBfp_decimate_do:nNnnnn \exp_after:wN {
\tex_the:D \etex_numexpr:D #3 - #5 }
\ZBfp_sub_back_mantissa:nnnnnnnn #4 #2 T {#3}
}
{
\exp_after:wN \ZBfp_decimate_do:nNnnnn \exp_after:wN {
\tex_the:D \etex_numexpr:D #5 - #3 }
\ZBfp_sub_back_mantissa:nnnnnnnn #2 #4 {#1} T {#5}
}
}
\cs_new:Npn \ZBfp_sub_exponent_aux:Nw #1 #2 T {#2 {#1} T}
% ===== Multiplication
% Syntax:
% "\ZBfp_mul:nn" \marg{ZBfp expr.} \marg{ZBfp expr.}
% First expand the first expression, then swap, and expand the
% second expression. Then treat exponents and place the sign at
% the right place.
\cs_new:Npn \ZBfp_mul:nn #1 {
\exp_after:wN \ZBfp_mul_aux:Nwn
\tex_romannumeral:D -`\0 #1 \q_stop
}
\cs_new:Npn \ZBfp_mul_aux:Nwn #1 \q_stop #2 {
\exp_after:wN \ZBfp_mul_signs_expo:NwnNwn
\tex_romannumeral:D -`\0 #2 #1
}
% The signs are "#1" and "#4", "#3" and "#6" are the exponents, and
% each "#2" and "#5" are 16 digits split in blocks of 4.
\cs_new:Npn \ZBfp_mul_signs_expo:NwnNwn #1 #2 T#3 #4 #5 T#6 {
\exp_after:wN \ZBfp_mul_mantissa:wnnnnnnnn
\exp_after:wN #1
\exp_after:wN #4
\exp_after:wN T
\exp_after:wN {\tex_the:D \etex_numexpr:D #3 + #6}
\q_stop
#2 #5
}
% Now we have
% "\ZBfp_mul_mantissa:wnnnnnnnn" <sign><sign> T {<exp>} \q_stop
% {<4d>}{<4d>}{<4d>}{<4d>} {<4d>}{<4d>}{<4d>}{<4d>}
% TODO: Round instead of truncating, by adapting one of the "99990000" shifts
% to something like "50000000" and review "\ZBfp_mul_mantissa_aux_signs"
% once that is done.
\cs_new:Npn \ZBfp_mul_mantissa:wnnnnnnnn #1\q_stop #2#3#4#5 #6#7#8#9 {
\exp_after:wN \ZBfp_mul_mantissa_aux:w
\tex_the:D \etex_numexpr:D -10000 +
\exp_after:wN \ZBfp_brace_last:nnnnnw
\tex_the:D \etex_numexpr:D 99990000 + #2*#6 +
\exp_after:wN \ZBfp_brace_last:nnnnnw
\tex_the:D \etex_numexpr:D 99990000 + #2*#7+#3*#6+
\exp_after:wN \ZBfp_brace_last:nnnnnw
\tex_the:D \etex_numexpr:D 99990000 + #2*#8+#3*#7+#4*#6+
\exp_after:wN \ZBfp_brace_last:nnnnnw
\tex_the:D \etex_numexpr:D 99990000 + #2*#9+#3*#8+#4*#7+#5*#6 +
\exp_after:wN \ZBfp_brace_last:nnnnnw
\tex_the:D \etex_numexpr:D 99990000 + #3*#9+#4*#8+#5*#7+
\exp_after:wN \ZBfp_brace_last:nnnnnw
\tex_the:D \etex_numexpr:D 99990000 + #4*#9+#5*#8 +
\exp_after:wN \ZBfp_brace_last:nnnnnw
\tex_the:D \etex_numexpr:D 100000000 + #5*#9
\ZBfp_mul_mantissa_aux_signs:NN #1
}
\cs_new:Npn \ZBfp_mul_mantissa_aux_signs:NN #1 #2 {
\if:w #1 #2
\exp_after:wN \use_i:nn
\else:
\exp_after:wN \use_ii:nn
\fi:
{ {} \q_stop {+} }
{ {} \q_stop {-} }
}
% We are left with
% "\ZBfp_mul_mantissa_aux:w
% <int>{4d}{4d}{4d}{4d}...{4d}{}\q_stop{<sign>}T{<exp>}"
% where <int> (positive) may range from 0 to 9999. If it is zero,
% we need to remove it and shift the exponent. But we know that
% <int> is in its simplest form, since it comes from a "numexpr"
% calculation, so <int>=0 is equivalent to having its first character
% equal to 0 (I guess "\if_meaning:w" is quicker than "\if_num:w"
% because it involves no expansion).
\cs_new:Npn \ZBfp_mul_mantissa_aux:w #1 #{
\if_meaning:w 0 #1
\exp_after:wN \ZBfp_mul_mantissa_aux_z:nnnnnnww
\else:
\exp_after:wN \ZBfp_mul_mantissa_aux_i:nnnnnww
\fi:
{#1}
}
\cs_new:Npn \ZBfp_mul_mantissa_aux_z:nnnnnnww #1 #2 #3 #4 #5 #6 \q_stop #7 T#8 {
#7 {#2}{#3}{#4}{#5}T{#8}}
\cs_new:Npn \ZBfp_mul_mantissa_aux_i:nnnnnww #1 #2 #3 #4 #5 \q_stop #6 T#7 {
#6 {#1}{#2}{#3}{#4}T{#7+1}}
% ===== Division [not tested properly!]
% Syntax:
% "\ZBfp_div:nn" \marg{ZBfp expr.} \marg{ZBfp expr.}
% First expand the first expression, then swap, and expand the
% second expression. Then treat exponents and place the sign at
% the right place.
\cs_new:Npn \ZBfp_div:nn #1 {
\exp_after:wN \ZBfp_div_aux:Nwn
\tex_romannumeral:D -`\0 #1 \q_stop
}
\cs_new:Npn \ZBfp_div_aux:Nwn #1 \q_stop #2 {
\exp_after:wN \ZBfp_div_check_zero:ww
\tex_romannumeral:D -`\0 #2 \q_stop #1 \q_stop
}
% First let's check if the numerator or denominator is zero.
% Note that "#1" is the denominator!
\cs_new:Npn \ZBfp_div_check_zero:ww #1 \q_stop #2 \q_stop {
\ZBfp_if_zero_aux:NnnnnwNN #1 \use_i_delimit_by_q_stop:nw \use_none:n
{\c_undefined_ZBfp}
\ZBfp_if_zero_aux:NnnnnwNN #2 \use_i_delimit_by_q_stop:nw \use_none:n
{\c_zero_ZBfp}
\ZBfp_div_i:Nn #1 #2
\q_stop
}
% We strip leading zeros from the denominator (specifically <B0>).
\cs_new:Npn \ZBfp_div_i:Nn #1 #2 {
\exp_after:wN \ZBfp_div_ii:NnnnnwNww
\exp_after:wN #1
\exp_after:wN {\tex_number:D #2}
}
% "\ZBfp_div_ii:NnnnnwNww"
% <Bsign> {<trimmed B0>}{<B1>}{<B2>}{<B3>} T {<Bexp>}
% <Asign> {<A0>}{<A1>}{<A2>}{<A3>} T {<Aexp>} \q_stop
% We grab the whole body of <A> as one argument.
\cs_new:Npn \ZBfp_div_ii:NnnnnwNww #1 #2#3#4#5T#6 #7 #8T#9 \q_stop{
\exp_after:wN \ZBfp_div_iii:nnnnnnnnn \tex_number:D #2 \exp_stop_f:
#3 #4 #5 0000\q_stop % <B>, unpacked
#8 % <A>, in blocks of 4
#1 #7 % signs
T {#9 - (#6)} % exponent
Z {\use_none:nnnn #2 000} % extra zeros compensating the power.
}
% Repack <B>.
\cs_new:Npn \ZBfp_div_iii:nnnnnnnnn #1#2#3#4#5#6#7#8#9 {
\ZBfp_div_iiii:nnnnnw {{#1#2#3#4#5}{#6#7#8#9}}
}
\cs_new:Npn \ZBfp_div_iiii:nnnnnw #1 #2#3#4#5 #6 \q_stop {
\ZBfp_div_v:nwnn #1 {#2#3#4#5} \q_stop
}
% Now we have
% "\ZBfp_div_v:nwnn" {<B'0>}{<B'1>}{<B'2>} \q_stop
% {<A0>}{<A1>}{<A2>}{<A3>} <Bsign><Asign> T{<exp>} Z{<zeros>}
%
% where each of <Ai> and <B'i> have 4 digits, except <B'0>, which
% has exactly 5 digits.
%
% We will now use as a quotient
% \[
% <Q0> = \operatorname{Round} \frac{<A0><A1>}{<B'0>+1} - 1.
% \]
% The extra $+1$ and $-1$ are needed to make sure that everything
% stays positive.\footnote{Better ideas are welcome!} The numerator
% of <Q0> is between $10^4$ and $10^8$. The denominator is
% between $10^4$ and $10^5$. So $-1\leq <Q0> \leq 10^4$.
%
\cs_new:Npn \ZBfp_div_v:nwnn #1 #2 \q_stop #3 #4 {
\exp_after:wN \ZBfp_div_vi:nnnnnnn \exp_after:wN
{
\tex_the:D \etex_numexpr:D #3#4 / (#1+1) - 1
}
{#1} #2 {#3 #4}
}
% We calculate $<A> - <Q0>*<B'>$.
%
% "\ZBfp_div_vi:nnnnnnn" {<Q0>} {<B'0>}{<B'1>}{<B'2>}
% {<A0><A1>}{<A2>}{<A3>} <Bsign><Asign> T{<exp>} Z{<zeros>}
%
\cs_new:Npn \ZBfp_div_vi:nnnnnnn #1 #2#3#4 #5#6#7 {
\exp_after:wN \ZBfp_brace_last:nw
\exp_after:wN \ZBfp_div_vii:nnn
\tex_the:D \etex_numexpr:D (#5 - #1*#2)*10000 + #6 - #1*#3 - 20000 +
\exp_after:wN
\ZBfp_brace_last:nnnnnw
\tex_the:D \etex_numexpr:D 200000000 + #7 - #1*#4
{#2}{#3}{#4} {#1}
}
% The term in parentheses is
% \begin{align*}
% <A0><A1> - <Q0>*<B'0>
% &= <A0><A1> - \frac{<B'0>}{<B'0>+1} <A0><A1> + \alpha <B'0>
% \\
% &= \frac{<A0><A1>}{<B'0>+1} + \alpha <B'0>
% \end{align*}
% where $1/2 \leq \alpha \leq 3/2$ (because of e\TeX's rounding
% and our $-1$). This is at most $10^4 + 1.5*<B'0> \leq 1.5\cdot 10^5$.
% Multiplying by $10^4$ won't overflow (but it can be quite close).
%
% All in all, the result should be some long number <s1s2> obeying
% $<s1s2> \leq 15000\cdot<B'0>$ (more or less).
% "\ZBfp_div_vii:nnn" {<s1s2>}{<s3>} {<B'0>}{<B'1>}{<B'2>} {<Q0>}
% <Bsign><Asign> T{<exp>} Z{<zeros>}
%
% We need to be very careful, because <s1s2> can be up to $1.5*10^9$,
% close to overflow. Calculate the next quotient
% \[
% <Q1> = \operatorname{Round} \frac{<s1s2>}{<B'0> + 1} - 1.
% \]
% A few lines above, we get $<s1s2> \leq 15000\cdot <B'0> + 10^8$,
% so $<Q1> \leq 15000 + 10^8/<B'0> \leq 25000$ or so.
%
\cs_new:Npn \ZBfp_div_vii:nnn #1 #2 #3 {
\exp_after:wN \ZBfp_div_viii:nnnnnnn \exp_after:wN
{
\tex_the:D \etex_numexpr:D #1/(#3 + 1) - 1
}
{#1} {#2} {#3}
}
% "\ZBfp_div_viii:nnnnnnn" {<Q1>} {<s1s2>}{<s3>} {<B'0>}{<B'1>}{<B'2>}
% {<Q0>} <Bsign><Asign> T{<exp>} Z{<zeros>}
%
% We now add one level down, to keep enough precision.
% Not sure I'm doing that rigt.
%
\cs_new:Npn \ZBfp_div_viii:nnnnnnn #1 #2#3 #4#5#6 #7 {
\exp_after:wN \ZBfp_brace_last:nw \exp_after:wN \ZBfp_div_ix:nnn
\tex_the:D \etex_numexpr:D (#2 - #1*#4)*10000 + #3 - #1*#5 - 30000 +
\exp_after:wN \ZBfp_brace_last:nnnnnw
\tex_the:D \etex_numexpr:D 300000000 - #1*#6
{#4} {#5} {#7} {#1}
}
% we won't need <B'2> anymore, so we drop "#6".
% "\ZBfp_div_ix:nnn" {<t1t2>}{<t3>} {<B'0>}{<B'1>} {<Q0>}{<Q1>}
% <Bsign><Asign> T{<exp>} Z{<zeros>}
\cs_new:Npn \ZBfp_div_ix:nnn #1#2 #3 {
\exp_after:wN \ZBfp_div_x:nnnnnnn \exp_after:wN
{
\tex_the:D \etex_numexpr:D #1/(#3 + 1) - 1
}
{#1} {#2} {#3}
}
% "\ZBfp_div_x:nnnnnnn" {<Q2>} {<t1t2>}{<t3>} {<B'0>}{<B'1>} {<Q0>}{<Q1>}
% <Bsign><Asign> T{<exp>} Z{<zeros>}
\cs_new:Npn \ZBfp_div_x:nnnnnnn #1 #2#3 #4#5 #6#7 {
\exp_after:wN \ZBfp_brace_last:nw \exp_after:wN \ZBfp_div_xi:nnnnnww
\tex_the:D \etex_numexpr:D (#2 - #1*#4)*10000 + #3 - #1*#5
{#4} {#6}{#7}{#1}
}
% "\ZBfp_div_xi:nnnnn" {<u>} {<B'0>} {<Q0>}{<Q1>}{<Q2>}
% <Bsign><Asign> T{<exp>} Z{<zeros>}
\cs_new:Npn \ZBfp_div_xi:nnnnnww #1 #2 #3 #4 #5 #6 T#7 {
\exp_after:wN \ZBfp_div_xii:nnnnw
\tex_romannumeral:D \etex_numexpr:D -1 + \exp_after:wN \ZBfp_brace_last:nw
\tex_the:D \etex_numexpr:D 9999 + #3 + \exp_after:wN \ZBfp_brace_last:nw
\tex_the:D \etex_numexpr:D 9999 + #4 + \exp_after:wN \ZBfp_brace_last:nw
\tex_the:D \etex_numexpr:D 9999 + #5 + \exp_after:wN \ZBfp_brace_last:nw
\tex_the:D \etex_numexpr:D 10000 + #1 / #2
\exp_after:wN {\tex_the:D \etex_numexpr:D #7\exp_after:wN} #6
}
% "\ZBfp_div_xii:nnnnw" {<Q0>}{<Q1>}{<Q2>}{<Q3>} {<exp>}
% <Bsign><Asign> Z{<zeros>}
% now the <zeros> are explicitly "000" or "00" or "0" or " ",
% and each <Qi> is 4 digits.
\cs_new:Npn \ZBfp_div_xii:nnnnw #1#2#3#4 #5 Z#6 {
\exp_after:wN \ZBfp_div_repack_i:nnnnnnnn #6 #1#2#3#4 0000\q_stop #5
}
% "\ZBfp_div_repack_i:nnnnnnnn" <20-23 digits> \q_stop
\cs_new:Npn \ZBfp_div_repack_i:nnnnnnnn #1#2#3#4 #5#6#7#8 {
\if_num:w #1#2#3#4 = \c_zero
\exp_after:wN \ZBfp_div_repack_ii_small:nnNNNNNNN
\else:
\exp_after:wN \ZBfp_div_repack_ii_large:nnNNNNNNN
\fi:
{#1#2#3#4} {#5#6#7#8}
}
\cs_new:Npn \ZBfp_div_repack_ii_large:nnNNNNNNN #1 #2 #3#4#5#6 #7#8#9 {
\ZBfp_div_repack_iii_large:nnnnNw {#1} {#2} {#3#4#5#6} {#7#8#9}
}
\cs_new:Npn \ZBfp_div_repack_iii_large:nnnnNw #1 #2 #3 #4#5 #6\q_stop #7 #8#9{
\if:w #8#9
\exp_after:wN +
\else:
\exp_after:wN -
\fi:
{#1}{#2}{#3}{#4#5} T{#7}
}
\cs_new:Npn \ZBfp_div_repack_ii_small:nnNNNNNNN #1 #2 #3#4#5#6 #7#8#9 {
\ZBfp_div_repack_iii_small:nnnNNNNNw {#2} {#3#4#5#6} {#7#8#9}
}
\cs_new:Npn \ZBfp_div_repack_iii_small:nnnNNNNNw #1 #2 #3#4 #5#6#7#8 #9\q_stop{
\ZBfp_div_repack_iv_small:nnNN { {#1} {#2} {#3#4} {#5#6#7#8} }
}
% "\ZBfp_div_repack_iv_small:nnNN" {<digits>} {<exp>} <sign><sign>
\cs_new:Npn \ZBfp_div_repack_iv_small:nnNN #1 #2 #3#4 {
\if:w #3#4
\exp_after:wN +
\else:
\exp_after:wN -
\fi:
#1 T{#2-1}
}
% ===== Absolute value
\cs_new:Npn \ZBfp_abs:n #1 {
\exp_after:wN \ZBfp_abs_aux:N \tex_romannumeral:D -`\0 #1
}
\cs_new:Npn \ZBfp_abs_aux:N #1 {+}
% ===== Negate
\cs_new:Npn \ZBfp_neg:n #1 {
\exp_after:wN \ZBfp_neg_aux:N \tex_romannumeral:D -`\0 #1
}
\cs_new:Npn \ZBfp_neg_aux:N #1 {
\if:w - #1
\exp_after:wN +
\else:
\exp_after:wN -
\fi:
}
% ===== Test for zero
% All of the functions below are based on one auxiliary function.
\cs_new:Npn \ZBfp_if_zero_aux:NnnnnwNN #1 #2#3#4#5 T#6 #7#8 {
\if_num:w \etex_numexpr:D #2#3+#4#5 = \c_zero
\exp_after:wN #7
\else:
\exp_after:wN #8
\fi:
}
\prg_new_conditional:Npnn \ZBfp_if_zero:n #1 {p,T,F,TF} {
\ZBfp_if_zero_aux:NnnnnwNN #1
\prg_return_true: \prg_return_false:
}
\prg_new_conditional:Npnn \ZBfp_if_zero:f #1 {p,T,F,TF} {
\exp_after:wN \ZBfp_if_zero_aux:NNNnnnnw
\tex_romannumeral:D -`\0 #1
\prg_return_true: \prg_return_false:
}
\begin{document}
\cs_generate_variant:Nn \iow_term:n {f}
\iow_term:f { \ZBfp_mul:nn
{\ZBfp_add:nn {+{1234}{0000}{0000}{0000}T{2}} {+{0034}{5678}{0000}{0000}T{3}}}
{\ZBfp_add:nn {+{1111}{1111}{1111}{1111}T{0}} {+{2222}{2222}{2222}{2222}T{1}}}
}
\iow_term:f {
\ZBfp_div:nn
{-{0001}{2309}{8124}{3772}T{2}}
{+{2384}{7623}{4098}{1230}T{-2}}
}
\end{document}
答案3
宏不能返回值，只能扩展为结果。这里的问题是 TeX 中的赋值必须是执行并且不是可扩展一种方法是在另一个宏中定义结果，然后可以将其扩展并作为某种形式的间接返回值。
pgfmath包的库就是pgf这样做的。表达式由 处理\pgfmathparse{...}，结果存储在 中\pgfmathresult。或者也\pgfmathsetmacro\mymacro{...}可以使用 。
一些值得注意的例外是 eTeX 原语\dimexpr、\numexpr和，\skipexpr它们可以进行一些算术运算并且可扩展。
答案4
正如已经评论过的，TeX 是一种宏扩展语言，这意味着除了原语之外，您无法获得许多其他语言中的函数行为。在引用的示例中，至少使用 e-TeX 可以实现所需的行为，因为它\numexpr是可扩展的：
\catcode`\@=11\relax
\def\Abs#1{\number\numexpr\ifnum#1<\z@ -\fi\numexpr#1\relax}
\Abs{-4}:%
\Abs{\Abs{-3}-\Abs{2}}
\bye
也可以通过使用\romannumeral或来实现完全扩展\csname（例如，参见可扩展的令牌完整扩展，保留 catcodes)，但确保获得类似函数的行为并不总是那么容易。还需要考虑性能，在大多数情况下，最好接受 TeX 是一种扩展语言。