我可以在 awk 中对复数进行算术运算，如果可以，怎么做？

Question

AWK 语言是否支持复数算术？如果是，我如何定义一个虚数单位？

Answer 1

您始终可以将复数定义为两个数字（实部和虚部）的数组。您需要手动定义所有算术运算符：

function cset(x, real, imaginary) {
  x["real"] = real
  x["imaginary"] = imaginary
}

function cadd(n1, n2, result) {
  result["real"] = n1["real"] + n2["real"]
  result["imaginary"] = n1["imaginary"] + n2["imaginary"]
}
function cmult(n1, n2, result) {
  result["real"] = n1["real"] * n2["real"] - n1["imaginary"] * n2["imaginary"]
  result["imaginary"] = n1["real"] * n2["imaginary"] + n2["real"] * n1["imaginary"]
}
function c2a(x, tmp) {
  if (x["real"]) {
    tmp = x["real"]
    if (x["imaginary"] > 0) tmp = tmp "+"
  }
  if (x["imaginary"]) {
    if (x["imaginary"] == -1) tmp = tmp "-i"
    else if (x["imaginary"] == 1) tmp = tmp "i"
    else tmp = tmp x["imaginary"] "i"
  }
  if (tmp == "") tmp = "0"
  return "(" tmp ")"
}

BEGIN {
  cset(i, 0, 1)
  cmult(i, i, i2)
  printf "%s * %s = %s\n", c2a(i), c2a(i), c2a(i2)
  cset(x, 1, 2)
  cset(y, 0, 4)
  cadd(x, y, xy)
  printf "%s + %s = %s\n", c2a(x), c2a(y), c2a(xy)
}

这将输出：

(i) * (i) = (-1)
(1+2i) + (4i) = (1+6i)

对于原生支持复数的语言，请参阅：

• python：

  $ python -c 'print(1j*1j)'
  (-1+0j)
  

• octave：

  $ octave --eval 'i*i'
  ans = -1
  

• calc（apcalcDebian 上的软件包）：

  $ calc '1i * 1i'
          -1
  

• R：

  $ $ Rscript -e '1i*1i'
  [1] -1+0i