FRFT程序运行时说Not enough input arguments. 是怎么回事？？？

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如题：FRFT程序运行时说Not enough input arguments. 是怎么回事？？？
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function Faf = frft(f, a)
6 T, y8 R* _) ^. A) a5 R& H, B! R% The fast Fractional Fourier Transform
% input: f = samples of the signal
& g$ K; B, `0 N3 ^% z# ?  K% a = fractional power
% output: Faf = fast Fractional Fourier transform

error(nargchk(2, 2, nargin));

f = f(;
1 @* p, d! A& c2 O. p/ A8 WN = length(f);
8 K$ y7 i) r1 Z% A% Ashft = rem((0:N-1)+fix(N/2),N)+1;
sN = sqrt(N);
! H9 n4 i) u7 B. z# M+ O3 W  da = mod(a,4);

% do special cases
if (a==0), Faf = f; return; end;
if (a==2), Faf = flipud(f); return; end;
if (a==1), Faf(shft,1) = fft(f(shft))/sN; return; end
: `8 F: h5 ~7 W0 i. k% Qif (a==3), Faf(shft,1) = ifft(f(shft))*sN; return; end

% reduce to interval 0.5 < a < 1.5
+ P  |5 i" \$ y: v' N' @if (a>2.0), a = a-2; f = flipud(f); end
if (a>1.5), a = a-1; f(shft,1) = fft(f(shft))/sN; end
if (a<0.5), a = a+1; f(shft,1) = ifft(f(shft))*sN; end

% the general case for 0.5 < a < 1.5
alpha = a*pi/2;
- D2 G  X, f% V, F8 `) ftana2 = tan(alpha/2);
sina = sin(alpha);
% ~- N# f8 E+ uf = [zeros(N-1,1) ; interp(f) ; zeros(N-1,1)];

% chirp premultiplication
/ L; G3 z, c7 O4 [1 v( P6 W  Qchrp = exp(-i*pi/N*tana2/4*(-2*N+2:2*N-2)'.^2);
f = chrp.*f;

% chirp convolution
% a  d3 c2 |! R6 g0 [c = pi/N/sina/4;
Faf = fconv(exp(i*c*(-(4*N-4):4*N-4)'.^2),f);
Faf = Faf(4*N-3:8*N-7)*sqrt(c/pi);

% chirp post multiplication
: n; U3 v- v3 {1 |$ ~Faf = chrp.*Faf;

% normalizing constant
Faf = exp(-i*(1-a)*pi/4)*Faf(N:2:end-N+1);

%%%%%%%%%%%%%%%%%%%%%%%%%
: C7 E; N+ ^- I0 h1 vfunction xint=interp(x)
- J8 J7 l' [. A0 D% sinc interpolation

N = length(x);
3 ~: D) {) c' N& L3 G& H+ H0 i1 R  `y = zeros(2*N-1,1);
2 A2 T+ E- p5 f. r  Q2 s& xy(1:2:2*N-1) = x;
xint = fconv(y(1:2*N-1), sinc([-(2*N-3)2*N-3)]'/2));
; y, X2 m  I! ?* K' @- l4 [( J; ]xint = xint(2*N-2:end-2*N+3);

%%%%%%%%%%%%%%%%%%%%%%%%%
9 {* B, h7 T* [/ I/ {function z = fconv(x,y)
% convolution by fft

N = length([x(;y(])-1;
P = 2^nextpow2(N);
z = ifft( fft(x,P) .* fft(y,P));
, E* N3 Q: M# e, Az = z(1:N);

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xixihahaheihei

好复杂