Error when I run function encode ADPCM

I write code about encode ADPCM
Then I run the program use this function, error appear like that:
"Improper assignment with rectangular empty matrix.
Error in ==> adpcmenc at 98

b(i)=(255.0/256.0)*b(i)+sign(dqq)*sgn(dq(i))/128.0;"

I don't know fix it
Help me, please
this is my function

function iiout=adpcmenc(input)
% This function performs ADPCM encoding
% function iiout=adpcmenc(input)
% Usage:
% input=input value
% iiout=output index
%
% Quantization tables
fitable=[0 0 0 1 1 1 1 3 7];
witable=[-0.75 1.13 2.56 4.00 7.00 12.38 22.19 70.13 ];
qtable=[-0.98 0.62 1.38 1.91 2.34 2.72 3.12];
invqtable=[0.031 1.05 1.66 2.13 2.52 2.91 3.32];
lgth=length(input);
sr=zeros(1,2);pk=zeros(1,2);
a=zeros(1,2);b=zeros(1,6);
dq=zeros(1,6);ii=zeros(1,lgth);
y=0; ap=0;al=0;yu=0;yl=0;dms=0;dml=0;tr=0;td=0;
for k=1:lgth
sl=input(k);
%
sez=b(1)*dq(1);
for i=2:6
sez=sez+b(i)*dq(i);
end
se=a(1)*sr(1)+a(2)*sr(2)+sez;
d=sl - se;
%
% Perform quantization
%
dqq=log10(abs(d))/log10(2.0)-y;
ik=0;
for i=1:7
if dqq>qtable(i)
ik=i;
end
end
if d<0
ik=-ik;
end
ii(k)=ik;
yu=(31.0/32.0)*y+witable(abs(ik)+1)/32.0;
if yu>10.0
yu=10.0;
end
if yu<1.06
yu=1.06;
end
yl=(63.0/64.0)*yl+yu/64.0;
%%Inverse quantization
%
if ik==0
dqq=2.^(-y);
else
dqq=2^(invqtable(abs(ik))+y);
end
if ik<0
dqq=-dqq;
end
srr=se+dqq;
dqsez=srr+sez-se;
%
% Update state
%
pk1=dqsez;
%
% Obtain adaptive predictor coefficients
%
if tr==1
a=zeros(1,2);
b=zeros(1,6);
tr=0;
td=0; %Set for the time being
else
% Update predictor poles
% Update a2 first
a2p=(127.0/128.0)*a(2);
if abs(a(1))<=0.5
fa1=4.0*a(1);
else
fa1=2.0*sgn(a(1));
end
a2p=a2p+(sign(pk1)*sgn(pk(2))-fa1*sign(pk1)*sgn(pk(1)))/128.0;
if abs(a2p)>0.75
a2p=0.75*sgn(a2p);
end
a2=a2p;
%
% Update a1
a1p=(255.0/256.0)*a(1);
a1p=a1p+3.0*sign(pk1)*sgn(pk(1))/256.0;
if abs(a1p)>15.0/16.0-a2p
a1p=15.0/16.0-a2p;
end
a1=a1p;
%
% Update b coefficients
%
for i=1:6
b(i)=(255.0/256.0)*b(i)+sign(dqq)*sgn(dq(i))/128.0;
end
if a2p<-0.7185
td=1;
else
td=0;
end
if a2p<-0.7185 && abs(dq(6))>24.0*2^(yl)
tr=1;
else
tr=0;
end
for i=6:-1:2
dq(i)=dq(i-1);
end
dq(1)=dqq;pk(2)=pk(1);pk(1)=pk1;sr(2)=sr(1);sr(1)=srr;
%
% Adaptive speed control
%
dms=(31.0/32.0)*dms;
dms=dms+fitable(abs(ik)+1)/32.0;
dml=(127.0/128.0)*dml;
dml=dml+fitable(abs(ik)+1)/128.0;
if ap>1.0
al=1:0;
else
al=ap;
end
ap=(15.0/16.0)*ap;
if abs(dms-dml)>=dml/8.0
ap=ap+1/8.0;
end
if y<3
ap=ap+1/8:0;
end
if td==1
ap=ap+1/8:0;
end
if tr==1
ap=1.0;
end
y=al*yu+(1.0-al)*yl;
end
end
iiout=ii;