Effacer les filtres
Effacer les filtres

Attempted to access X1(1); index out of bounds because numel(X1)=0. Error in (line 108)

1 vue (au cours des 30 derniers jours)
yuqi
yuqi le 25 Mar 2014
Commenté : Walter Roberson le 25 Mar 2014
Hi, here is my script code, I just copy it here.
clear all;
close all;
[x1,x2,x3,x4,x5,x6]=nbc(6);
func=(~x1*~x2*x3)+(x1*~x2*x4)+(~x1*x2*x5)+(x1*x2*x6);
btab(func,x1,x2,x3,x4,x5,x6);
Function = [ ];
X1 = [ ];
X2 = [ ];
X3 = [ ];
X4 = [ ];
X5 = [ ];
X6 = [ ];
for i=1:16
if (b2d(func(i))==0)
Function(i)=0;
else
Function(i)=1;
end
end
for i=1:16
if (b2d(x1(i))==0)
X1 (i)=0;
else
X1 (i)=1;
end
end
for i=1:16
if (b2d(x2(i))==0)
X2 (i)=0;
else
X2 (i)=1;
end
end
for i=1:16
if (b2d(x3(i))==0)
X3 (i)=0;
else
X3 (i)=1;
end
end
for i=1:16
if (b2d(x4(i))==0)
X4 (i)=0;
else
X4 (i)=1;
end
end
for i=1:16
if (b2d(x5(i))==0)
X5 (i)=0;
else
X5 (i)=1;
end
end
for i=1:16
if (b2d(x6(i))==0)
X6 (i)=0;
else
X6 (i)=1;
end
end
for i=1:15
for j = (i+1):16
if ((X1(i) ~= X1(j)) && (X2(i) == X2(j)) && (X3(i) == X3(j)) && (X4(i) == X4(j)) && (X5(i) == X5(j)) && (X6(i) == X6(j)) && (Function(i) ~=Function(j)))
Temp1 = [X1(i) X2(i) X3(i) X4(i) X5(i) X6(i)];
Temp2 = [X1(j) X2(j) X3(j) X4(j) X5(j) X6(j)];
Store1 = [Temp1; Temp2];
end
if ((X1(i) == X1(j)) && (X2(i) ~= X2(j)) && (X3(i) == X3(j)) && (X4(i) == X4(j)) && (X5(i) == X5(j)) && (X6(i) == X6(j)) && (Function(i) ~=Function(j)))
Temp1 = [X1(i) X2(i) X3(i) X4(i) X5(i) X6(i)];
Temp2 = [X1(j) X2(j) X3(j) X4(j) X5(j) X6(j)];
Store2 = [X1(i) X2(i) X3(i) X4(i) X5(i) Function(i);X1(j) X2(j) X3(j) X4(j) X5(j) X6(j)];
end
if ((X1(i) == X1(j)) && (X2(i) == X2(j)) && (X3(i) ~= X3(j)) && (X4(i) == X4 (j)) && (X5(i) == X5(j)) && (X6(i) == X6(j)) && (Function(i) ~=Function(j)))
Temp1 = [X1(i) X2(i) X3(i) X4(i) X5(i) X6(i)];
Temp2 = [X1(j) X2(j) X3(j) X4(j) X5(j) X6(j)];
Store3 = [X1(i) X2(i) X3(i) X4(i) X5(i) X6(i);X1(j) X2(j) X3(j) X4(j) X5(j) X6(j)];
end
if ((X1(i) == X1(j)) && (X2(i) == X2(j)) && (X3(i) == X3(j)) && (X4(i) ~= X4 (j)) && (X5(i) == X5(j)) && (X6(i) == X6(j)) && (Function(i) ~=Function(j)))
Temp1 = [X1(i) X2(i) X3(i) X4(i) X5(i) X6(i)];
Temp2 = [X1(j) X2(j) X3(j) X4(j) X5(j) X6(j)];
Store4 = [X1(i) X2(i) X3(i) X4(i) X5(i) X6(i);X1(j) X2(j) X3(j) X4(j) X5(j) X6(j)];
end
if ((X1(i) == X1(j)) && (X2(i) == X2(j)) && (X3(i) == X3(j)) && (X4(i) == X4 (j)) && (X5(i) ~= X5(j)) && (X6(i) == X6(j)) && (Function(i) ~=Function(j)))
Temp1 = [X1(i) X2(i) X3(i) X4(i) X5(i) X6(i)];
Temp2 = [X1(j) X2(j) X3(j) X4(j) X5(j) X6(j)];
Store5 = [X1(i) X2(i) X3(i) X4(i) X5(i) X6(i);X1(j) X2(j) X3(j) X4(j) X5(j) X6(j)];
end
if ((X1(i) == X1(j)) && (X2(i) == X2(j)) && (X3(i) == X3(j)) && (X4(i) == X4 (j)) && (X5(i) == X5(j)) && (X6(i) ~= X6(j)) && (Function(i) ~=Function(j)))
Temp1 = [X1(i) X2(i) X3(i) X4(i) X5(i) X6(i)];
Temp2 = [X1(j) X2(j) X3(j) X4(j) X5(j) X6(j)];
Store6 = [X1(i) X2(i) X3(i) X4(i) X5(i) X6(i);X1(j) X2(j) X3(j) X4(j) X5(j) X6(j)];
end
end
end
Store = [Store1;Store2;Store3;Store4;Store5;Store6];
This is code that works under the boolean algebra toolbox environment. The output shuuld display a truth table and a complete test set. But there is an error at line 108. I don't know how to fix it. Can someone help me to fix it?
  1 commentaire
Walter Roberson
Walter Roberson le 25 Mar 2014
I'm not going to try to debug that. Use variable names that are less confusing than x1 being different than X1 . Naming a variable "Function" is a bad idea as well; it is going to confuse people as people know that "function" is a reserved word.
Any program that has "clear all" in it is broken. If you want to make sure the workspace has nothing in it then use a function.

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