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

Question

yuqi 2014-3-25

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/122946-attempted-to-access-x1-1-index-out-of-bounds-because-numel-x1-0-error-in-line-108

评论： Walter Roberson 2014-3-25

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?

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Walter Roberson 2014-3-25

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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