error when using horzcat

2015 1 25
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更新时间：2015 1 25
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Hi for all...
i have error when using horzcat >>> i crate two vectors randomly ( count1=randi(0:1 ,[1,32]); ) and ( count2=randi(0:1 ,[1,32]); ) and i have this vector number ( d0 = dec2bin( 17623,15); ) the problem is : when i using horzcat ( count1,count2,d0) the result appear with some symbols
can anyone help me

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Star Strider
Star Strider 2015-1-25

0 个投票

The problem is that the output from dec2bin is a string, so horizcat considers everything a string. The ‘symbols’ you see are therefore ‘d0’. To see all of the variables as a string of [0 1] ‘bits’ (strings in this instance), use num2str:
count1=num2str(randi(0:1 ,[1,32]),'%d');
count2=num2str(randi(0:1 ,[1,32]),'%d');
d0 = dec2bin( 17623,15);
result = horzcat(count1,count2,d0)

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mays afif
mays afif 2015-1-25
编辑：Star Strider 2015-1-25
thank you .. but i cann't using (xor)operation with the result
count1=num2str(randi(0:1 ,[1,32]),'%d');
count2=num2str(randi(0:1 ,[1,32]),'%d');
d0 = dec2bin( 17623,15);
result1 = horzcat(count1,count2,d0)
count4=num2str(randi(0:1 ,[1,32]),'%d');
count3=num2str(randi(0:1 ,[1,32]),'%d');
d0 = dec2bin( 17623,15);
result2 = horzcat(count3,count4,d0)
did you have another suggestion
Star Strider
Star Strider 2015-1-25
编辑：Star Strider 2015-1-25
My pleasure.
You didn’t mention using xor!
It’s necessary to be creative for this to work:
count1=num2str(randi(0:1 ,[1,32]),'%d');
count2=num2str(randi(0:1 ,[1,32]),'%d');
d0 = dec2bin( 17623,15);
result1 = horzcat(count1,count2,d0);
count4=num2str(randi(0:1 ,[1,32]),'%d');
count3=num2str(randi(0:1 ,[1,32]),'%d');
d0 = dec2bin( 17623,15);
result2 = horzcat(count3,count4,d0);
result1n = uint8(result1)-48;
result2n = uint8(result2)-48;
xor_result = xor(result1n, result2n);
fprintf(1, ['\nxor_result = ' repmat('% d', 1, length(xor_result)) '\n'], xor_result);
produces (in this instance, results will differ between runs):
xor_result = 1 0 1 0 0 1 1 1 0 1 1 1 0 1 1 0 0 1 1 1 1 0 0 1 1 1 0 0 0 1 0 0 1 0 1 1 1 1 1 1 0 0 0 0 0 0 0 1 0 1 1 0 0 0 1 0 0 0 0 1 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
There might be more efficient ways to do this (since I don’t usually work with binary strings). Even if inefficient, this has the virtue of producing the correct result.
mays afif
mays afif 2015-1-25
thank you very much

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