floor((I3(i,j-1)+I3(i-1,j))/2)
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the expression evaluates to 192.5, but the function returns 193 instead of 192. Please help me out with this problem.
채택된 답변
Bruno Luong
2023년 8월 15일
이동: Bruno Luong
2023년 8월 15일
The above is a FALSE statement since (I3(i,j-1)+I3(i-1,j))/2 cannot have fractional digit provided that I3 is of class uint32.
Solution: convert your array to double to do floating point arithmetics
I3d = double(I3);
floor((I3d(i,j-1)+I3d(i-1,j))/2)
추가 답변 (3개)
Sulaymon Eshkabilov
2023년 8월 14일
To get the displayed data w.r.t your format specs:
ANS = 192.5;
% One decimal point
fprintf('Solution = %.1f \n', ANS)
% OR two decimal points
fprintf('Solution = %.2f \n', ANS)
% OR round up towards to -inf as Fangjun suggested
fprintf('Solution = %d \n', floor(ANS))
댓글 수: 0
Bruno Luong
2023년 8월 15일
편집: Bruno Luong
2023년 8월 15일
Set a breat point in your program, when it breaks type this and report the result
I3(i,j-1)
I3(i-1,j)
(I3(i,j-1)+I3(i-1,j))/2
floor(ans)
exist('floor')
which floor((I3(i,j-1)+I3(i-1,j))/2)
댓글 수: 0
Walter Roberson
2023년 8월 15일
You values are integer data class, such as uint8() . When you do calculations on integer data type, the result is calculated as if the values were converted to double, then the value was calculated, and then the result was cast to the integer data type. But when you cast a floating point number to an integer data type, the value is rounded. So the internal 192.5 that is calculated from the / operation is uint8() and converting to integer class rounds so it is rounded to uint8(193) . Then you are applying floor() to the uint8 193 which of course is still 193.
댓글 수: 4
Walter Roberson
2023년 8월 15일
floor((double(I3(i,j-1))+double(I3(i-1,j)))/2)
but if you were doing a bunch of these, convert the entire I3 array to double and index the double array.
D3 = double(I3);
floor((D3(i,j-1)+D3(i-1,j))/2)
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