How to round the number to the integer? ( having other's code)

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WEN SHIN LU on 2 Feb 2022
Edited: Max Heimann on 2 Feb 2022
Hi!
The code is about the rotation of an image. I don't get the rounded part like [ x=int16(x-min(min(x))+1) ]. I can't figure out because it's not intuitive for me. BTW, is there another way to round the number?
theta = 0.25*pi;
R = [cos(theta) -sin(theta);sin(theta) cos(theta)];
for i = 1:size(I,1)
for j = 1:size(I,2)
a = R*[i-1;j-1];
x(i,j)=a(1); %new x
y(i,j)=a(2); %new y
end
end
x=int16(x-min(min(x))+1);%rounded
y=int16(y-min(min(y))+1);%rounded
for i = 1:size(I,1)
for j = 1:size(I,2)
I_new(x(i,j),y(i,j)) = I(i,j);
end
end
subplot(1,2,1); imshow(I);
subplot(1,2,2); imshow(I_new);

Max Heimann on 2 Feb 2022
Edited: Max Heimann on 2 Feb 2022
int16() is a cast into an integer datatype. Those datatypes only support whole numbers and thus the data is rounded via some internal logic to the nearest integer value. Its not the cleanest way to round a number, but in MATLAB it works.
Other ways to round would be round(),ceil() and floor(). They differ in when they round up or down. Their exact behaviour is found in their help texts.
Rik on 2 Feb 2022
One more relevant function to mention: fix. It doesn't round in the usual sense of the word, but it has its uses. It rounds towards zero, essentially replacing sign(x)*floor(abs(x)).
Walter Roberson on 2 Feb 2022
Technically, int16() is a cast rather than a typecast . A cast changes representation while attempting to preserve value, whereas a typecast preserves representation while changing interpretation.
Calculations on integer datatypes (except for the 64 bit ones) are defined to be "as-if" the operands were promoted to double precision, and the operation done in double precision, and then the values were cast back to the integer data types. The conversion of double to the integer data types is defined as using round() -- this is documented.