How to add integers without correction?

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Dejia Kong
Dejia Kong am 14 Sep. 2021
Kommentiert: Jan am 23 Sep. 2021
when using integer class such as int8 in matlab, and some calculation result is bigger than the upper range (for int8 , the upper range is 127), matlab would automatically set the value to the maximum.
eg.
a=int8(126);
b=int8(2);
c=a+b; % matlab would gives c=127, since 128 exceed the upper range of int8
but in some cases, i want to get a negetive result, like many programming languages do
i want c to be -128 because 01111110+00000010=10000000 and it's -128 for int8
this could be done by a "bitadd" like
function int1 = bitadd(int1,int2)
while (int2 ~=0)
carry=bitand(int1,int2);
int1=bitxor(int1,int2);
int2=bitshift(carry,1);
end
end
however this is very inefficient, since a while loop in the function
Setting the result to maximum should be additional correction, so is there a faster method to do bitwise add without such correction?

Akzeptierte Antwort

Jan
Jan am 14 Sep. 2021
a = int8(126);
b = int8(2);
tic
for k = 1:1e4
c = bitadd1(a, b);
end
toc
Elapsed time is 0.141335 seconds.
tic
for k = 1:1e4
c = bitadd2(a, b);
end
toc
Elapsed time is 0.068172 seconds.
function int1 = bitadd1(int1,int2)
while (int2 ~=0)
carry = bitand(int1,int2);
int1 = bitxor(int1,int2);
int2 = bitshift(carry,1);
end
end
function c = bitadd2(a, b)
m = typecast(int16(a) + int16(b), 'int8');
c = m(1);
end
  3 Kommentare
Dejia Kong
Dejia Kong am 15 Sep. 2021
thanks for your bitadd2 and introducing typecast to me
in fact i want some code as fast as
tic;
for i=1:1e4
c=a+b;
end
toc;
since i think setting the result to maximum is additional correction and would take extra time.
But now i think when i have such performance requirement, it's better to use c++/fortran instead of matlab.
Jan
Jan am 23 Sep. 2021
It will be much faster to use int16 in general and crop the higher bits during the addition. In my tests this is 10 times faster, but I did not get the correct results when the overflow appears.

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