How to add integers without correction?

5 ビュー (過去 30 日間)
Dejia Kong
Dejia Kong 2021 年 9 月 14 日
コメント済み: Jan 2021 年 9 月 23 日
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?

採用された回答

Jan
Jan 2021 年 9 月 14 日
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 件のコメント
Dejia Kong
Dejia Kong 2021 年 9 月 15 日
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 2021 年 9 月 23 日
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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