a problem with a function
5 ビュー (過去 30 日間)
古いコメントを表示
hi all ! i wrote this simple function
%
function [ output_args ] = fract_int16( input_args )
if input_args<0
output_args=round(2^16+input_args*2^15);
else
output_args=round(input_args*2^15);
end
end
when i use it for negative number for exemple
%
fract_int16(-0.9133)=35609
fract_int16(0.3481)=11409,
both values ARE TRUE
but when use this function in an array i get a false value
%
A=[-0.9133 0.3481]
fract_int16(A)=[-29927 11407]
this is wrong!!
did i make i mistake in writing this function?
thank you all
0 件のコメント
採用された回答
David Sanchez
2013 年 7 月 23 日
Try this out:
input_args = [-0.9133 0.3481];
output_args = zeros(size(input_args));
for k=1:length(input_args)
if input_args(k)<0
output_args(k)=round(2^16+input_args(k)*2^15);
else
output_args(k)=round(input_args(k)*2^15);
end
end
0 件のコメント
その他の回答 (2 件)
Narges M
2013 年 7 月 23 日
編集済み: Narges M
2013 年 7 月 23 日
Yes, your function is designed to work with one value, and see if that's less than zero or not. So if you want to use it for arrays you need to use a loop either in the function or when calling it. Right now, if any of the array values are less than zero, if would fire TRUE, and then the output would be calculated for the whole array.
0 件のコメント
Jan
2013 年 7 月 23 日
And a faster vectorized version:
function output_args = fract_int16(input_args)
output_args= 2^16 * (input_args < 0) + round(input_args * 2^15);
end
0 件のコメント
参考
カテゴリ
Help Center および File Exchange で Data Type Conversion についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!