error with sqrt in Matlab

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EDISON TOAPANTA IZA 2020-12-11

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https://ww2.mathworks.cn/matlabcentral/answers/690870-error-with-sqrt-in-matlab

回答： ramadurai ramapragasam 2021-7-29

hi, help please when I put this formula GEH2= sqrt(2*int64((flujo_sim2-FlujoReal2)^2)/int64(flujo_sim2+FlujoReal2)) in matlab, this returns me the following error: Check for missing argument or incorrect argument data type in call to function 'sqrt'.

why does this error give me?

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回答（4 个）

Steven Lord 2020-12-11

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The sqrt function is not defined for variables of the int64 data type. The square root of an integer value is not necessarily itself an integer value. If it were defined, what exactly would you expect x to be in the following code?

five = int64(5)
five = int64
    5
x = sqrt(five)
Check for missing argument or incorrect argument data type in call to function 'sqrt'.

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EDISON TOAPANTA IZA 2020-12-14

Hi, of course.

flujo_sim2=datapoint2.get('AttValue', 'Vehs(Current, Total, All)'); %give me value the software external (simulation)

FlujoReal2 =(flujo_med233 + flujo_med234 + flujo_med235); %gives me the total field value (real life) obtained from an excel table.

Steven Lord 2020-12-14

在 MATLAB Online 中打开

The lines of code you posted doesn't tell me the class of the variables, which is the important information.

But I repeat my previous question. If you have an int64 number that is not a perfect square, would you be okay with getting an answer that when squared does not give you back the original number?

sqrtDouble = sqrt(5)
sqrtDouble = 2.2361
sqrtInt = int64(sqrt(5))
sqrtInt = int64
    2
sqrtDouble.^2 - 5 % close to 0
ans = 8.8818e-16
sqrtInt.^2 - 5 % not really close to 0
ans = int64
    -1