MATLAB: Int2str returns the wrong result

doublefloating pointint2strr2014bsignificant figures

>> int2str(73167176531330624919225119674426574742355349194934)
ans =
73167176531330625585928818171611107149744948903936

Anyone know what's going on here? I'm guessing the number is too because when I evaluate the difference between the input and output, the answer is 0.

What causes Matlab to fill in nonsense numbers though?

Best Answer

Your original number has too many digits to store in a double precision number. E.g.,

num2strexact(73167176531330624919225119674426574742355349194934)
ans =
7.3167176531330625585928818171611107149744948903936e49

So what you are seeing as output is exactly what is being stored (i.e., the exact decimal conversion of the binary bit pattern being stored), and there aren't enough mantissa bits in double precision to store more precision.

In fact, the nearest numbers in the double precision set to your original number are:

>> x = 73167176531330624919225119674426574742355349194934
>> num2strexact(x)
ans =
7.3167176531330625585928818171611107149744948903936e49
>> num2strexact(x+eps(x))
ans =
7.3167176531330635970522535241266364210737607344128e49
>> num2strexact(x-eps(x))
ans =
7.3167176531330615201335101101955850088752290463744e49

Related Solutions

MATLAB: Does “mat2str(0.005,90)” return 0.0050000000000000001 in MATLAB

Double-precision numbers are stored in MATLAB using 64 bits as per the IEEE standard for floating-point arithmetic. With 64 bits in the binary representation, the maximum possible precision in base 10 is 15 to 17 digits. Therefore, it should not matter if 0.05 is represented as a string with more than 15 decimals.

The "mat2str" function uses "sprintf" to print a number with n digits. The 1 that appears near the end with "mat2str(0.005,90)" is due to the fact that 0.005 cannot be exactly represented on a computer with 64 bits. However, if we look at the hexadecimal representation of the output of "mat2str", we get the same number:

% Set format to hexadecimal
>> format hex
% Convert 0.005 to a string with precision 15 digits
>> s = mat2str(0.005,15)
s =
0.005
>> eval(s)
ans =
3f747ae147ae147b
% Convert 0.05 to a string with precision 90 digits
>> s = mat2str(0.005,90)
s =
0.0050000000000000001
>> eval(s)
ans =
3f747ae147ae147b

The string "s" might be different, but the garbage digit at the end does not matter. Only the first 15 digits after the decimal point matter and this is shown by evaluating "s": we obtain the same binary number.

Instead of using "mat2str" you could store the exact representation of the number in hexadecimal form. The "num2hex" function converts a double-precision number into its hexadecimal representation as a string. This might be a better solution if you only plan to use it on scalars -- "num2hex" does not work on arrays unlike "mat2str".

MATLAB: How to avoid the loss of the last digit (it gives a zero instead of a one)

num2str(uint64(14472334024676221),'%u')

You have exceeded flintmax:

flintmax Largest consecutive integer in floating point format.
  flintmax returns the largest consecutive integer in IEEE double
  precision, which is 2^53. Above this value, double precision format
  does not have integer precision, and not all integers can be represented 
  exactly.

Best Answer

Related Solutions

MATLAB: Does “mat2str(0.005,90)” return 0.0050000000000000001 in MATLAB

MATLAB: How to avoid the loss of the last digit (it gives a zero instead of a one)

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