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I want to show my float number with just 1 digit and avoid to show this case -0.0.
My program is a C/C++ development for an Arduino board.
How can I do that?
(Applies to both C and C++).
IEEE754 floating point defines a signed zero. This is the effect you're observing here. One may obtain negative zero as the result of certain computations, for instance as the result of arithmetic underflow on a negative number, or −1.0 * 0.0, or simply as −0.0.
-0.0 is defined to equal 0.0.
One solution would be to analyse it as a special case: x == -0.0 ? /*handle -0.0 and 0.0 here: e.g. ::fabs(x)*/ : /*non-zero cases here*/.
See http://en.wikipedia.org/wiki/Signed_zero
you can try printing out this way using the c ternary operator.
printf("%2.1f",(f<0 &&f >-1)?-f:f);
here f is an floating point variable and if u get value of f as -0.0001 , the function will print out as 0.0 .
To avoid showing the negative sign you will need to actually round of the number before displaying it since the +/- of a number is irrespective of the precision
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I am calculating a floating point number by formula: number=1/(n-2.001)
Where n is any integer from 1 to infinite.
But it give me different answer in laptop and scientific calculator.
C++ calculation : 0.333444
Calculator answer: 0.3334444815
I have to get all digits in c++. How i get this.
Decimal equivalent of 1/3 is 0.33333333333333….
An infinite length number would require infinite memory to store, and we typically have 4 or 8 bytes. Therefore, Floating point numbers store only a certain number of significant digits, and the rest are lost.
NOTE : When outputting floating point numbers, cout has a default precision of 6 and it truncates anything after that.
The precision of a floating point number defines how many significant digits it can represent without information loss.
Therefore in your case only 6 decimals points are outputted and rest are turncated
To change the Precision of floating-point data types in C++ check this
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I have a vector that contains non-negative doubles. I want to distinguish the cases when an entry is equal to zero and when an entry is greater than zero.
Is it numerically safe to just check if(a>0.0) or can this cause problems? I have no a-priori lower bound for the non-zero values, except machine precision. Should I create a helper-vector containing integers to mark the zero-values for safe checking?
For better understanding: The entries of the vector are something like weights on a graph, and I figured I don't need the adjacency matrix to keep track of the graph topology.
EDIT: My question is: Can and will 0.0 be exactly represented in doubles?
Floating point numbers aren't literally evil. Nor are they designed by stupid people. The one and only issue you need to concern yourself with here, is that of rounding.
A number which is set to zero, will be zero. There would be no reason to design a computational system which did not behave this way.
A number which is set to 0.1 will not be 0.1, because 0.1 is not exactly representable and is therefore rounded to the nearest representable number; see Is floating point math broken? for details. But if you set two variables to 0.1 they will compare equal to each other, because 0.1 is rounded the same way each time. (In fact the rounding happens during compilation; at runtime you're just setting the variable to the pre-rounded value.)
Similarly, a number which is set to 0.1 * 3 - 0.3 may not be equal to zero, because 0.1 was rounded, and then the rounded result was multiplied by 3 and that result was rounded, and so on.
So the issue is not one of representation, but of computation. If you set something to a particular value, that's the value it has. If it got there through a sequence of inexact computations, you can't rely on exact equality.
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How do I design an algorithm that takes two floats and multiplies them using only addition, bit shifting and bitwise operations?
I have already found one like this for integers, but that doesn't work with floats.
I have also found another that is much like what I need but log is also prohibited in my case.
The floats are stored according to the IEEE754 standard. I have also tried to keep their exponent part, and bitwise multiply their fractional part with no luck.
According to http://en.wikipedia.org/wiki/IEEE_floating_point, an IEEE754 number x = (-1)^s * c * b^q is represented by s,c,b,q , all are integers. for Two floating point numbers with the same base b is the same.
So the multiplication of two floating point numbers x and y is:
(-1)^(s1+s2)*c1*c2*b^(q1+q2) so the new floating point is represented by: s1+s2, c1*c2, b q1+q2 so you only have left to deal with multiplication of c1 and c2, both are integers so you are done.
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In C++ are the following statements when comparing two integer values for equality the same? If not, why?
if(a == b)
...do
if(!(a ^ b))
...do
For integer values, yes. Obviously the xor operator will return not-zero if there are any bit differences between A and B, and ! will invert that. For integer data types, the conditions are equivalent.
For floating point values, because of how you can perform two mathematical operations that "should" give the same result, but they may be represented differently as floats, you should not use either of these to compare floats for equality, you should check whether they are the same to within a small margin of error (an "epsilon").
For pointers...I have no idea why you would want to do this to pointers. But if you really want to do it, then yes, they are the same.
However, there is no reason to do this. With optimizations enabled, they will compile to the same code, without, the first will likely be faster. Why would you use the less-clear !(a^b)?
The two comparisons are equivalent: a^b is 0 if and only if a==b, so !(a^b) is true if and only if a and b have the same value.
Whether you can call them "the same" depends on what you mean by two different operations being "same." They probably will not be compiled into the same code, and a==b is definitely easier to read.
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I am using openCV C++ libraries and calculated a double. It does the arithmetic but when I read out the number, prints out -1.#QNAN on the command prompt. What does that mean?
I am using a 64-bit i3 processor.
It means you got a quiet NAN, probably by dividing -Inf / Inf or multiplying something with -Inf, or perhaps casing a non-double into a double. It's not so much a precision error as much as it's an arithmetic exception.
EDIT: or adding/substracting Inf ... read more on NaNs here
That's not an error, read more about floating point here