c++,class,floating-point-conversion

Change 1 / 125 to, say, 1.0 / 125. 1 / 125 uses integer division, so the result is 0. Or change this expression ((inputPos + 125)*(1 / 125)) to (inputPos + 125) / 125 Since inputPos is floating point, so is inputPos + 125, and then dividing a float...

python-2.7,matplotlib,computer-science,floating-point-conversion

You can't append to a tuple at all (tuples are immutable), and extending to a list with + requires another list. Make curveList a list by declaring it with: curveList = [] and use: curveList.append(curve) to add an element to the end of it. Or (less good because of the...

c++,floating-point,memcpy,floating-point-precision,floating-point-conversion

Your assumption is wrong. Look at this: float f = -.022119; std::cout << std::setprecision(20) << f << std::endl; Prints: -0.022119000554084778...

java,floating-point,primitive,floating-point-conversion

If you only want to round when rand==1.0f, then I second @paxdiablo's answer. However, it sounds like you're okay with always rounding, and simply want to always round down. If so: float rand = Math.nextDown((float)Math.random()); from the javadoc: nextDown(float f) Returns the floating-point value adjacent to f in the direction...

c,floating-point,floating-point-conversion

This is the actual article. The blog post is referring to the results in section 7 (in other words, “they” are not comparing anything in the blog post, “they” are regurgitating the information from the actual article, omitting crucial details): Implementations of Dragon4 or Grisu3 can be found in implementations...

java,floating-point-conversion

You are trying to set a double to a float variable To fix, change this line m=km*0.621371; to m=km*0.621371f; ...

c#,.net,winforms,floating-point-conversion,string-conversion

Your floating point value ggg has a very small value. When you convert it to a string, as happens in this call String.Format("{0}\r\n{1}", ggg, txtDebugLog.Text); it will be converted to a string that uses an exponential format to represent that value. You can read about what an exponential format is,...

c,floating-point-conversion,math.h

They are fundamentally different. Cast to int will truncate the non-integer part toward 0. floor will do the same, but toward, -infinity. Example: double d = -1.234; printf("%d %d\n", (int)d, (int)floor(d)); Yields: -1 -2 ...

This is a very inefficient, probably bugged implementation that should give a good start: template<class T> std::string convert(T value, int precision) { std::stringstream ss; ss << std::setprecision(100) << value; return ss.str(); } template<class T> std::string find_shortest(T value) { auto lower = convert(std::nextafter(value, -std::numeric_limits<T>::infinity()), 100); auto higher = convert(std::nextafter(value, std::numeric_limits<T>::infinity()), 100);...

c++,language-lawyer,c++14,floating-point-conversion

No. It's possible that i == std::numeric_limits<I>::max(), but that i is not exactly representable in F. If the value being converted is in the range of values that can be represented but the value cannot be represented exactly, it is an implementation-defined choice of either the next lower or higher...

javascript,floating-point,floating-point-precision,floating-point-conversion

As pointed out by Mark Dickinson in a comment on the question, the ECMA-262 ECMAScript Language Specification requires the use of IEEE 754 64-bit binary floating point to represent the Number Type. The relevant rounding rules are "Choose the member of this set that is closest in value to x....

c,floating-point-precision,c11,floating-point-conversion

All that you need to know to which limits you may go and still have integer precision should be available to you through the macros defined in <float.h>. There you have the exact description of the floating point types, FLT_RADIX for the radix, FLT_MANT_DIG for the number of the digits,...

c++,types,int,floating-point-conversion,promotions

When int is promoted to unsigned in the integral promotions, negative values are also lost (which leads to such fun as 0u < -1 being true). Like most mechanisms in C (that are inherited in C++), the usual arithmetic conversions should be understood in terms of hardware operations. The makers...

c++,c,floating-point,floating-point-precision,floating-point-conversion

The reason is that unsigned long long will store exact integers whereas double stores a mantissa (with limited 52-bit precision) and an exponent. This allows double to store very large numbers (around 10308) but not exactly. You have about 15 (almost 16) valid decimal digits in a double, and the...

floating-point,intel,floating-point-precision,floating-point-conversion

Problem was mxcsr register's DAZ bit. One machine A had it set so it was treating denormals as 0 . Where as on machine 2 DAZ was reset and hence it was normalizing it and treating denormals as different values , which resulted in comparison result as not equal. But...

mysql,floating-point,floating-accuracy,floating-point-conversion

Decisions and consequences This is the consequences that you've got because you decided to use floating-point data type. Floats are not precise. And that means: yes, you can result in a>a = true For instance, your fourth row: mysql> SELECT * FROM t WHERE id=4; +------+--------+ | id | rating...

ios,swift,floating-point,floating-point-precision,floating-point-conversion

This is due to the way the floating-point format works. A Float is a 32-bit floating-point number, stored in the IEEE 754 format, which is basically scientific notation, with some bits allocated to the value, and some to the exponent (in base 2), as this diagram from the single-precision floating-point...

c,binary,floating-point-conversion

You seem to have the correct concept, including an understanding of the excess-N representation, but you're missing a crucial point. The 3 bits used to encode the fractional part of the magnitude are 001, but there is an implicit 1. preceding the fraction bits, so the full magnitude is actually...

go,floating-point,type-conversion,floating-point-conversion

The problem here is the representation of constants and floating point numbers. Constants are represented in arbitrary precision. Floating point numbers are represented using the IEEE 754 standard. Spec: Constants: Numeric constants represent values of arbitrary precision and do not overflow. Spec: Numeric types: float64 the set of all IEEE-754...

java,double,byte,floating-point-conversion

The problem is that (bytes[0] & 0xFF) is still a 32-bit integer value. If you bit-shift it 56 bits to the left on a 32-bit value, Java shifts by 56 % 32 = 24 bits instead of 56 bits. You first need to promote the value to 64-bit long before...

math,floating-point,integer,floating-point-conversion,integer-arithmetic

pseudocode: n / pow(10, floor(log10(n))+1) ...

c,floating-point,avr,floating-point-conversion

You can use printf("%07.3f", value); where 07 means printing at least 7 characters with padding zero. e.g. printf("%07.3f", 3.3); prints 003.300...

hash,floating-point,64bit,ieee-754,floating-point-conversion

Apart from +0/-0, are there denormalized floats, which really translate to the same discrete value, or can I just hash their binary representation as is, without fear of generating different hashes for identical values? There is no translation from the representation to something else that would be the “real”...

ruby-on-rails,postgresql,activerecord,database-normalization,floating-point-conversion

Empty strings cannot be converted to a number for obvious reasons. You have to account for that. Replace all occurrences with NULL or 0 or something compatible. Also for the number of fractional digits in your example you want the data type numeric, not float - neither real (float4) nor...

java,floating-point,int,floating-accuracy,floating-point-conversion

number2 should be promoted to float type Number 2 got promoted to float. and I should get a 0.96 You should get 1.0, because: 2 / 12 * 6 = 2 / 2 = 1 0.16 * 6 = 0.96 ... and that's why. It has nothing to do...

floating-point,ocaml,floating-point-conversion

float_of_string should be able to parse them: # float_of_string "0x1.199999999999Ap1";; - : float = 2.2 However as Alain Frisch noted on OCaml's bug tracker the support actually depends on the underlying libc and won't currently work on the MSVC ports....

python,python-2.7,floating-point,floating-point-conversion

isinstance(next, (float, int)) will do the trick simply if next is already converted from a string. It isn't in this case. As such you would have to use re to do the conversion if you want to avoid using try..except. I would recommend using the try..except block that you had...

floating-point,floating-point-conversion

I post my code, which is not optimized, but works. If you have further suggestion please... let me know, especially for some optimization. void cast(uint64_t __x, uint64_t* __y, uint64_t __exs, uint64_t __mxs, uint64_t __eys, uint64_t __mys) { /* Input : __x -> number to be casted __y -> casted value...

c++,floating-point,floating-point-conversion

The problem is that floating point is inexact by nature when talking about decimal numbers. A decimal number can be rounded either up or down when converted to binary, depending on which value is closest. In this case you just want to make sure that if the number was rounded...

python,string,floating-point,floating-point-conversion,floating-point-exceptions

Ok, how about def fn(x): try: return float(x) except (ValueError, TypeError): return 0.0 ...

python,ieee-754,floating-point-conversion

As you can see in Java they are using structure to do the trick. /* * Find the float corresponding to a given bit pattern */ JNIEXPORT jfloat JNICALL Java_java_lang_Float_intBitsToFloat(JNIEnv *env, jclass unused, jint v) { union { int i; float f; } u; u.i = (long)v; return (jfloat)u.f; }...

python,list,formatting,floating-point-precision,floating-point-conversion

Use string formatting to get the desired number of decimal places. >>> nums = [1883.95, 1878.3299999999999, 1869.4300000000001, 1863.4000000000001] >>> ['{:.2f}'.format(x) for x in nums] ['1883.95', '1878.33', '1869.43', '1863.40'] The format string {:.2f} means "print a fixed-point number (f) with two places after the decimal point (.2)". str.format will automatically round...

bash,floating-point,floating-point-precision,floating-point-conversion

What about a=`echo "5+50*3/20 + (19*2)/7" | bc -l` a_rounded=`printf "%.3f" $a` echo "a = $a" echo "a_rounded = $a_rounded" which outputs a = 17.92857142857142857142 a_rounded = 17.929 ?...

c#,visual-studio-2012,floating-point-conversion

You're hitting inherent limtations in the System.Single type (float is a name-alias of Single). According to MSDN ( http://msdn.microsoft.com/en-us/library/system.single.aspx ) the Single type has 22 bits dedicated to the mantissa (also known as the significand), these are the bits that store the actual 'numbers' in the value (the other 10...

floating-point,double,ieee-754,floating-point-conversion,significant-digits

First, for this question it is better to use the total significand sizes 24 and 53. The fact that the leading bit is not represented is just an aspect of the encoding. If you are interested only in a vague explanation, one decimal digits contains exactly log2(10) (about 3.32) bits...