floating-point

问题 (Assume no bits are implied, there is no biasing, exponents use two’s complement notation, and exponents of all zeros and all ones are allowed.) I am trying to find the largest and smallest number that can be represented if the system is normalized. I thought that the largest number would be: .1111 x 2^4 = 0 100 1111 = 15 and the smallest: 1.0 x 2^-4 = 0 000 0001 = 0.0625 But the answers that I saw were: Largest: .1111 x 2^3 = 111.1 = 7.5 Smallest: 0.1 x 2^-4 = .00001 = 0.03125 I do not

问题 I don't understand why Single epsilon value is 1.401298E-45 and not 1E-126, if internally has an exponent of -126 and a mantissa of 1. 回答1: The smallest positive Single value has an exponent of −126 with a base of two and a binary significand of .00000000000000000000001 (2 −23 ), so its value is 2 −149 , which is approximately 1.4• 10 −45 . 回答2: Remember that IEEE-754 single-precision floats are stored in base 2-representation. This is how the smallest possible positive denormal value for

问题 This may be a duplicate of this question, but I don't feel it was actually answered correctly. Observe: #include <iostream> #include <iomanip> using namespace std; int main () { float p = 1.00; cout << showpoint << setprecision(3) << p << endl; } Output: 1.00 Now if we change that line to: cout << fixed << showpoint << setprecision(3) << p << endl; we get: 1.000 And if we use the "opposite" of fixed we get something totally different: cout << scientific << showpoint << setprecision(3) << p <<

问题 This may be a duplicate of this question, but I don't feel it was actually answered correctly. Observe: #include <iostream> #include <iomanip> using namespace std; int main () { float p = 1.00; cout << showpoint << setprecision(3) << p << endl; } Output: 1.00 Now if we change that line to: cout << fixed << showpoint << setprecision(3) << p << endl; we get: 1.000 And if we use the "opposite" of fixed we get something totally different: cout << scientific << showpoint << setprecision(3) << p <<

问题 If I have 2 denormal floating point numbers with different bit patterns and compare them for equality, can the result be affected by the Denormals-Are-Zero flag, the Flush-to-Zero flag, or other flags on commonly used processors? Or do these flags only affect computation and not equality checks? 回答1: DAZ (Denormals Are Zero) affects reading input, so DAZ affects compares . All denormals are literally treated as -0.0 or +0.0 , according to their sign. FTZ (Flush To Zero) affects only writing

问题 If I have 2 denormal floating point numbers with different bit patterns and compare them for equality, can the result be affected by the Denormals-Are-Zero flag, the Flush-to-Zero flag, or other flags on commonly used processors? Or do these flags only affect computation and not equality checks? 回答1: DAZ (Denormals Are Zero) affects reading input, so DAZ affects compares . All denormals are literally treated as -0.0 or +0.0 , according to their sign. FTZ (Flush To Zero) affects only writing

问题 <?php $x=PHP_INT_MAX; echo ((float)($x+1026)==(float)($x))?'EQUAL':'Not Equal'; I know floating point arithmetic is not exact and $x and $x+1 are so close together that they are rounded to the same floating point value and it shows the output as EQUAL if you use any number between 1 and 1025 but its only after you use value beyond 1025 it will start giving output as 'Not Equal'. I want to know why? What's the reason behind it? Why only after 1025? 回答1: With float, your assumption $x == $x + 1

问题 I'm a bit surprised with MSVC ldexp behavior (it happens in Visual Studio 2013, but also with all older versions at least down to 2003...). For example: #include <math.h> #include <stdio.h> int main() { double g=ldexp(2.75,-1074); double e=ldexp(3.0,-1074); printf("g=%g e=%g \n",g,e); return 0; } prints g=9.88131e-324 e=1.4822e-323 The first one g is strangely rounded... It is 2.75 * fmin_denormalized, so i definitely expect the second result e. If I evaluate 2.75*ldexp(1.0,-1074) I correctly

问题 32 bits are represented in binary using the IEEE format. So how can I extract those bits? Bitwise operations like & and | do not work on them! what i basically want to do is extract the LSB from 32 bit float images in opencv thanx in advance! 回答1: uint32_t get_float_bits(float f) { assert(sizeof(float) == sizeof(uint32_t)); // or static assert uint32_t bits; memcpy(&bits, &f, sizeof f); return bits; } As of C99, the standard guarantees that the union trick works (provided the sizes match),

问题 As per the C99 standard: 6.3.1.8.2 : The values of floating operands and of the results of floating expressions may be represented in greater precision and range than that required by the type; the types are not changed thereby.52)> However, outside the scope of Annex F, we have: 5.2.4.2.2.7 : The values of operations with floating operands and values subject to the usual arithmetic conversions and of floating constants are evaluated to a format whose range and precision may be greater than