ieee

Dart Convert IEEE-11073 32-bit FLOAT to a simple double

由 纵然是瞬间 提交于 2021-02-16 15:34:07
问题 I don't have much experience working with these low level bytes and numbers, so I've come here for help. I'm connecting to a bluetooth thermometer in my Flutter app, and I get an array of numbers formatted like this according to their documentation. I'm attempting to convert these numbers to a plain temperature double, but can't figure out how. This is the "example" the company gives me. However when I get a reading of 98.5 on the thermometer I get a response as an array of [113, 14, 0, 254]

Casting between int, float and double in C

由 江枫思渺然 提交于 2020-05-13 17:26:23
问题 I don't really understand casting in C. Can anyone help me with a question in the book Computer Systems: A Programmer's Perspective: We generate arbitrary integer values x, y, and z, and convert them to values of type double as follows: int x = random(); int y = random(); int z = random(); double dx = (double) x; double dy = (double) y; double dz = (double) z; For each of the following C expressions, you are to indicate whether or not the expression always yields 1. If it always yields 1,

Casting between int, float and double in C

由 折月煮酒 提交于 2020-05-13 17:26:03
问题 I don't really understand casting in C. Can anyone help me with a question in the book Computer Systems: A Programmer's Perspective: We generate arbitrary integer values x, y, and z, and convert them to values of type double as follows: int x = random(); int y = random(); int z = random(); double dx = (double) x; double dy = (double) y; double dz = (double) z; For each of the following C expressions, you are to indicate whether or not the expression always yields 1. If it always yields 1,

Casting between int, float and double in C

由 大城市里の小女人 提交于 2020-05-13 17:26:01
问题 I don't really understand casting in C. Can anyone help me with a question in the book Computer Systems: A Programmer's Perspective: We generate arbitrary integer values x, y, and z, and convert them to values of type double as follows: int x = random(); int y = random(); int z = random(); double dx = (double) x; double dy = (double) y; double dz = (double) z; For each of the following C expressions, you are to indicate whether or not the expression always yields 1. If it always yields 1,

Casting between int, float and double in C

由 ﹥＞﹥吖頭↗ 提交于 2020-05-13 17:24:51
问题 I don't really understand casting in C. Can anyone help me with a question in the book Computer Systems: A Programmer's Perspective: We generate arbitrary integer values x, y, and z, and convert them to values of type double as follows: int x = random(); int y = random(); int z = random(); double dx = (double) x; double dy = (double) y; double dz = (double) z; For each of the following C expressions, you are to indicate whether or not the expression always yields 1. If it always yields 1,

Issue with Square Root in C Algorithm

由 故事扮演 提交于 2020-01-04 15:58:56
问题 I have a bit of code that finds a point on a unit sphere. Recall, for a unit sphere: 1 = sqrt( x^2 + y^2 + z^2 ) The algorithm picks two random points (the x and y coordinates) between zero and one. Provided their magnitude is less than one we have room to define a third coordinate by solving the above equation for z. void pointOnSphere(double *point){ double x, y; do { x = 2*randf() - 1; y = 2*randf() - 1; } while (x*x + y*y > 1); double mag = sqrt(fabs(1 - x*x - y*y)); point[0] = 2*(x*mag);

Highest (existing) number in half precision IEEE 754

由 邮差的信 提交于 2019-12-13 02:45:56
问题 Why is 0 11110 1111111111 and not 0 11111 1111111111 the highest half precision number? 回答1: Because an exponent field of 11111 2 is reserved for infinities and NaNs. Section 3.4 of the IEEE 754-2008 standard says: The range of the encoding’s biased exponent E shall include: every integer between 1 and 2 w − 2, inclusive, to encode normal numbers the reserved value 0 to encode ±0 and subnormal numbers the reserved value 2 w − 1 to encode ±∞ and NaNs. Here "w" is the width of the exponent

Half precision conversion

由 笑着哭i 提交于 2019-12-11 12:04:17
问题 How comes that 0 11110 1111111111 is equal to the half precision format 1.1111111111 * 2^15 ? Both should be 65504. The sign bit here is a 0. The exponent would be 11101 and the fractional part 1111111111. But that doesn't look like 1.1111111111 * 2^15 at all. Can someone explain that to me? 回答1: Here is the layout of your half-precision number: The exponent's value is 11110 2 , which is 30 10 . Half-precision numbers have exponent bias of 15 10 , so we need to subtract 15 10 from 30 10 to

ARM number conversion program

由 你离开我真会死。 提交于 2019-12-02 15:41:10
问题 I am trying to write a program that will convert a number from ieee to TNS (big endian), and vice versa. I'm very new to ARM and assembly in general. I'm not getting an error, its just not working as intended and I'd appreciate it if anyone could look at it. Every line is commented, the sub routine actually in question here is the "unpack ieee" procedure, it is not working as intended and I cannot see why. EDIT: IEEE and TNS numbers have 3 parts, I am attempting to "grab" or seperate the 3

how many whole numbers in IEEE 754

由 久未见 提交于 2019-11-28 13:59:45
I am trying to figure out how many different whole numbers exist in the ieee 754. The number I got was 1778384895 but I couldn't find any resource to check myself. Thanks a lot in advance. I will assume single precision floats. We got the zero, which although can be represented as negative zero, is still just zero as an integer so I count it as one. Numbers with exponent less than 127 are not integers. Exponent Free bits # Numbers 127 0 1 128 1 2 129 2 4 ... 149 22 2^22 These sum up to 2^23-1 . If exponent is greater than 149, all the numbers are integers. So that's an additional 105*2^23
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