Approximating cosine on [0,pi] using only single precision floating point
来源： https://stackoverflow.com/questions/63918873/approximating-cosine-on-0-pi-using-only-single-precision-floating-point
single-precision
Add Two 32 bit Floating Point Numbers with AVR-Assembler
问题 Im trying to use AVR Studio to add two 32bit floating point numbers together. I know that I will need to store the 32bit number in 4 separate 8bit registers. I'll then need to add the registers together using the carry flag. This is what I have so far. Im adding 5.124323 and 2.2134523. ;5.124323 (01000000101000111111101001110100) ;Store hex value (40A3FA74) ldi r21,$40 ldi r22,$A3 ldi r23,$FA ldi r24,$74 ;2.2134523 (01000000000011011010100100110100) ;Store hex value (400DA934) ldi r25,$40 ldi
How are double-precision floating-point numbers converted to single-precision floating-point format?
问题 Converting numbers from double-precision floating-point format to single-precision floating-point format results in loss of precision. What's the algorithm used to achieve this conversion? Are numbers greater than 3.4028234e+38 or lesser than -3.4028234e+38 simply reduced to the respective limits? I feel that the conversion process is a bit more involved than this but I couldn't find documentation for it. 回答1: The most common floating-point formats are the binary floating-point formats
How do you determine how many integers are in set S of all in 32-bit IEEE floating-point values [duplicate]
问题 This question already has answers here : how many whole numbers in IEEE 754 (2 answers) Closed 2 years ago . Could anybody explain to me what it is stating exactly? I know this basically means that it's single precision with 1bit sign, 8bit exponents and 23bit mantissa. Shouldn't the answer is just be 2 * 2^8-2 * 2^23? Edit:does 2 * 2^8-2 * 2^23 determine all 32-bit IEEE floating-point values 回答1: The finite positive floating-point numbers range from 2 -149 (the smallest subnormal) to 2 128
How do you determine how many integers are in set S of all in 32-bit IEEE floating-point values [duplicate]
This question already has an answer here: how many whole numbers in IEEE 754 2 answers Could anybody explain to me what it is stating exactly? I know this basically means that it's single precision with 1bit sign, 8bit exponents and 23bit mantissa. Shouldn't the answer is just be 2 * 2^8-2 * 2^23? Edit:does 2 * 2^8-2 * 2^23 determine all 32-bit IEEE floating-point values The finite positive floating-point numbers range from 2 -149 (the smallest subnormal) to 2 128 -2 104 (the number with the largest exponent for finite values and a significand of all one bits). We can group them into three
Building a 32-bit float out of its 4 composite bytes
I'm trying to build a 32-bit float out of its 4 composite bytes. Is there a better (or more portable) way to do this than with the following method? #include <iostream> typedef unsigned char uchar; float bytesToFloat(uchar b0, uchar b1, uchar b2, uchar b3) { float output; *((uchar*)(&output) + 3) = b0; *((uchar*)(&output) + 2) = b1; *((uchar*)(&output) + 1) = b2; *((uchar*)(&output) + 0) = b3; return output; } int main() { std::cout << bytesToFloat(0x3e, 0xaa, 0xaa, 0xab) << std::endl; // 1.0 / 3.0 std::cout << bytesToFloat(0x7f, 0x7f, 0xff, 0xff) << std::endl; // 3.4028234 × 10^38 (max single
Building a 32-bit float out of its 4 composite bytes
问题 I\'m trying to build a 32-bit float out of its 4 composite bytes. Is there a better (or more portable) way to do this than with the following method? #include <iostream> typedef unsigned char uchar; float bytesToFloat(uchar b0, uchar b1, uchar b2, uchar b3) { float output; *((uchar*)(&output) + 3) = b0; *((uchar*)(&output) + 2) = b1; *((uchar*)(&output) + 1) = b2; *((uchar*)(&output) + 0) = b3; return output; } int main() { std::cout << bytesToFloat(0x3e, 0xaa, 0xaa, 0xab) << std::endl; // 1