long-double

Long double overflows but value smaller than maximum representable value

六月ゝ 毕业季﹏ 提交于 2021-02-11 09:18:48
问题 I'm trying to compute a series using C++. The series is: (for those wondering) My code is the following: #include <iostream> #include <fstream> #include <cmath> // exp #include <iomanip> //setprecision, setw #include <limits> //numeric_limits (http://en.cppreference.com/w/cpp/types/numeric_limits) long double SminOneCenter(long double gamma) { using std::endl; using std::cout; long double result=0.0l; for (long double k = 1; k < 1000 ; k++) { if(isinf(pow(1.0l+pow(gamma,k),6.0l/4.0l))) { cout

Long double overflows but value smaller than maximum representable value

给你一囗甜甜゛ 提交于 2021-02-11 09:18:31
问题 I'm trying to compute a series using C++. The series is: (for those wondering) My code is the following: #include <iostream> #include <fstream> #include <cmath> // exp #include <iomanip> //setprecision, setw #include <limits> //numeric_limits (http://en.cppreference.com/w/cpp/types/numeric_limits) long double SminOneCenter(long double gamma) { using std::endl; using std::cout; long double result=0.0l; for (long double k = 1; k < 1000 ; k++) { if(isinf(pow(1.0l+pow(gamma,k),6.0l/4.0l))) { cout

Why are double and long double completely the same on my 64 bit machine?

房东的猫 提交于 2021-02-04 15:01:29
问题 This question may sound like for beginners, however when I found that out I thought I'm either a beginner or my comp is missing something: int main() { cout << sizeof(double) << endl; cout << sizeof(long double) << endl; cout << DBL_DIG << endl; cout << LDBL_DIG << endl; return 0; } PROGRAM OUTPUT: 8 8 15 15 I thought long double is 10 bytes and has 18 decimal digits while double is 8 bytes and has 15 digits but it seems I was wrong. Why is that so? Using MSVC 2010 on 64bit machine. 回答1: In

Why are double and long double completely the same on my 64 bit machine?

|▌冷眼眸甩不掉的悲伤 提交于 2021-02-04 15:00:52
问题 This question may sound like for beginners, however when I found that out I thought I'm either a beginner or my comp is missing something: int main() { cout << sizeof(double) << endl; cout << sizeof(long double) << endl; cout << DBL_DIG << endl; cout << LDBL_DIG << endl; return 0; } PROGRAM OUTPUT: 8 8 15 15 I thought long double is 10 bytes and has 18 decimal digits while double is 8 bytes and has 15 digits but it seems I was wrong. Why is that so? Using MSVC 2010 on 64bit machine. 回答1: In

Format specifier %Lf is giving errors for `long double` variables

家住魔仙堡 提交于 2020-06-28 03:25:39
问题 I am getting the following errors: In function 'main': [Warning] unknown conversion type character 'L' in format [-Wformat=] [Warning] too many arguments for format [-Wformat-extra-args] In function 'error_user': [Warning] unknown conversion type character 'L' in format [-Wformat=] [Warning] too many arguments for format [-Wformat-extra-args] In the below code: #include <stdio.h> #include <stdlib.h> void error_user (long double *error); int main(void) { long double error; printf("What error

Format specifier %Lf is giving errors for `long double` variables

早过忘川 提交于 2020-06-28 03:23:11
问题 I am getting the following errors: In function 'main': [Warning] unknown conversion type character 'L' in format [-Wformat=] [Warning] too many arguments for format [-Wformat-extra-args] In function 'error_user': [Warning] unknown conversion type character 'L' in format [-Wformat=] [Warning] too many arguments for format [-Wformat-extra-args] In the below code: #include <stdio.h> #include <stdlib.h> void error_user (long double *error); int main(void) { long double error; printf("What error

machine epsilon - long double in c++

守給你的承諾、 提交于 2020-04-17 22:55:46
问题 I wanted to calculate the machine Epsilon, the smallest possible number e that gives 1 + e > 1 using different data types of C++: float , double and long double . Here's my code: #include <cstdio> template<typename T> T machineeps() { T epsilon = 1; T expression; do { epsilon = epsilon / 2; expression = 1 + epsilon; } while(expression > 1); return epsilon; } int main() { auto epsf = machineeps<float>(); auto epsd = machineeps<double>(); auto epsld = machineeps<long double>(); std::printf(

C long double in golang

余生长醉 提交于 2020-01-06 05:35:22
问题 I am porting an algorithm from C to Go. And I got a little bit confused. This is the C function: void gauss_gen_cdf(uint64_t cdf[], long double sigma, int n) { int i; long double s, d, e; //Calculations ... for (i = 1; i < n - 1; i++) { cdf[i] = s; } } And in the for loop value "s" is assigned to element "x" the array cdf. How is this possible? As far as I know, a long double is a float64 (in the Go context). So I shouldn't be able to compile the C code because I am assigning an long double

Substitutions for Eigen::MatrixXd typedefs

。_饼干妹妹 提交于 2019-12-31 04:38:09
问题 What is the simplest way to replace all Eigen::MatrixXd s and Eigen::VectorXd s with Vectors and Matrices that have long double elements? Every basic floating point variable in my code is of type long double . Also, everytime I use a matrix or vector, I use the following typedefs. typedef Eigen::VectorXd Vec; typedef Eigen::MatrixXd Mat; What's the best thing to switch these typedefs to? What happens if I leave them as they are? 回答1: Simply define your own typedefs based on Eigen's own global

Cout long double issue

旧街凉风 提交于 2019-12-30 10:38:50
问题 So, I'm working on a C++ project. I have a var of long double type and assigned it a value like "1.02" Then, I try to use cout to print it and the result is: -0 I already tried to use setprecision and all I found googling the problem. What is the solution for this? Example code: #include <cstdlib> #include <iomanip> using namespace std; int main(int argc, char** argv) { cout.precision(15); long double var = 1.2; cout << var << endl; return 0; } OS: Windows 8.1 64 bits Compiler: minGW IDE: