Is there a way to calculate between long double types in C language?

#include <stdio.h>
#include <math.h>

int main(void)
{
    double x,y;
    printf("Please enter x and y : ");
    scanf("%lf %lf", &x, &y);
    long double z = ((x + sqrt(x*x+y*y))/2.0);
    printf("The value of the expression is %25.20Lf\n", z);
    return 0;
}

Currently, I made the code as below, but if you print it out as lf after setting z as double type, it comes out well, but if you use long double type, you get 0. But the problem condition is that z must be a long double type, so how can I divide the long double right away in the calculation?

c long-double type calculate

2022-09-22 18:40

1 Answers

When std=c++11 in clang, there is no problem as shown below (cling is a clang-based interpreter).)

What is the compiler and version you used?

allinux@ip-10-0-0-29:~$ cling -std=c++11

****************** ****************** CLING ******************
* * Type C++ code and press enter to run it *
*             *             Type .q to exit             *
*******************************************
[cling]$ #include <math.h>
[cling]$ double x = 2.0
(double) 2.0000000
[cling]$ double y = 2.0
(double) 2.0000000
[cling]$ long double z = ((x + sqrt(x*x+y*y))/2.0);
[cling]$ z
(long double) 2.4142136L

For your information, I will also upload the results of gcc (5.4.0).

allinux@ip-10-0-0-29:~$ cat long_test.c
#include <stdio.h>
#include <math.h>

int main(){
        double x = 2.0;
        double y = 2.0;
        long double z = ((x + sqrt(x*x+y*y))/2.0);
        printf("The value of the expression is %25.20Lf\n", z);
        return 0;
}

allinux@ip-10-0-0-29:~$ gcc -std=c99 -o long_test long_test.c -lm
allinux@ip-10-0-0-29:~$ ./long_test
The value of the expression is    2.41421356237309492343

2022-09-22 18:40

If you have any answers or tips

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