我应该在我的函数中改变什么才能计算实数？

Question

i am trying to write a function that repeatedly adds 0.001 to 't' and then plugs it into 'y' until 't' reaches 0.3 however the numbers come out wrong, but i've noticed that if i change float to int and change the numbers to integer, the fuction works.. what should i change so the function works properly

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

void main(void)
{
    float t,y,dt;

    dt = 0.001;
    y = 1;
    t = 0;

    while (t <= 0.3)
    {
        y = y + dt*(sin(y)+(t)*(t)*(t));
        t = t + dt;
    }

    printf("y is %d when t is 0.3\n" , y);
    return 0;
}

Answer 1

i've noticed that if i change float to int and change the numbers to integer, the fuction works.. what should i change so the function works properly

as said in a remark the problem is the way you (try to) print the value in

printf("y is %d when t is 0.3\\n" , y);

%d suppose the corresponding argument is an int and prints it as an int , but y is a float . Note that there is no conversion from float to int in that case because the arguments are managed through a varargs

just do

 printf("y is %f when t is 0.3\n" , y);

Also change

void main(void)

to

int main()

After the changes, compilation and execution :

/tmp % gcc -pedantic -Wall -Wextra f.c -lm
/tmp % ./a.out
y is 1.273792 when t is 0.3

---

Note that all the calculations are done in double , so better to replace float to double to type your vars

---

(edit) Compiling your initial code with gcc and the option -Wall signals your problems :

/tmp % gcc -Wall f.c -lm
f.c:4: warning: return type of 'main' is not 'int'
f.c: In function 'main':
f.c:18: warning: format '%d' expects type 'int', but argument 2 has type 'double'
f.c:19: warning: 'return' with a value, in function returning void

To use both -Wall and -Wextra is the better option