Matrix of structure C

Question

So I have this code where I'm trying to define a dynamic matrix with n*n size. It turns out that, instead of the output printing the value for G, it prints 0 instead. Any idea why?

#define G 6.67408e-11

typedef struct matrix
{
    double mass;
    double cmx;
    double cmy;

}MATRIX;

MATRIX **mtr;

void main(int argc, char** argv){
    const long n = atoi(argv[1]);

    mtr = (MATRIX**)calloc(n,sizeof(MATRIX*));
    for (int i=0; i<n; ++i)
    {
        mtr[i]=(MATRIX*)calloc(n,sizeof(MATRIX));
    }
    mtr[0][0].cmx=G;
    printf("%f\n", mtr[0][0].cmx);
}

This is not a [mcve]. A [mcve] would be `#include int main(void) { printf("%f\n", 6.67408e-11); }` — Antti Haapala -- Слава Україні, Mar 19 '19 at 09:28
In a quick glance, it is probable that 6.67408e-11 is a really small value and it gets rounded to 0. — kyriakosSt, Mar 19 '19 at 09:31
https://stackoverflow.com/questions/54162152/what-precisely-does-the-g-printf-specifier-mean — Antti Haapala -- Слава Україні, Mar 19 '19 at 09:34

dteod · Answer 1 · 2019-03-19T09:49:58.673

2

Just use %e or %g to display the output. If you use %f, you're telling the compiler to print the number in standard format, that would be 0.0000000000667408 but since %f is formatted as default with 6 decimal ciphers the compiler rounds it to 0.000000 .

edited Mar 19 '19 at 09:49

answered Mar 19 '19 at 09:40

dteod

195
4

1

"%.16f\n" would be another way if OP has severe allergies to scientific notation. – visibleman Mar 19 '19 at 09:46

Matrix of structure C

1 Answers1