C: Multidimensional array implementatio not working, giving different values

Question

I am trying to implement a multidimensional array. Below, you will see for one example in jArray[2][2] that I assign 20.0, clearly. However, both printf statements don't yield the same result. Thanks for your help!

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

#define M_PI 3.14159265358979323846

int main(){ 

    float x1 = 0.1;
    float x2 = 0.1;
    float x3 = -0.1;

    float jArray [3][3] = {
        {3.0, x3*sin(x2*x3), x2*sin(x2*x3)}, 
        {2*x1, -162*(x2+0.1), cos(x3)}, 
        {-x2*exp(-x1*x2), -x1*exp(x1*x2), 20.0}
    };

    float matrix0 [3][3] = {
        {jArray[0][0], jArray[0][1], jArray[0][2]},
        {jArray[1][0], jArray[1][1], jArray[1][2]},
        {jArray[2][0], jArray[2][1], jArray[2][2]},     
    };      

    printf("%f\n\n", jArray[2][2]);

    printf("[%f\t%f\t%f]\n[%f\t%f\t%f]\n[%f\t%f\t%f]\n\n",
        matrix0[0][0], matrix0[0][1], matrix0[0,2],
        matrix0[1][0], matrix0[1][1], matrix0[1,2],
        matrix0[2][0], matrix0[2][1], matrix0[2,2]);    

    return 1;

}

Output:

20.000000

[3.000000   0.001000    0.200000]
[-32.400002 -0.099005   -0.101005]
[0.000000   0.010000    0.000000]

matrix0[0,2]? If your compiler isn't giving you warnings, then you should enable more warnings. — Some programmer dude
– Some programmer dude, Commented Mar 17, 2015 at 6:37
@JoachimPileborg Wow ... just, wow. I need to sleep. Thanks. — capcom
– capcom, Commented Mar 17, 2015 at 6:39

Spikatrix · Accepted Answer · 2015-03-17 06:39:57Z

3

Replace

    printf("[%f\t%f\t%f]\n[%f\t%f\t%f]\n[%f\t%f\t%f]\n\n",
    matrix0[0][0], matrix0[0][1], matrix0[0,2],
    matrix0[1][0], matrix0[1][1], matrix0[1,2],
    matrix0[2][0], matrix0[2][1], matrix0[2,2]);

with

    printf("[%f\t%f\t%f]\n[%f\t%f\t%f]\n[%f\t%f\t%f]\n\n",
    matrix0[0][0], matrix0[0][1], matrix0[0][2],
    matrix0[1][0], matrix0[1][1], matrix0[1][2],
    matrix0[2][0], matrix0[2][1], matrix0[2][2]);

Your compiler would've emitted a warning for that because %f expects a float(or a double), not a float*.

answered Mar 17, 2015 at 6:39

Spikatrix

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