Assigning arrays to function pointer arguments in C

Question

I'm declaring pointers before my function call and want to get arrays when the function is executed. I followed the question answered here to obtain the following code

double *x;
double *y;
myfunction(1.0,&x,&y);
printf("x[0] = %1.1f\n",x[0] );

Where myfunction is

void myfunction(double data, double **x, double **y){
    /* Some code */
    int calculated_size = 10;
    *x = malloc(calculated_size*sizeof(double));
    *y = malloc(calculated_size*sizeof(double));
    int k;
    for (k = 0;k < calculated_size; k++)
    {
        *x[k] = k ;
        *y[k] = k ;
    }

}

Which gives me a segmentation fault as soon as it tries to assign

*x[k] = k;

Could someone indicate the exact way I should be doing this?

sideshowbarker · Accepted Answer · 2023-01-27 11:56:09Z

7

Replace *x[k] = k ; with (*x)[k] = k ; or better use as shown below

int *xa = *x;
xa[k] = k ;

[] binds tighter than * in C Operator precedence table causing the fault.

You expect : (*x)[k]
But you get: *(x[k])

Live example

edited Jan 27, 2023 at 11:56

sideshowbarker

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answered Nov 5, 2014 at 11:20

Mohit Jain

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2 Comments

struct Over a year ago

You can also do *(*x + k) = k and *(*y + k) = k as an alternative syntax.

Brieuc de R Over a year ago

Wonderful! Drives me crazy how long I was looking for a mistake like this... Thank you

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