I am a bit confused with this type of C code:
void double_function(double **arr){
printf("Value at 1: %f \n", arr[1]);
}
int main() {
double arr[3] = {0.11,1.2,2.56};
double_function(&arr);
}
This does not print the value 1.2. I tried *(arr)[1] and (*arr[1]) as well, and I can't seem to access it. I keep getting:
Process finished with exit code 139 (interrupted by signal 11: SIGSEGV)
Can someone help clarify this notation on how to access the array?
Please note that the specifications require that the function take a double **arr.
The variable arr is an array of doubles and you clarified that you wanted to pass it in a pointer to pointer. Using temporary (double *) { arr } so we pass in its address with &(double *) { arr }:
#include <stdio.h>
void double_function(double **arr){
printf("Value at 1: %f \n", (*arr)[1]);
}
int main() {
double arr[3] = {0.11,1.2,2.56};
double_function(&(double *) {arr});
}