Initialize a double array with int values in C

Question

I'm learning some c now, and must do an exercise for the university. I need an double array which should be filld with int values. So I've this.

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

int main( int argc, char *argv[] ) {

int size=0, i=0;

printf("Enter size of array: ");
scanf("%d", &size);

double* field = malloc(size * sizeof(double));
double value;

for(i=0; i<size; i++) {
    field[i] = i;;
}

for(i=0; i<size; i++) {
    printf("Field%i: %d\n",i, field[i]);
}

free(field);

return 0;

}

But, when I execute it all values are set to 0.

Enter size of array: 10
Field0: 0
Field1: 0
Field2: 0
Field3: 0
Field4: 0
Field5: 0
Field6: 0
Field7: 0
Field8: 0
Field9: 0

Answer 1

printf's %d stands for decimal integer, not double.

So this line:

printf("Field%i: %d\n",i, field[i]);

should be:

printf("Field%i: %f\n",i, field[i]);

Using the wrong format specifier is undefined behavior which you had earlier.

From standard

If any argument is not the correct type for the corresponding conversion specification, the behavior is undefined.

---

Also check the return value of malloc. It let's you know whether the allocation failed (returns NULL) or not.

_{A bit more insight from David C. Rankin's comment}

_{The %d format specifier was looking for 4-bytes (sizeof int bytes) in memory ordered based on the endianess of your machine. When you provided an 8-byte (sizeof double) value comprised of a sign-bit, normalized exponent and mantissa, it had no idea what to do with it.}

_{Well here in printf we can use %f instead of %lf also which has added benefit of compatibility with pre-C99 versions and also it is shorter. (chux pointed this).}

Answer 2

I don't think whether malloc function allocated the memory for the array. Check the syntax of malloc and also the return type.