Typecasting int pointer to float pointer

Question

I'm trying to do the following

int a[8]={1,2,3,4,5,6,7,8};

printf("%f\n", *(float *)a);
printf("%f\n", *((float *)a+1));
printf("%f\n", *((float *)a+2));
printf("%f\n", *((float *)a+3));
printf("%f\n", *((float *)a+4));
printf("%f\n", *((float *)a+5));
printf("%f\n", *((float *)a+6));
printf("%f\n", *((float *)a+7));

I get

0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000

The reason why I'm trying to print the elements in this way is because, I want to cast the int pointer to the array to the float pointer and pass it as a parameter for another function which only takes float *.

It seems that this does not work well. Can someone explain why this is not working?

int *ptr;
function((float *)ptr);

If I do this the function does not read the values the pointer is pointing to properly.. just returning 0.0000.

Answer 1

This is not correct. int and float are not guaranteed to have the same alignment.

Remember: Casting a value and casting a pointer are different scenarios. Casting a pointer changes the way to refer to the type value, which can almost certainly result in a mis-alignment in most of the cases.

As per C11 standard document, chapter §6.3.2.3

A pointer to an object type may be converted to a pointer to a different object type. If the resulting pointer is not correctly aligned⁶⁸⁾ for the referenced type, the behavior is undefined.

In your case, a work-around may be

printf("%f\n", (float)*a);  //cast the value, not the pointer

Answer 2

You cannot cast a pointer to int to a pointer to float, and expect to get your value converted to the corresponding number in floating point representation. Casting a single value works, but casting by changing a pointer type does not alter the representation.

If you need an array of floats, declare an array of floats, and cast one element at a time:

float b[8];
for (int i = 0 ; i != 8 ; i++) {
    b[i] = a[i];
}
func_expects_float(b, 8);