If you know that sizeof(float) >= sizeof(int) the memory alignment should be fine. Also in this case * (int* )&a does not access any bytes outside of an defined data object.

To predict the outcome one have to know the representation of float (IEEE 754) and the big/little endian issues which could be different between floats and ints.

printf("%d\n", a) does promote the value a to a double first (promotion rules) and then either the first or last part i(assuming sizeof(double) == 2*sizeof(int)) is shown depending on little or big endian architecture.