When a 64bit int is cast to 64bit float in C/C++ and doesn't have an exact match, will it always land on a non-fractional number?

When int64_t is cast to double and doesn't have an exact match, to my knowledge I get a sort of best-effort-nearest-value equivalent in double. For example, 9223372036854775000 in int64_t appears to end up as 9223372036854774784.0 in double:

#include <stdio.h>

int main(int argc, const char **argv) {
    printf("Corresponding double: %f\n", (double)9223372036854775000LL);
    // Outputs: 9223372036854774784.000000
    return 0;
}

It appears to me as if an int64_t cast to a double always ends up on as a clean non-fractional number, even in this higher number range where double has really low precision. However, I just observed this from random attempts. Is this guaranteed to happen for any value of int64_t cast to a double?

And if I cast this non-fractional double back to int64_t, will I always get the exact corresponding 64bit int with the .0 chopped off? (Assuming it doesn't overflow during the conversion back.) Like here:

#include <inttypes.h>
#include <stdio.h>

int main(int argc, const char **argv) {
    printf("Corresponding double: %f\n", (double)9223372036854775000LL);
    // Outputs: 9223372036854774784.000000
    printf("Corresponding int to corresponding double: %" PRId64 "\n",
           (int64_t)((double)9223372036854775000LL));
    // Outputs: 9223372036854774784
    return 0;
}

Or can it be imprecise and get me the "wrong" int in some corner cases?

Intuitively and from my tests the answer to both points appears to be "yes", but if somebody with a good formal understanding of the floating point standards and the maths behind it could confirm this that would be really helpful to me. I would also be curious if any known more aggressive optimizations like gcc's -Ofast are known to break any of this.