Can a IEEE 754 real number “cover” all integers within its range?

Question

The original question was edited (shortened) to focus on a problem of precision, not range.

Single, or double precision, every representation of real number

Answer 1

No.

For example, on my system, the type float can represent values up to approximately 3.40282e+38. As an integer, that would be approximately 340282000000000000000000000000000000000, or about 2¹²⁸.

The size of float is 32 bits, so it can exactly represent at most 2³² distinct numbers.

An integer object generally uses all of its bits to represent values (with 1 bit dedicated as a sign bit for signed types). A floating-point object uses some of its bits to represent an exponent (8 bits for IEEE 32-bit float); this increases its range at the cost of losing precision.

A concrete example (1267650600228229401496703205376.0 is 2¹⁰⁰, and is exactly representable as a float):

#include 
#include 
#include 
int main(void) {
    float x = 1267650600228229401496703205376.0;
    float y = nextafterf(x, FLT_MAX);
    printf("x = %.1f\n", x);
    printf("y = %.1f\n", y);

    return 0;
}

The output on my system is:

x = 1267650600228229401496703205376.0
y = 1267650751343956853325350043648.0

Another way to look at it:

A 32-bit object can represent at most 2³² distinct values.

A 32-bit signed integer can represent all integer values in the range -2147483648 .. 2147483647 (-2³¹ .. +2³¹-1).

A 32-bit float can represent many values that a 32-bit signed integer can't, either because they're fractional (0.5) or because they're too big (2.0¹⁰⁰). Since there are values that can be represented by a 32-bit float but not by a 32-bit int, there must be other values that can be represented by a 32-bit int but not by a 32-bit float. Those values are integers that have more significant digits than a float can handle, because the int has 31 value bits but the float has only about 24.