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

No, not all, but there exists a range within which you can represent all integers accurately.

Structure of 32bit floating point numbers

The 32bit floating point type uses

• 1 bit for the sign
• 8 bits for the exponent
• 23 bits for the fraction (leading 1 implied)

Representing numbers

Basically, you have a number in the form

(-)1.xxxx_xxxx_xxxx_xxxx_xxxx_xxx (binary)

which you then shift left/right with the (unbiased) exponent.

To have it represent an integer requiring n bits, you need to shift it by n-1 bits to the left. (All xes beyond the floating point are simply zero)

Representing integers with 24 bits

It is easy to see, that we can represent all integers requiring 24 bits (and less)

1xxx_xxxx_xxxx_xxxx_xxxx_xxxx.0 (unbiased exponent = 23)

since we can set the xes at will to either 1 or 0.

The highest number we can represent in this fashion is:

1111_1111_1111_1111_1111_1111.0

or 2^24 - 1 = 16777215

The next higher integer is 1_0000_0000_0000_0000_0000_0000. Thus, we need 25 bits.

Representing integers with 25 bits

If you try to represent a 25 bit integer (unbiased exponent = 24), the numbers have the following form:

1_xxxx_xxxx_xxxx_xxxx_xxxx_xxx0.0

The twenty-three digits that are available to you have all been shifted past the floating point. The leading digit is always a 1. In total, we have 24 digits. But since we need 25, a zero is appended.

A maximum is found

We can represent ``1_0000_0000_0000_0000_0000_0000with the form1_xxxx_xxxx_xxxx_xxxx_xxxx_xxx0.0, by simply assigning 1to allxes. The next higher integer from that is: 1_0000_0000_0000_0000_0000_0001. It's easy to see that this number cannot be represented accurately, because the form does not allow us to set the last digit to 1: It is always 0`.

It follows, that the 1 followed by 24 zeroes is an upper bound for the integers we can accurately represent. The lower bound simply has its sign bit flipped.

Range within which all integers can be represented (including boundaries)

• 2²⁴ as an upper bound
• -2²⁴ as a lower bound

Structure of 64bit floating point numbers

• 1 bit for the sign
• 11 exponent bits
• 52 fraction bits

Range within which all integers can be represented (including boundaries)

• 2⁵³ as an upper bound
• -2⁵³ as a lower bound

This easily follows by applying the same argumentation to the structure of 64bit floating point numbers.

Note: That is not to say these are all integers we can represent, but it gives you a range within which you can represent all integers. Beyond that range, we can only represent a power of two multiplied with an integer from said range.

Combinatorial argument

Simply convincing ourselves that it is impossible for 32bit floating point numbers to represent all integers a 32bit integer can represent, we need not even look at the structure of floating point numbers.

1. With 32 bits, there are 2³² different things we can represent. No more, no less.
2. A 32bit integer uses all of these "things" to represent numbers (pairwise different).
3. A 32bit floating point number can represent at least one number with a fractional part.

Thus, it is impossible for the 32bit floating point number to be able to represent this fractional number in addition to all 2³² integers.

macias, to add to the already excellent answer by phant0m (upvoted; I suggest you accept it), I'll use your own words.

"No it cannot be done, for example number 16777217 is not represented exactly by float number."

Also, "for example number 9223372036854775809 is not represented exactly by double number".

This is assuming your computer is using the IEEE floating point format, which is a pretty strong bet.

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 <stdio.h>
#include <float.h>
#include <math.h>
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.

Apparently you are asking whether a Real data type can represent all of the integer values in its range (absolute values up to FLT_MAX or DBL_MAX, in C, or similar constants in other languages).

The largest numbers representable by floating point numbers stored in K bits typically are much larger than the 2^K number of integers that K bits can represent, so typically the answer is no. 32-bit C floats exceed 10^37, 32-bit C integers are less than 10^10. To find out the next representable number after some number, use nextafter() or nextafterf(). For example, the code

printf ("%20.4f %20.4f\n", nextafterf(1e5,1e9), nextafterf(1e6,1e9));
printf ("%20.4f %20.4f\n", nextafterf(1e7,1e9), nextafterf(1e8,1e9));

prints out

     100000.0078         1000000.0625
   10000001.0000       100000008.0000

You might be interested in whether an integer J that is between two nearby fractional floating values R and S can be represented exactly, supposing S-R < 1 and R < J < S. Yes, such a J can be represented exactly. Every float value is the ratio of some integer and some power of 2. (Or is the product of some integer and some power of 2.) Let the power of 2 be P, and suppose R = U/P, S = V/P. Now U/P < J < V/P so U < J*P < V. More of J*P's low-order bits are zero than are those of U, V (because V-U < P, due to S-R < 1), so J can be represented exactly.

I haven't filled in all the details to show that J*P-U < P and V-J*P < P, but under the assumption S-R < 1 that's straightforward. Here is an example of R,J,S,P,U,V value computations: Let R=99999.9921875 = 12799999/128, (ie P=128); let S=100000.0078125 = 12800001/128; we have U=0xc34fff and V=0xc35001 and there is a number between them that has more low-order zeroes than either; to wit, J = 0xc35000/128 = 12800000/128 = 100000.0. For the numbers in this example, note that U and V require 24 bits for their exact representations (6 ea. 4-bit hex digits). Note that 24 bits is the number of bits of precision in IEEE 754 single-precision floating point numbers. (See table in wikipedia article.)

That each floating point number is a product or ratio of some integer and some power of 2 (as mentioned two paragraphs above) also is discussed in that floating point article, in a paragraph that begins:

By their nature, all numbers expressed in floating-point format are rational numbers with a terminating expansion in the relevant base (for example, ... a terminating binary expansion in base-2). Irrational numbers, such as π or √2, or non-terminating rational numbers, must be approximated. The number of digits (or bits) of precision also limits the set of rational numbers that can be represented exactly.