*Quiz:* Does this loop end, and if it does, what is the value of `k`

after the loop? (assuming IEEE-754 and usual 32-bit semantics for `int/float`

)

Running this code in VS you get `k=16777217`

which equals `2^24+1`

.

In other words, in line with the true intent of the quiz, `float`

can encode -exactly- (without any precission loss) all natural numbers up to `2^24`

(included). Because `float`

encodes sign in a dedicated bit, this property holds true for negative values as well. So you could *hypothetically* encode any `[-2^24..+2^24]`

`int`

in a `float`

with a static/implicit cast.

I said *hypothetically* because generally speaking I would never do such thing. Carelessly casting between `int`

and `float`

(without an explicit `floor`

, `ceil`

, `trunc`

, `round`

…) is often a sign of lousy coding, unless performance is at stake and you know well what you are doing.

However, I came across the situation in Unity recently, where the `Vertex ID`

node in their Shader Graph hands over `SV_VertexID`

as a `float`

(disappointed face :-P). I would’ve expected an `uint/int`

output or a `float`

that you could reinterpret-cast to retrieve the raw `SV_VertexID`

bits with asuint(). But nope. You get a `float`

which seemingly carries `(float)SV_VertexID`

.

One may recklessly use this `float`

to index a `StructuredBuffer`

ignoring the way static-casting works. This is one case where ignorance is bliss, because the `float`

received exactly matches the original `int`

as long as the original value is `<=2^24`

. That is, as long as you are dealing with fewer than (roughly) 16.7 million vertices, which is usually the case.

I believe that Unity’s ShaderGraph doesn’t support `int/uint`

as In/Out values between nodes, so I guess that the `Vertex ID`

and `Instance ID`

nodes just static-cast the corresponding `SV_...`

value to a float. But it would be better (in the *pedantic* sense) to reinterpret the bit pattern of the raw values with `asfloat`

and then let the user retrieve them with `asint()/asuint()`

.

–

**reinterpret-cast between int/float in HSLS:**

–

**A (loose) explanation of the 2^24 limit:**

This is the IEEE-754 standard for floating-point numbers, as described in Wikipedia:

The value of a float-encoded number is reconstructed as `(-1)^S * M * 2^E`

.

`S`

is the 1-bit sign.`M`

is the 23-bit mantissa, interpreted as`1.xxxxx`

(in binary).`E`

is the 8-bit exponent, used as`E-127`

where 127 is often called*bias*.

*e.g.,* a power-of-2 integer number like 4 would be encoded as:

`M=1.00`

which is the binary representation of 4, with the point at the left-most 1.`E=2`

(+bias).

The restored number is `1.00 shl 2=4`

.

We should be able to do the same for all power-of-2 integers until we max out `E`

.

For non-power-of-2 integers the process is similar. *e.g.,* number 5:

`M=1.01...00`

.`E=2`

(+bias).

The restored number is now `1.01 shl 2=5`

.

This works the same for all integer numbers as long as `M`

can hold the raw binary representation of the number. The tallest binary number that `M`

can hold is 23 consecutive 1s. That is: 1.11…11 (24x1s in total). With `E=23`

(+bias) this equals `2^24-1`

.

The next integer `2^24`

would be encoded as `1.00...00`

(24x0s, clamped to 23, but the trailing 0s are meaningless here). With `E=24`

(+bias) this equals `2^24`

(**the answer provided above!!**).

But the next integer `1.{23x0s.1}`

can’t be encoded in a 23-bit `M`

. So from 2^24 onwards, there is necessarily a loss. Some integers beyond 2^24 (such as powers-of-2) will be luckily encoded exactly by `float`

. But not all consecutive numbers will. Actually, `2^24+1`

is the first integer that won’t.

As always, *apologies for any mistakes in any of my posts*.

–

**[EDIT]** The same reasoning can be applied to double, where the mantisa `M`

is 52-bit long.

Hence double can encode exactly all integers in the range `[-2^53..+2^53]`

.