How to convert float to uint by float representation?

Question

In C I will do this to convert float representation of number into DWORD. Take the value from the address and cast the content to DWORD.

dwordVal = *(DWORD*)&floatVal;

So for example 44.54321 will become 0x42322C3F.

How can I do the same in C#?

Actually *(DWORD*)&floatVal is the wrong way to do it in C as it breaks strict aliasing rules that C compilers rely on for optimisation. Correct ways involve a union (see C99tc3 footnote 82) or memcpy(). — Pascal Cuoq
– Pascal Cuoq, Commented Feb 15, 2014 at 22:53

Community · Accepted Answer · 2017-05-23 12:09:14Z

You can use the BitConverter class:

uint value = BitConverter.ToUInt32(BitConverter.GetBytes(44.54321F), 0); Console.WriteLine("{0:x}", value); // 42322c3f

You could also do this more directly using an unsafe context:

float floatVal = 44.54321F; uint value; unsafe { value = *((uint*)(&floatVal)); } Console.WriteLine("{0:x}", value); // 42322c3f

However, I'd strongly recommend avoiding this. See Should you use pointers (unsafe code) in C#?

MarcinJuraszek · Accepted Answer · 2014-02-15 17:47:55Z

Use the BitConverter class:

float f = 44.54321f; uint u = BitConverter.ToUInt32(BitConverter.GetBytes(f), 0); System.Diagnostics.Debug.Assert(u == 0x42322C3F);

phuclv · Accepted Answer · 2022-06-10 14:41:03Z

.NET 6 added 2 more utilities:

Klaus Kobald · Accepted Answer · 2022-04-16 13:57:07Z

bitconvert creates a new byte array on every call. this is the solution:

 unsafe public static float ToDecibel(this float x) { uint* y = (uint*)&x; (*y) &= 0x7fffffff; return (*y) * 7.17711438e-7f - 764.6161886f; }

+1 for remembering unsafe. BUT ... This changes x! Better to use uint y = *(uint*)&x; y &= 0x7fffffff; return y * 7.17711438e-7f - 764.6161886f; to avoid changing x.

4 Answers 4