In a comment I mentioned in passing that you could switch to representing 1024ths of meters instead of millimeters. That also helps with representability generally. Your code would look like this:

int i = 1234567890;
float f = ((float)i) / 1024.f;

You don't need to worry about representability at all here. The temporary float value will already be the best representation of i possible in a single precision float, and the divisor 1024.f is not only a perfectly representable floating point number (feel free to write it with the wonky hexadecimal floating literal notation if you prefer) but will actually reduce to a trivial subtraction of the exponent of your floating point value when executed.

In a comment I mentioned in passing that you could switch to representing 1024ths of meters instead of millimeters. That also helps with representability generally. Your code would look like this:

int i = 1234567890;
float f = ((float)i) / 1024.f;

You don't need to worry about representability at all here. The temporary float value will already be the best representation of i possible in a single precision float, and the divisor 1024.f is not only a perfectly representable floating point number (feel free to write it with the wonky hexadecimal floating literal notation if you prefer) but will actually reduce to a trivial subtraction of the exponent of your floating point value when executed.

Of course you (and your compiler) will probably notice that a division isn't necessary at all at this point. The above code could we written with a similar multiplication (or even maybe some fancy hardware instructions):

int i = 1234567890;
float f = ((float)i) * 0x1p-10f;

(0x1p-10f is just a direct representation of the value 1.f / 1024.f)