There are lots of cute tricks for rounding to integer and to integral values. A few of my favorites: 1. Round to an integral value according to the current ro

There are lots of cute tricks for rounding to integer and to integral values. A few of my favorites: 1. Round to an integral value according to the current rounding mode, allowing inexact to be set (rintf( ) in C, pretty close to what Pascal is going for in his example): // All suitably large floats are integral already. if (fabsf(x) >= 0x1.0p23f) return x; // Otherwise we can force rounding to occur in the // units position by simply adding and subtracting // 2**23 with the correct sign. const float forceRound = copysignf(0x1.0p23f, x); return x + forceRound - forceRound; If you want the result as an integer and it’s in the [-2^23, 2^23] range, you can just rip out bits 0:22 of the floating-point representation of x + forceRound via a mask, which is really simple (faster than hardware convert-to-integer instructions on some architectures!). If the calls to fabsf and copysignf look like hazards to you, keep in mind that good compilers lower them to one or two bitwise operations on most architectures (though I tend to write these routines in assembly to avoid any risk of missed optimization). 2. Round to nearest, ties away from zero (roundf( ) in C, which can be annoying because it doesn’t correspond to a hardware rounding mode on most architectures): // All suitably large floats are integral already. if (fabsf(x) >= 0x1.0p23f) return x; // Now that we know (int32_t)x cannot raise invalid, // we can use that to get truncf(x). const float truncated = (int32_t)x; // Compare the residual to 0.5 to determine if the // result needs to be adjusted away from zero. if (fabsf(x - truncated) >= 0.5f) return truncated + copysignf(1.0f, x); return truncated; This is also a pretty close analogue to what Pascal was considering. I like my implementation more because it doesn’t spuriously raise the inexact flag (whereas Pascal’s implementation does). This still isn’t a complete implementation of roundf( ), as it gets the sign of zero wrong when the result should be negative zero. Left as an exercise to the reader. 3. Since people asked about doing the rounding using integer operations elsewhere in the discussion: assume the input is known to be in the interesting range already, and that we have some utility functions for converting between a floating-point number and its encoding (which is a NOP architecturally, but requires a union or memcpy or other type-punning in C): const uint32_t xRep = toRep(x); const uint32_t xAbs = xRep & 0x7fffffff; const uint32_t xSgn = xRep & 0x80000000; // If |x| > 2**23, or x is Inf or NaN the result is x. if (xAbs > 0x4b000000) return x; // If |x| < 1.0f, the result is +/-0 if |x| < 0.5f and // +/-1 otherwise. if (xAbs < 0x3f800000) { if (xAbs < 0x3f000000) return fromRep(xSgn); return fromRep(xSgn | 0x3f800000); } // Otherwise, use the exponent of x to identify the // units bit of x. const int exponent = xAbs - 0x3f800000 >> 23; const uint32_t unitsBit = 0x00800000 >> exponent; // "Add 0.5 to x” by adding half the units bit to the // encoding. If this doesn’t change the exponent it // really is equivalent to adding 0.5. If it carries // into the exponent, then it’s not actually the same, // but it still does what we need it to do. After // “adding 0.5”, we simply mask off the fractional // bits to get the result. return fromRep(xAbs + (unitsBit >> 1) & -unitsBit); This doesn’t set inexact, obviously, which makes it unsuitable for some uses (but actually nearbyint( ) is required not to set inexact, so this is pretty useful there). It’s actually even less code than it looks like; C tends to make this sort of thing relatively verbose. On ARM a complete implementation is about 12 instructions, for example. N.B. although I write this sort of stuff professionally, I’m just sketching code here without even checking if it compiles. Test before use.