How to obtain hexadecimal value of -nan - c++

Question

I am writing some code in intel intrinsics and did this:
#include <iostream>
#include <xmmintrin.h>
float data[4];
__m128 val1 = _mm_set_ps1(2);
__m128 val2 = _mm_set_ps1(1);
val1 = _mm_cmpgt_ps(val1, val2);
_mm_store_ps(data, val1);
std::cout << std::hex << data[0];
I am trying to get the hexadecimal value of "true" in SSE intrinsics (which is -nan), but only keep getting -nan as "the hexadecimal value" whenever I try to print the hexadecimal value of -nan.
I also tried using std::oct and std::dec and neither of those worked.
I also tried comparing 0xFFFFFFFF and data[0] in different combinations and got this:
float data[4];
__m128 val1 = _mm_set_ps1(2);
__m128 val2 = _mm_set_ps1(1);
val1 = _mm_cmpgt_ps(val1, val2);
_mm_store_ps(data, val1);
float f = 0xFFFFFFFF;
float g = 0xFFFFFFFF;
std::cout << std::dec << (data[0] == f) << "\n"; // Prints "0"
std::cout << std::dec << (data[0] == data[0]) << "\n"; // Prints "0"
std::cout << std::dec << (f == g); // Prints "1"
Is there any way for me to print the hexadecimal value of -nan and if not, can somebody please tell me the binary, hexadecimal, etc. value of -nan?

Answer 1

Per IEEE specification, NaN is a floating-point value that has all of its exponent bits set to "1".
So a value with all the bits set to "1" would also be a NaN.
If you want to see the raw bytes, just print the raw bytes:
#include <cmath>
#include <iomanip>
#include <iostream>
#include <sstream>
template<typename T>
std::string get_hex_bytes(T x) {
std::stringstream res;
auto p = reinterpret_cast<const unsigned char*>(&x);
for (int i = 0; i < sizeof(x); ++i) {
if (i)
res << ' ';
res << std::setfill('0') << std::setw(2) << std::hex << (int)p[i];
}
return res.str();
}
int main() {
float data = NAN;
std::cout << get_hex_bytes(data) << std::endl;
}
On a little-endian machine will print something like:
00 00 c0 ff
P.S. float f = 0xFFFFFFFF; will not set all of the bits to "1", it simply converts an integer 0xFFFFFFFF to a floating point representation (perfectly representable with some loss of precision).

Answer 2

As the manual says, _mm_cmpgt_ps (which is really cmpps with a specific comparison predicate),
Performs a SIMD compare of the packed single-precision floating-point values in the source operand (second
operand) and the destination operand (first operand) and returns the results of the comparison to the destination
operand. The comparison predicate operand (third operand) specifies the type of comparison performed on each of
the pairs of packed values. The result of each comparison is a doubleword mask of all 1s (comparison true) or all
0s (comparison false). The sign of zero is ignored for comparisons, so that –0.0 is equal to +0.0.
(emphasis added)
"All 1s", or 0xFFFFFFFF in hexadecimal (since it's 32 bits per element), has the sign bit set (so there is a legitimate reason to print a - sign in front of whatever else this number might be) and since the exponent is all ones and the significand is not zero, it is also a NaN. The NaN-ness usually isn't very relevant, the main intended use for this result is as a mask in bitwise operations (eg _mm_and_ps, _mm_blendv_ps, etc), which do not care about the special semantics of NaN.

Answer 3

First of all, there's no such thing as a "negative nan". nan is, by definition, Not a Number. You can't negate it. -nan is the same sort of thing as nan.
There's no exactly standards-compliant way to get the underlying bits comprised by a floating-point value, but the closest thing is memcpy. Simply copy from a pointer to float or double to a pointer to an equivalently-sized unsigned integer type, then print that with std::hex active.