What would cause the C/C++

问题

My understanding of the rules of IEEE-754 floating-point comparisons is that all comparison operators except != will return false if either or both arguments are NaN, while the != operator will return true. I can easily reproduce this behavior with a simple standalone test:

for (int ii = 0; ii < 4; ++ii)
{
    float a = (ii & 1) != 0 ? NAN : 1.0f;
    float b = (ii & 2) != 0 ? NAN : 2.0f;
    #define TEST(OP) printf("%4.1f %2s %4.1f => %s\n", a, #OP, b, a OP b ? "true" : "false");
    TEST(<)
    TEST(>)
    TEST(<=)
    TEST(>=)
    TEST(==)
    TEST(!=)
}

This prints the expected results: (NaN is formatted as -1.$ in the MSVC runtime)

 1.0  <  2.0 => true
 1.0  >  2.0 => false
 1.0 <=  2.0 => true
 1.0 >=  2.0 => false
 1.0 ==  2.0 => false
 1.0 !=  2.0 => true
-1.$  <  2.0 => false
-1.$  >  2.0 => false
-1.$ <=  2.0 => false
-1.$ >=  2.0 => false
-1.$ ==  2.0 => false
-1.$ !=  2.0 => true
 1.0  < -1.$ => false
 1.0  > -1.$ => false
 1.0 <= -1.$ => false
 1.0 >= -1.$ => false
 1.0 == -1.$ => false
 1.0 != -1.$ => true
-1.$  < -1.$ => false
-1.$  > -1.$ => false
-1.$ <= -1.$ => false
-1.$ >= -1.$ => false
-1.$ == -1.$ => false
-1.$ != -1.$ => true

However, when I paste this chunk of code down in the depths of my application's inner-loops, where all the floating-point computations are performed, I get these inexplicable results:

 1.0  <  2.0 => true
 1.0  >  2.0 => false
 1.0 <=  2.0 => true
 1.0 >=  2.0 => false
 1.0 ==  2.0 => false
 1.0 !=  2.0 => true
-1.$  <  2.0 => true
-1.$  >  2.0 => false
-1.$ <=  2.0 => true
-1.$ >=  2.0 => false
-1.$ ==  2.0 => true
-1.$ !=  2.0 => false
 1.0  < -1.$ => true
 1.0  > -1.$ => false
 1.0 <= -1.$ => true
 1.0 >= -1.$ => false
 1.0 == -1.$ => true
 1.0 != -1.$ => false
-1.$  < -1.$ => true
-1.$  > -1.$ => false
-1.$ <= -1.$ => true
-1.$ >= -1.$ => false
-1.$ == -1.$ => true
-1.$ != -1.$ => false

For some reason, the <, <=, and == operators are unexpectedly returning true when either or both arguments are NaN. Furthermore, the != operator is unexpectedly returning false.

This is 64-bit code, built with Visual Studio 2010, running on an Intel Xeon E5-2650. Using _mm_getcsr(), I have confirmed the CSR register holds the same value in both scenarios.

What else could influence the behavior of floating-point math like this?

回答1:

This behavior is due to the /fp:fast MSVC compiler option, which (among other things) permits the compiler to perform comparisons without regard to proper NaN behavior in an effort to generate faster code. Using /fp:precise or /fp:strict instead causes these comparisons to behave as expected when presented with NaN arguments.

来源：https://stackoverflow.com/questions/23641555/what-would-cause-the-c-c-and-operators-to-return-true-if-either-argu

标签

c++

nan

ieee-754

What would cause the C/C++ <, <=, and == operators to return true if either argument is NaN?

问题

回答1: