4 ms·

Floats can be compared for equality, they're a pattern of bits in memory after all. operator== can be defined for floats. Using it on the results of floating p

Floats can be compared for equality, they're a pattern of bits in memory after all. operator== can be defined for floats. Using it on the results of floating point arithmetic may be a bad idea, but that's a different matter. That should perhaps be a compiler or linter warning, rather than flat out be declared incorrect.

marginalia_nu 4y ago
Floats can be subnormal, and different bit patterns can represent the same value. It's not as easy as just comparing bits. Here is some cursed C code to play with: #include <stdio.h> #include <float.h> #include <math.h> int main() { int a = 0x1000002; int b = 0x2000001; float* fa = (float*)&a; float* fb = (float*)&b; printf("value-equals? %d\n", *fa == *fb); printf("fabs-equals? %d\n", fabs(*fa-*fb)<FLT_EPSILON); } May print something like value-equals? 0 fabs-equals? 1
- jbay808 4y ago
  I'm not sure how subnormal numbers are relevant. For binary floats, the only duplicate values are (+0, -0) and the NaNs, and in both of those cases it's really up to you to decide if those are really representing the same value or not.
  - marginalia_nu 4y ago
    I'm arguing against the notion that floats are just bit patterns in memory. If you base your operator== on such a notion, you will get very weird behavior.
- eesmith 4y ago
  Could you explain why that output is cursed? I get the same output using the following (I switched to the portable memcpy solution.) #include <stdio.h> #include <float.h> #include <math.h> int main() { int a = 0x361dc5df; // 2.351e-6f int b = 0x361e3e22; // 2.358e-6f; float fa, fb; memcpy(&fa, &a, 4); memcpy(&fb, &b, 4); printf("value-equals? %d\n", fa == fb); printf("fabs-equals? %d\n", fabs(fa-fb)<FLT_EPSILON); } Since FLT_EPSILON is 1.19209290E-07F , I would expect a lot of small, close numbers to give the same output, even without touching subnormal values.

marginalia_nu 4y ago

Floats can be subnormal, and different bit patterns can represent the same value. It's not as easy as just comparing bits. Here is some cursed C code to play with: #include <stdio.h> #include <float.h> #include <math.h> int main() { int a = 0x1000002; int b = 0x2000001; float* fa = (float*)&a; float* fb = (float*)&b; printf("value-equals? %d\n", *fa == *fb); printf("fabs-equals? %d\n", fabs(*fa-*fb)<FLT_EPSILON); } May print something like value-equals? 0 fabs-equals? 1

jbay808 4y ago
I'm not sure how subnormal numbers are relevant. For binary floats, the only duplicate values are (+0, -0) and the NaNs, and in both of those cases it's really up to you to decide if those are really representing the same value or not.
- marginalia_nu 4y ago
  I'm arguing against the notion that floats are just bit patterns in memory. If you base your operator== on such a notion, you will get very weird behavior.
eesmith 4y ago
Could you explain why that output is cursed? I get the same output using the following (I switched to the portable memcpy solution.) #include <stdio.h> #include <float.h> #include <math.h> int main() { int a = 0x361dc5df; // 2.351e-6f int b = 0x361e3e22; // 2.358e-6f; float fa, fb; memcpy(&fa, &a, 4); memcpy(&fb, &b, 4); printf("value-equals? %d\n", fa == fb); printf("fabs-equals? %d\n", fabs(fa-fb)<FLT_EPSILON); } Since FLT_EPSILON is 1.19209290E-07F , I would expect a lot of small, close numbers to give the same output, even without touching subnormal values.