Why does setting a const variable (which will be stored with the same value) lead to a different result once...
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}
Pretty basic code:
#include <iostream>
int main() {
std::cout.precision(100);
double a = 9.79999999999063220457173883914947509765625;
double b = 0.057762265046662104872599030613855575211346149444580078125;
const double bConst = 0.057762265046662104872599030613855575211346149444580078125;
double c = a * b;
std::cout << " a: " << a << std::endl;
std::cout << " b: " << b << std::endl;
std::cout << " bConst: " << bConst << std::endl;
std::cout << " c: " << c << std::endl << std::endl;
std::cout << " c/b: " << c / b << std::endl;
std::cout << " c/bConst: " << c / bConst << std::endl;
}
Which outputs:
a: 9.79999999999063220457173883914947509765625
b: 0.057762265046662104872599030613855575211346149444580078125
bConst: 0.057762265046662104872599030613855575211346149444580078125
c: 0.5660701974567474703547986791818402707576751708984375
c/b: 9.7999999999906304282148994388990104198455810546875
c/bConst: 9.79999999999063220457173883914947509765625
As you can see, b
and bConst
seem to be treated using the same value - i.e. it prints for both the same 0.057762265046662104872599030613855575211346149444580078125
value.
So I guess they are "stored" both the same. The only difference is that b
is not const
.
Then, I do the same c / b
operation twice: one time using b
, another time using bConst
.
As you can see, it leads to two different results. And this makes me wonder.
Can you explain technically why this happens?
c++ optimization floating-point const
add a comment |
Pretty basic code:
#include <iostream>
int main() {
std::cout.precision(100);
double a = 9.79999999999063220457173883914947509765625;
double b = 0.057762265046662104872599030613855575211346149444580078125;
const double bConst = 0.057762265046662104872599030613855575211346149444580078125;
double c = a * b;
std::cout << " a: " << a << std::endl;
std::cout << " b: " << b << std::endl;
std::cout << " bConst: " << bConst << std::endl;
std::cout << " c: " << c << std::endl << std::endl;
std::cout << " c/b: " << c / b << std::endl;
std::cout << " c/bConst: " << c / bConst << std::endl;
}
Which outputs:
a: 9.79999999999063220457173883914947509765625
b: 0.057762265046662104872599030613855575211346149444580078125
bConst: 0.057762265046662104872599030613855575211346149444580078125
c: 0.5660701974567474703547986791818402707576751708984375
c/b: 9.7999999999906304282148994388990104198455810546875
c/bConst: 9.79999999999063220457173883914947509765625
As you can see, b
and bConst
seem to be treated using the same value - i.e. it prints for both the same 0.057762265046662104872599030613855575211346149444580078125
value.
So I guess they are "stored" both the same. The only difference is that b
is not const
.
Then, I do the same c / b
operation twice: one time using b
, another time using bConst
.
As you can see, it leads to two different results. And this makes me wonder.
Can you explain technically why this happens?
c++ optimization floating-point const
Here (mingw-w64 with g++ 8.1) both results are the same when optimized with -O2 and I get your result with -Ofast.
– Benjamin Bihler
Feb 15 at 10:12
1
I thought you got a grasp of the perverse effects of-Ofast
on floatting operation determinism in your other question stackoverflow.com/q/54694274/5470596
– YSC
Feb 15 at 14:33
add a comment |
Pretty basic code:
#include <iostream>
int main() {
std::cout.precision(100);
double a = 9.79999999999063220457173883914947509765625;
double b = 0.057762265046662104872599030613855575211346149444580078125;
const double bConst = 0.057762265046662104872599030613855575211346149444580078125;
double c = a * b;
std::cout << " a: " << a << std::endl;
std::cout << " b: " << b << std::endl;
std::cout << " bConst: " << bConst << std::endl;
std::cout << " c: " << c << std::endl << std::endl;
std::cout << " c/b: " << c / b << std::endl;
std::cout << " c/bConst: " << c / bConst << std::endl;
}
Which outputs:
a: 9.79999999999063220457173883914947509765625
b: 0.057762265046662104872599030613855575211346149444580078125
bConst: 0.057762265046662104872599030613855575211346149444580078125
c: 0.5660701974567474703547986791818402707576751708984375
c/b: 9.7999999999906304282148994388990104198455810546875
c/bConst: 9.79999999999063220457173883914947509765625
As you can see, b
and bConst
seem to be treated using the same value - i.e. it prints for both the same 0.057762265046662104872599030613855575211346149444580078125
value.
So I guess they are "stored" both the same. The only difference is that b
is not const
.
Then, I do the same c / b
operation twice: one time using b
, another time using bConst
.
As you can see, it leads to two different results. And this makes me wonder.
Can you explain technically why this happens?
c++ optimization floating-point const
Pretty basic code:
#include <iostream>
int main() {
std::cout.precision(100);
double a = 9.79999999999063220457173883914947509765625;
double b = 0.057762265046662104872599030613855575211346149444580078125;
const double bConst = 0.057762265046662104872599030613855575211346149444580078125;
double c = a * b;
std::cout << " a: " << a << std::endl;
std::cout << " b: " << b << std::endl;
std::cout << " bConst: " << bConst << std::endl;
std::cout << " c: " << c << std::endl << std::endl;
std::cout << " c/b: " << c / b << std::endl;
std::cout << " c/bConst: " << c / bConst << std::endl;
}
Which outputs:
a: 9.79999999999063220457173883914947509765625
b: 0.057762265046662104872599030613855575211346149444580078125
bConst: 0.057762265046662104872599030613855575211346149444580078125
c: 0.5660701974567474703547986791818402707576751708984375
c/b: 9.7999999999906304282148994388990104198455810546875
c/bConst: 9.79999999999063220457173883914947509765625
As you can see, b
and bConst
seem to be treated using the same value - i.e. it prints for both the same 0.057762265046662104872599030613855575211346149444580078125
value.
So I guess they are "stored" both the same. The only difference is that b
is not const
.
Then, I do the same c / b
operation twice: one time using b
, another time using bConst
.
As you can see, it leads to two different results. And this makes me wonder.
Can you explain technically why this happens?
c++ optimization floating-point const
c++ optimization floating-point const
edited Feb 15 at 15:47
sepp2k
301k39602617
301k39602617
asked Feb 15 at 10:09
markzzzmarkzzz
19.3k94234408
19.3k94234408
Here (mingw-w64 with g++ 8.1) both results are the same when optimized with -O2 and I get your result with -Ofast.
– Benjamin Bihler
Feb 15 at 10:12
1
I thought you got a grasp of the perverse effects of-Ofast
on floatting operation determinism in your other question stackoverflow.com/q/54694274/5470596
– YSC
Feb 15 at 14:33
add a comment |
Here (mingw-w64 with g++ 8.1) both results are the same when optimized with -O2 and I get your result with -Ofast.
– Benjamin Bihler
Feb 15 at 10:12
1
I thought you got a grasp of the perverse effects of-Ofast
on floatting operation determinism in your other question stackoverflow.com/q/54694274/5470596
– YSC
Feb 15 at 14:33
Here (mingw-w64 with g++ 8.1) both results are the same when optimized with -O2 and I get your result with -Ofast.
– Benjamin Bihler
Feb 15 at 10:12
Here (mingw-w64 with g++ 8.1) both results are the same when optimized with -O2 and I get your result with -Ofast.
– Benjamin Bihler
Feb 15 at 10:12
1
1
I thought you got a grasp of the perverse effects of
-Ofast
on floatting operation determinism in your other question stackoverflow.com/q/54694274/5470596– YSC
Feb 15 at 14:33
I thought you got a grasp of the perverse effects of
-Ofast
on floatting operation determinism in your other question stackoverflow.com/q/54694274/5470596– YSC
Feb 15 at 14:33
add a comment |
2 Answers
2
active
oldest
votes
The "issue" is due to the -freciprocal-math
switch (implied by -Ofast
):
Allow the reciprocal of a value to be used instead of dividing by the value if this enables optimizations. For example
x / y
can be replaced withx * (1/y)
, which is useful if(1/y)
is subject to common subexpression elimination. Note that this loses precision and increases the number of flops operating on the value.
The compiler can calculate d = 1/bConst
at compile time and change from:
c/bConst
to
c * d
but multiplication and division are different instructions with different performance and precision.
See: http://coliru.stacked-crooked.com/a/ba9770ec39ec5ac2
Thank you for that explanation... interesting!
– Benjamin Bihler
Feb 15 at 12:27
Uhm... not sure that's the behaviour: coliru.stacked-crooked.com/a/79e5a3501af9a217 . Usingc * 1 / b
=c / b)
(givingb
=bConst
). Instead,c * 1 / bConst
give the same asc / bConst
... so it should be somethings different.
– markzzz
Feb 15 at 13:05
Uhm... wait! If I didc * (1 / b)
(instead ofc * 1 / b
) works as you said.
– markzzz
Feb 15 at 13:08
add a comment |
You are using -Ofast
in your link, which enables all -O3
optimizations and includes both -ffast-math
, which in turns includes -funsafe-math-optimizations
.
From what I could glean, with optimizations enabled, -funsafe-math-optimizations
allows the compiler to reduce the precision of some computations. This seems to be what happens in the c/bConst
case.
The anwer is correct. Any idea why this is the case?
– Benjamin Bihler
Feb 15 at 10:15
What do you mean with "leads to a different result depending on the const-ness"? Shouldn't I get the same results starting with the same quotient and dividend?
– markzzz
Feb 15 at 10:16
@BenjaminBihler I'm looking into it.
– Nelfeal
Feb 15 at 10:16
2
@markzzz Unsafe optimizations tend to not care about this kind of logic.
– Nelfeal
Feb 15 at 10:17
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The "issue" is due to the -freciprocal-math
switch (implied by -Ofast
):
Allow the reciprocal of a value to be used instead of dividing by the value if this enables optimizations. For example
x / y
can be replaced withx * (1/y)
, which is useful if(1/y)
is subject to common subexpression elimination. Note that this loses precision and increases the number of flops operating on the value.
The compiler can calculate d = 1/bConst
at compile time and change from:
c/bConst
to
c * d
but multiplication and division are different instructions with different performance and precision.
See: http://coliru.stacked-crooked.com/a/ba9770ec39ec5ac2
Thank you for that explanation... interesting!
– Benjamin Bihler
Feb 15 at 12:27
Uhm... not sure that's the behaviour: coliru.stacked-crooked.com/a/79e5a3501af9a217 . Usingc * 1 / b
=c / b)
(givingb
=bConst
). Instead,c * 1 / bConst
give the same asc / bConst
... so it should be somethings different.
– markzzz
Feb 15 at 13:05
Uhm... wait! If I didc * (1 / b)
(instead ofc * 1 / b
) works as you said.
– markzzz
Feb 15 at 13:08
add a comment |
The "issue" is due to the -freciprocal-math
switch (implied by -Ofast
):
Allow the reciprocal of a value to be used instead of dividing by the value if this enables optimizations. For example
x / y
can be replaced withx * (1/y)
, which is useful if(1/y)
is subject to common subexpression elimination. Note that this loses precision and increases the number of flops operating on the value.
The compiler can calculate d = 1/bConst
at compile time and change from:
c/bConst
to
c * d
but multiplication and division are different instructions with different performance and precision.
See: http://coliru.stacked-crooked.com/a/ba9770ec39ec5ac2
Thank you for that explanation... interesting!
– Benjamin Bihler
Feb 15 at 12:27
Uhm... not sure that's the behaviour: coliru.stacked-crooked.com/a/79e5a3501af9a217 . Usingc * 1 / b
=c / b)
(givingb
=bConst
). Instead,c * 1 / bConst
give the same asc / bConst
... so it should be somethings different.
– markzzz
Feb 15 at 13:05
Uhm... wait! If I didc * (1 / b)
(instead ofc * 1 / b
) works as you said.
– markzzz
Feb 15 at 13:08
add a comment |
The "issue" is due to the -freciprocal-math
switch (implied by -Ofast
):
Allow the reciprocal of a value to be used instead of dividing by the value if this enables optimizations. For example
x / y
can be replaced withx * (1/y)
, which is useful if(1/y)
is subject to common subexpression elimination. Note that this loses precision and increases the number of flops operating on the value.
The compiler can calculate d = 1/bConst
at compile time and change from:
c/bConst
to
c * d
but multiplication and division are different instructions with different performance and precision.
See: http://coliru.stacked-crooked.com/a/ba9770ec39ec5ac2
The "issue" is due to the -freciprocal-math
switch (implied by -Ofast
):
Allow the reciprocal of a value to be used instead of dividing by the value if this enables optimizations. For example
x / y
can be replaced withx * (1/y)
, which is useful if(1/y)
is subject to common subexpression elimination. Note that this loses precision and increases the number of flops operating on the value.
The compiler can calculate d = 1/bConst
at compile time and change from:
c/bConst
to
c * d
but multiplication and division are different instructions with different performance and precision.
See: http://coliru.stacked-crooked.com/a/ba9770ec39ec5ac2
edited Feb 15 at 13:32
answered Feb 15 at 11:34
manliomanlio
14.3k105182
14.3k105182
Thank you for that explanation... interesting!
– Benjamin Bihler
Feb 15 at 12:27
Uhm... not sure that's the behaviour: coliru.stacked-crooked.com/a/79e5a3501af9a217 . Usingc * 1 / b
=c / b)
(givingb
=bConst
). Instead,c * 1 / bConst
give the same asc / bConst
... so it should be somethings different.
– markzzz
Feb 15 at 13:05
Uhm... wait! If I didc * (1 / b)
(instead ofc * 1 / b
) works as you said.
– markzzz
Feb 15 at 13:08
add a comment |
Thank you for that explanation... interesting!
– Benjamin Bihler
Feb 15 at 12:27
Uhm... not sure that's the behaviour: coliru.stacked-crooked.com/a/79e5a3501af9a217 . Usingc * 1 / b
=c / b)
(givingb
=bConst
). Instead,c * 1 / bConst
give the same asc / bConst
... so it should be somethings different.
– markzzz
Feb 15 at 13:05
Uhm... wait! If I didc * (1 / b)
(instead ofc * 1 / b
) works as you said.
– markzzz
Feb 15 at 13:08
Thank you for that explanation... interesting!
– Benjamin Bihler
Feb 15 at 12:27
Thank you for that explanation... interesting!
– Benjamin Bihler
Feb 15 at 12:27
Uhm... not sure that's the behaviour: coliru.stacked-crooked.com/a/79e5a3501af9a217 . Using
c * 1 / b
= c / b)
(giving b
= bConst
). Instead, c * 1 / bConst
give the same as c / bConst
... so it should be somethings different.– markzzz
Feb 15 at 13:05
Uhm... not sure that's the behaviour: coliru.stacked-crooked.com/a/79e5a3501af9a217 . Using
c * 1 / b
= c / b)
(giving b
= bConst
). Instead, c * 1 / bConst
give the same as c / bConst
... so it should be somethings different.– markzzz
Feb 15 at 13:05
Uhm... wait! If I did
c * (1 / b)
(instead of c * 1 / b
) works as you said.– markzzz
Feb 15 at 13:08
Uhm... wait! If I did
c * (1 / b)
(instead of c * 1 / b
) works as you said.– markzzz
Feb 15 at 13:08
add a comment |
You are using -Ofast
in your link, which enables all -O3
optimizations and includes both -ffast-math
, which in turns includes -funsafe-math-optimizations
.
From what I could glean, with optimizations enabled, -funsafe-math-optimizations
allows the compiler to reduce the precision of some computations. This seems to be what happens in the c/bConst
case.
The anwer is correct. Any idea why this is the case?
– Benjamin Bihler
Feb 15 at 10:15
What do you mean with "leads to a different result depending on the const-ness"? Shouldn't I get the same results starting with the same quotient and dividend?
– markzzz
Feb 15 at 10:16
@BenjaminBihler I'm looking into it.
– Nelfeal
Feb 15 at 10:16
2
@markzzz Unsafe optimizations tend to not care about this kind of logic.
– Nelfeal
Feb 15 at 10:17
add a comment |
You are using -Ofast
in your link, which enables all -O3
optimizations and includes both -ffast-math
, which in turns includes -funsafe-math-optimizations
.
From what I could glean, with optimizations enabled, -funsafe-math-optimizations
allows the compiler to reduce the precision of some computations. This seems to be what happens in the c/bConst
case.
The anwer is correct. Any idea why this is the case?
– Benjamin Bihler
Feb 15 at 10:15
What do you mean with "leads to a different result depending on the const-ness"? Shouldn't I get the same results starting with the same quotient and dividend?
– markzzz
Feb 15 at 10:16
@BenjaminBihler I'm looking into it.
– Nelfeal
Feb 15 at 10:16
2
@markzzz Unsafe optimizations tend to not care about this kind of logic.
– Nelfeal
Feb 15 at 10:17
add a comment |
You are using -Ofast
in your link, which enables all -O3
optimizations and includes both -ffast-math
, which in turns includes -funsafe-math-optimizations
.
From what I could glean, with optimizations enabled, -funsafe-math-optimizations
allows the compiler to reduce the precision of some computations. This seems to be what happens in the c/bConst
case.
You are using -Ofast
in your link, which enables all -O3
optimizations and includes both -ffast-math
, which in turns includes -funsafe-math-optimizations
.
From what I could glean, with optimizations enabled, -funsafe-math-optimizations
allows the compiler to reduce the precision of some computations. This seems to be what happens in the c/bConst
case.
edited Feb 15 at 10:34
answered Feb 15 at 10:14
NelfealNelfeal
5,2321825
5,2321825
The anwer is correct. Any idea why this is the case?
– Benjamin Bihler
Feb 15 at 10:15
What do you mean with "leads to a different result depending on the const-ness"? Shouldn't I get the same results starting with the same quotient and dividend?
– markzzz
Feb 15 at 10:16
@BenjaminBihler I'm looking into it.
– Nelfeal
Feb 15 at 10:16
2
@markzzz Unsafe optimizations tend to not care about this kind of logic.
– Nelfeal
Feb 15 at 10:17
add a comment |
The anwer is correct. Any idea why this is the case?
– Benjamin Bihler
Feb 15 at 10:15
What do you mean with "leads to a different result depending on the const-ness"? Shouldn't I get the same results starting with the same quotient and dividend?
– markzzz
Feb 15 at 10:16
@BenjaminBihler I'm looking into it.
– Nelfeal
Feb 15 at 10:16
2
@markzzz Unsafe optimizations tend to not care about this kind of logic.
– Nelfeal
Feb 15 at 10:17
The anwer is correct. Any idea why this is the case?
– Benjamin Bihler
Feb 15 at 10:15
The anwer is correct. Any idea why this is the case?
– Benjamin Bihler
Feb 15 at 10:15
What do you mean with "leads to a different result depending on the const-ness"? Shouldn't I get the same results starting with the same quotient and dividend?
– markzzz
Feb 15 at 10:16
What do you mean with "leads to a different result depending on the const-ness"? Shouldn't I get the same results starting with the same quotient and dividend?
– markzzz
Feb 15 at 10:16
@BenjaminBihler I'm looking into it.
– Nelfeal
Feb 15 at 10:16
@BenjaminBihler I'm looking into it.
– Nelfeal
Feb 15 at 10:16
2
2
@markzzz Unsafe optimizations tend to not care about this kind of logic.
– Nelfeal
Feb 15 at 10:17
@markzzz Unsafe optimizations tend to not care about this kind of logic.
– Nelfeal
Feb 15 at 10:17
add a comment |
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Here (mingw-w64 with g++ 8.1) both results are the same when optimized with -O2 and I get your result with -Ofast.
– Benjamin Bihler
Feb 15 at 10:12
1
I thought you got a grasp of the perverse effects of
-Ofast
on floatting operation determinism in your other question stackoverflow.com/q/54694274/5470596– YSC
Feb 15 at 14:33