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Is the inverse of std::numeric_limits::infinity() zero?
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Is there anything in the C++ standard (or the IEEE 754 floating-point standard) that guarantees that 1./std::numeric_limits<double>::infinity() is zero (or at least a small number)?
c++ floating-point infinity
edited Aug 26 at 11:03
Peter Mortensen
12.9k1983111
asked Aug 25 at 9:25
davidhigh
8,65411843
add a comment |Â
up vote
17
down vote
favorite
Is there anything in the C++ standard (or the IEEE 754 floating-point standard) that guarantees that 1./std::numeric_limits<double>::infinity() is zero (or at least a small number)?
c++ floating-point infinity
edited Aug 26 at 11:03
Peter Mortensen
12.9k1983111
asked Aug 25 at 9:25
davidhigh
8,65411843
Look into floating-point-gui.de
– Basile Starynkevitch
Aug 25 at 9:27
Very related.
– user202729
Aug 25 at 14:46
add a comment |Â
up vote
17
down vote
favorite
up vote
17
down vote
favorite
Is there anything in the C++ standard (or the IEEE 754 floating-point standard) that guarantees that 1./std::numeric_limits<double>::infinity() is zero (or at least a small number)?
c++ floating-point infinity
edited Aug 26 at 11:03
Peter Mortensen
12.9k1983111
asked Aug 25 at 9:25
davidhigh
8,65411843
Is there anything in the C++ standard (or the IEEE 754 floating-point standard) that guarantees that 1./std::numeric_limits<double>::infinity() is zero (or at least a small number)?
c++ floating-point infinity
edited Aug 26 at 11:03
Peter Mortensen
12.9k1983111
asked Aug 25 at 9:25
davidhigh
8,65411843
edited Aug 26 at 11:03
Peter Mortensen
12.9k1983111
edited Aug 26 at 11:03
Peter Mortensen
12.9k1983111
edited Aug 26 at 11:03
Peter Mortensen
12.9k1983111
12.9k1983111
asked Aug 25 at 9:25
davidhigh
8,65411843
asked Aug 25 at 9:25
davidhigh
8,65411843
asked Aug 25 at 9:25
davidhigh
8,65411843
8,65411843
Look into floating-point-gui.de
– Basile Starynkevitch
Aug 25 at 9:27
Very related.
– user202729
Aug 25 at 14:46
add a comment |Â
Look into floating-point-gui.de
– Basile Starynkevitch
Aug 25 at 9:27
Very related.
– user202729
Aug 25 at 14:46
Look into floating-point-gui.de
– Basile Starynkevitch
Aug 25 at 9:27
Look into floating-point-gui.de
– Basile Starynkevitch
Aug 25 at 9:27
Very related.
– user202729
Aug 25 at 14:46
Very related.
– user202729
Aug 25 at 14:46
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
11
down vote
accepted
Yes, according to the GNU C library reference manual (assuming IEEE 754):
Infinities propagate through calculations
as one would expect: for example, 2 + ∞ = ∞, 4/∞ =
0
https://www.gnu.org/software/libc/manual/html_node/Infinity-and-NaN.html
You may want to check if your C++ compiler uses IEEE 754:
How to check if C++ compiler uses IEEE 754 floating point standard
edited Aug 26 at 11:05
Peter Mortensen
12.9k1983111
answered Aug 25 at 9:43
Gonen I
1,4841025
add a comment |Â
up vote
15
down vote
Any finite number divided by infinity results in zero under IEEE 754 (and therefore the same in most typical C++ implementations).
If the sign of the of numerator and denominator differ, the result will be negative zero, which is equal to zero.
answered Aug 25 at 9:34
John Zwinck
141k16171279
3
Uhm, no, not exactly: if the finite number is negative it returns -0, which is not the same thing as 0 (although it compares equal).
– Federico Poloni
Aug 25 at 12:31
6
Yeah, the proper wording would be "results in a zero" :)
– Ruslan
Aug 25 at 13:09
2
@FedericoPoloni: Both +0 and −0 in IEEE 754 specification level 2 (floating-point data) represent zero in level 1 (extended real numbers).
– Eric Postpischil
Aug 25 at 17:46
add a comment |Â
up vote
4
down vote
IEEE 754-2008 6.1 says:
The behavior of infinity in floating-point arithmetic is derived from the limiting cases of real arithmetic with operands of arbitrarily large magnitude, when such a limit exists. Infinities shall be interpreted in the affine sense, that is: −∞ < every finite number < +∞.
Operations on infinite operands are usually exact and therefore signal no exceptions,…
Since the limit of 1/x as x increases without bound is zero, a consequence of this clause is that 1/∞ is zero.
Clause 6.3 tells us the sign of the result is +:
When neither the inputs nor result are NaN, the sign of a product or quotient is the exclusive OR of the operands’ signs;…
answered Aug 25 at 17:52
Eric Postpischil
64.8k872145
add a comment |Â
up vote
0
down vote
if(std::numeric_limits<double>::is_iec559)
yes();
else
no();
(see 18.3.2.4)
IEC 559, which is identical with IEEE 754, guarantees that to be the case. However, C++ does not guarantee in any way that IEC 559 is in place (although 99.99% of the time that's just what happens to be the case, you still need to verify to be sure).
answered Aug 25 at 16:50
Damon
49.2k1492151
1
Since nobody is likely to bother implementing and testing non-IEEE 754 platforms, one might as wellstatic_assert(std::numeric_limits<double>::is_iec559).
– John Zwinck
Aug 26 at 2:01
@JohnZwinck: That's actually a good point. If your code relies on IEEE 754 particulars, then it won't work if these aren't present. Sostatic_assertis a very valid option (possibly, probably, even the better one).
– Damon
Aug 26 at 12:26
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
accepted
Yes, according to the GNU C library reference manual (assuming IEEE 754):
Infinities propagate through calculations
as one would expect: for example, 2 + ∞ = ∞, 4/∞ =
0
https://www.gnu.org/software/libc/manual/html_node/Infinity-and-NaN.html
You may want to check if your C++ compiler uses IEEE 754:
How to check if C++ compiler uses IEEE 754 floating point standard
edited Aug 26 at 11:05
Peter Mortensen
12.9k1983111
answered Aug 25 at 9:43
Gonen I
1,4841025
add a comment |Â
up vote
11
down vote
accepted
Yes, according to the GNU C library reference manual (assuming IEEE 754):
Infinities propagate through calculations
as one would expect: for example, 2 + ∞ = ∞, 4/∞ =
0
https://www.gnu.org/software/libc/manual/html_node/Infinity-and-NaN.html
You may want to check if your C++ compiler uses IEEE 754:
How to check if C++ compiler uses IEEE 754 floating point standard
edited Aug 26 at 11:05
Peter Mortensen
12.9k1983111
answered Aug 25 at 9:43
Gonen I
1,4841025
add a comment |Â
up vote
11
down vote
accepted
up vote
11
down vote
accepted
Yes, according to the GNU C library reference manual (assuming IEEE 754):
Infinities propagate through calculations
as one would expect: for example, 2 + ∞ = ∞, 4/∞ =
0
https://www.gnu.org/software/libc/manual/html_node/Infinity-and-NaN.html
You may want to check if your C++ compiler uses IEEE 754:
How to check if C++ compiler uses IEEE 754 floating point standard
edited Aug 26 at 11:05
Peter Mortensen
12.9k1983111
answered Aug 25 at 9:43
Gonen I
1,4841025
Yes, according to the GNU C library reference manual (assuming IEEE 754):
Infinities propagate through calculations
as one would expect: for example, 2 + ∞ = ∞, 4/∞ =
0
https://www.gnu.org/software/libc/manual/html_node/Infinity-and-NaN.html
You may want to check if your C++ compiler uses IEEE 754:
How to check if C++ compiler uses IEEE 754 floating point standard
edited Aug 26 at 11:05
Peter Mortensen
12.9k1983111
answered Aug 25 at 9:43
Gonen I
1,4841025
edited Aug 26 at 11:05
Peter Mortensen
12.9k1983111
edited Aug 26 at 11:05
Peter Mortensen
12.9k1983111
edited Aug 26 at 11:05
Peter Mortensen
12.9k1983111
12.9k1983111
answered Aug 25 at 9:43
Gonen I
1,4841025
answered Aug 25 at 9:43
Gonen I
1,4841025
answered Aug 25 at 9:43
Gonen I
1,4841025
1,4841025
add a comment |Â
add a comment |Â
up vote
15
down vote
Any finite number divided by infinity results in zero under IEEE 754 (and therefore the same in most typical C++ implementations).
If the sign of the of numerator and denominator differ, the result will be negative zero, which is equal to zero.
answered Aug 25 at 9:34
John Zwinck
141k16171279
3
Uhm, no, not exactly: if the finite number is negative it returns -0, which is not the same thing as 0 (although it compares equal).
– Federico Poloni
Aug 25 at 12:31
6
Yeah, the proper wording would be "results in a zero" :)
– Ruslan
Aug 25 at 13:09
2
@FedericoPoloni: Both +0 and −0 in IEEE 754 specification level 2 (floating-point data) represent zero in level 1 (extended real numbers).
– Eric Postpischil
Aug 25 at 17:46
add a comment |Â
up vote
15
down vote
Any finite number divided by infinity results in zero under IEEE 754 (and therefore the same in most typical C++ implementations).
If the sign of the of numerator and denominator differ, the result will be negative zero, which is equal to zero.
answered Aug 25 at 9:34
John Zwinck
141k16171279
3
Uhm, no, not exactly: if the finite number is negative it returns -0, which is not the same thing as 0 (although it compares equal).
– Federico Poloni
Aug 25 at 12:31
6
Yeah, the proper wording would be "results in a zero" :)
– Ruslan
Aug 25 at 13:09
2
@FedericoPoloni: Both +0 and −0 in IEEE 754 specification level 2 (floating-point data) represent zero in level 1 (extended real numbers).
– Eric Postpischil
Aug 25 at 17:46
add a comment |Â
up vote
15
down vote
up vote
15
down vote
Any finite number divided by infinity results in zero under IEEE 754 (and therefore the same in most typical C++ implementations).
If the sign of the of numerator and denominator differ, the result will be negative zero, which is equal to zero.
answered Aug 25 at 9:34
John Zwinck
141k16171279
Any finite number divided by infinity results in zero under IEEE 754 (and therefore the same in most typical C++ implementations).
If the sign of the of numerator and denominator differ, the result will be negative zero, which is equal to zero.
answered Aug 25 at 9:34
John Zwinck
141k16171279
edited Aug 26 at 2:03
answered Aug 25 at 9:34
John Zwinck
141k16171279
answered Aug 25 at 9:34
John Zwinck
141k16171279
answered Aug 25 at 9:34
John Zwinck
141k16171279
141k16171279
3
Uhm, no, not exactly: if the finite number is negative it returns -0, which is not the same thing as 0 (although it compares equal).
– Federico Poloni
Aug 25 at 12:31
6
Yeah, the proper wording would be "results in a zero" :)
– Ruslan
Aug 25 at 13:09
2
@FedericoPoloni: Both +0 and −0 in IEEE 754 specification level 2 (floating-point data) represent zero in level 1 (extended real numbers).
– Eric Postpischil
Aug 25 at 17:46
add a comment |Â
3
Uhm, no, not exactly: if the finite number is negative it returns -0, which is not the same thing as 0 (although it compares equal).
– Federico Poloni
Aug 25 at 12:31
6
Yeah, the proper wording would be "results in a zero" :)
– Ruslan
Aug 25 at 13:09
2
@FedericoPoloni: Both +0 and −0 in IEEE 754 specification level 2 (floating-point data) represent zero in level 1 (extended real numbers).
– Eric Postpischil
Aug 25 at 17:46
3
3
Uhm, no, not exactly: if the finite number is negative it returns -0, which is not the same thing as 0 (although it compares equal).
– Federico Poloni
Aug 25 at 12:31
Uhm, no, not exactly: if the finite number is negative it returns -0, which is not the same thing as 0 (although it compares equal).
– Federico Poloni
Aug 25 at 12:31
6
6
Yeah, the proper wording would be "results in a zero" :)
– Ruslan
Aug 25 at 13:09
Yeah, the proper wording would be "results in a zero" :)
– Ruslan
Aug 25 at 13:09
2
2
@FedericoPoloni: Both +0 and −0 in IEEE 754 specification level 2 (floating-point data) represent zero in level 1 (extended real numbers).
– Eric Postpischil
Aug 25 at 17:46
@FedericoPoloni: Both +0 and −0 in IEEE 754 specification level 2 (floating-point data) represent zero in level 1 (extended real numbers).
– Eric Postpischil
Aug 25 at 17:46
add a comment |Â
up vote
4
down vote
IEEE 754-2008 6.1 says:
The behavior of infinity in floating-point arithmetic is derived from the limiting cases of real arithmetic with operands of arbitrarily large magnitude, when such a limit exists. Infinities shall be interpreted in the affine sense, that is: −∞ < every finite number < +∞.
Operations on infinite operands are usually exact and therefore signal no exceptions,…
Since the limit of 1/x as x increases without bound is zero, a consequence of this clause is that 1/∞ is zero.
Clause 6.3 tells us the sign of the result is +:
When neither the inputs nor result are NaN, the sign of a product or quotient is the exclusive OR of the operands’ signs;…
answered Aug 25 at 17:52
Eric Postpischil
64.8k872145
add a comment |Â
up vote
4
down vote
IEEE 754-2008 6.1 says:
The behavior of infinity in floating-point arithmetic is derived from the limiting cases of real arithmetic with operands of arbitrarily large magnitude, when such a limit exists. Infinities shall be interpreted in the affine sense, that is: −∞ < every finite number < +∞.
Operations on infinite operands are usually exact and therefore signal no exceptions,…
Since the limit of 1/x as x increases without bound is zero, a consequence of this clause is that 1/∞ is zero.
Clause 6.3 tells us the sign of the result is +:
When neither the inputs nor result are NaN, the sign of a product or quotient is the exclusive OR of the operands’ signs;…
answered Aug 25 at 17:52
Eric Postpischil
64.8k872145
add a comment |Â
up vote
4
down vote
up vote
4
down vote
IEEE 754-2008 6.1 says:
The behavior of infinity in floating-point arithmetic is derived from the limiting cases of real arithmetic with operands of arbitrarily large magnitude, when such a limit exists. Infinities shall be interpreted in the affine sense, that is: −∞ < every finite number < +∞.
Operations on infinite operands are usually exact and therefore signal no exceptions,…
Since the limit of 1/x as x increases without bound is zero, a consequence of this clause is that 1/∞ is zero.
Clause 6.3 tells us the sign of the result is +:
When neither the inputs nor result are NaN, the sign of a product or quotient is the exclusive OR of the operands’ signs;…
answered Aug 25 at 17:52
Eric Postpischil
64.8k872145
IEEE 754-2008 6.1 says:
The behavior of infinity in floating-point arithmetic is derived from the limiting cases of real arithmetic with operands of arbitrarily large magnitude, when such a limit exists. Infinities shall be interpreted in the affine sense, that is: −∞ < every finite number < +∞.
Operations on infinite operands are usually exact and therefore signal no exceptions,…
Since the limit of 1/x as x increases without bound is zero, a consequence of this clause is that 1/∞ is zero.
Clause 6.3 tells us the sign of the result is +:
When neither the inputs nor result are NaN, the sign of a product or quotient is the exclusive OR of the operands’ signs;…
answered Aug 25 at 17:52
Eric Postpischil
64.8k872145
answered Aug 25 at 17:52
Eric Postpischil
64.8k872145
answered Aug 25 at 17:52
Eric Postpischil
64.8k872145
answered Aug 25 at 17:52
Eric Postpischil
64.8k872145
64.8k872145
add a comment |Â
add a comment |Â
up vote
0
down vote
if(std::numeric_limits<double>::is_iec559)
yes();
else
no();
(see 18.3.2.4)
IEC 559, which is identical with IEEE 754, guarantees that to be the case. However, C++ does not guarantee in any way that IEC 559 is in place (although 99.99% of the time that's just what happens to be the case, you still need to verify to be sure).
answered Aug 25 at 16:50
Damon
49.2k1492151
1
Since nobody is likely to bother implementing and testing non-IEEE 754 platforms, one might as wellstatic_assert(std::numeric_limits<double>::is_iec559).
– John Zwinck
Aug 26 at 2:01
@JohnZwinck: That's actually a good point. If your code relies on IEEE 754 particulars, then it won't work if these aren't present. Sostatic_assertis a very valid option (possibly, probably, even the better one).
– Damon
Aug 26 at 12:26
add a comment |Â
up vote
0
down vote
if(std::numeric_limits<double>::is_iec559)
yes();
else
no();
(see 18.3.2.4)
IEC 559, which is identical with IEEE 754, guarantees that to be the case. However, C++ does not guarantee in any way that IEC 559 is in place (although 99.99% of the time that's just what happens to be the case, you still need to verify to be sure).
answered Aug 25 at 16:50
Damon
49.2k1492151
1
Since nobody is likely to bother implementing and testing non-IEEE 754 platforms, one might as wellstatic_assert(std::numeric_limits<double>::is_iec559).
– John Zwinck
Aug 26 at 2:01
@JohnZwinck: That's actually a good point. If your code relies on IEEE 754 particulars, then it won't work if these aren't present. Sostatic_assertis a very valid option (possibly, probably, even the better one).
– Damon
Aug 26 at 12:26
add a comment |Â
up vote
0
down vote
up vote
0
down vote
if(std::numeric_limits<double>::is_iec559)
yes();
else
no();
(see 18.3.2.4)
IEC 559, which is identical with IEEE 754, guarantees that to be the case. However, C++ does not guarantee in any way that IEC 559 is in place (although 99.99% of the time that's just what happens to be the case, you still need to verify to be sure).
answered Aug 25 at 16:50
Damon
49.2k1492151
if(std::numeric_limits<double>::is_iec559)
yes();
else
no();
(see 18.3.2.4)
IEC 559, which is identical with IEEE 754, guarantees that to be the case. However, C++ does not guarantee in any way that IEC 559 is in place (although 99.99% of the time that's just what happens to be the case, you still need to verify to be sure).
answered Aug 25 at 16:50
Damon
49.2k1492151
answered Aug 25 at 16:50
Damon
49.2k1492151
answered Aug 25 at 16:50
Damon
49.2k1492151
answered Aug 25 at 16:50
Damon
49.2k1492151
49.2k1492151
1
Since nobody is likely to bother implementing and testing non-IEEE 754 platforms, one might as wellstatic_assert(std::numeric_limits<double>::is_iec559).
– John Zwinck
Aug 26 at 2:01
@JohnZwinck: That's actually a good point. If your code relies on IEEE 754 particulars, then it won't work if these aren't present. Sostatic_assertis a very valid option (possibly, probably, even the better one).
– Damon
Aug 26 at 12:26
add a comment |Â
1
Since nobody is likely to bother implementing and testing non-IEEE 754 platforms, one might as wellstatic_assert(std::numeric_limits<double>::is_iec559).
– John Zwinck
Aug 26 at 2:01
@JohnZwinck: That's actually a good point. If your code relies on IEEE 754 particulars, then it won't work if these aren't present. Sostatic_assertis a very valid option (possibly, probably, even the better one).
– Damon
Aug 26 at 12:26
1
1
Since nobody is likely to bother implementing and testing non-IEEE 754 platforms, one might as well
static_assert(std::numeric_limits<double>::is_iec559).– John Zwinck
Aug 26 at 2:01
Since nobody is likely to bother implementing and testing non-IEEE 754 platforms, one might as well
static_assert(std::numeric_limits<double>::is_iec559).– John Zwinck
Aug 26 at 2:01
@JohnZwinck: That's actually a good point. If your code relies on IEEE 754 particulars, then it won't work if these aren't present. So
static_assert is a very valid option (possibly, probably, even the better one).– Damon
Aug 26 at 12:26
@JohnZwinck: That's actually a good point. If your code relies on IEEE 754 particulars, then it won't work if these aren't present. So
static_assert is a very valid option (possibly, probably, even the better one).– Damon
Aug 26 at 12:26
add a comment |Â
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Look into floating-point-gui.de
– Basile Starynkevitch
Aug 25 at 9:27
Very related.
– user202729
Aug 25 at 14:46