Mixing
46- and 64-bit integers
$begingroup$
Some Cray supercomputers used to support 46-bit and 64-bit integer
data types. What are the maximum and minimum values that we could
express in a 46-bit integer? in a 64-bit integer?
Is my thinking correct that the ranges would be $pm 2^{46}$ and $pm 2^{64}$, correspondingly?
computer-science binary computational-mathematics floating-point
asked Jan 11 at 23:04
sequencesequence
4,24031336
$endgroup$
add a comment |
$begingroup$
Some Cray supercomputers used to support 46-bit and 64-bit integer
data types. What are the maximum and minimum values that we could
express in a 46-bit integer? in a 64-bit integer?
Is my thinking correct that the ranges would be $pm 2^{46}$ and $pm 2^{64}$, correspondingly?
computer-science binary computational-mathematics floating-point
asked Jan 11 at 23:04
sequencesequence
4,24031336
$endgroup$
1
$begingroup$
Those ranges would require 47 and 65 bits, respectively (one extra for the sign)
$endgroup$
– Morgan Rodgers
Jan 11 at 23:08
add a comment |
$begingroup$
Some Cray supercomputers used to support 46-bit and 64-bit integer
data types. What are the maximum and minimum values that we could
express in a 46-bit integer? in a 64-bit integer?
Is my thinking correct that the ranges would be $pm 2^{46}$ and $pm 2^{64}$, correspondingly?
computer-science binary computational-mathematics floating-point
asked Jan 11 at 23:04
sequencesequence
4,24031336
$endgroup$
Some Cray supercomputers used to support 46-bit and 64-bit integer
data types. What are the maximum and minimum values that we could
express in a 46-bit integer? in a 64-bit integer?
Is my thinking correct that the ranges would be $pm 2^{46}$ and $pm 2^{64}$, correspondingly?
computer-science binary computational-mathematics floating-point
computer-science binary computational-mathematics floating-point
asked Jan 11 at 23:04
sequencesequence
4,24031336
asked Jan 11 at 23:04
sequencesequence
4,24031336
asked Jan 11 at 23:04
sequencesequence
4,24031336
asked Jan 11 at 23:04
sequencesequence
4,24031336
asked Jan 11 at 23:04
sequencesequence
4,24031336
4,24031336
1
$begingroup$
Those ranges would require 47 and 65 bits, respectively (one extra for the sign)
$endgroup$
– Morgan Rodgers
Jan 11 at 23:08
add a comment |
1
$begingroup$
Those ranges would require 47 and 65 bits, respectively (one extra for the sign)
$endgroup$
– Morgan Rodgers
Jan 11 at 23:08
1
1
$begingroup$
Those ranges would require 47 and 65 bits, respectively (one extra for the sign)
$endgroup$
– Morgan Rodgers
Jan 11 at 23:08
$begingroup$
Those ranges would require 47 and 65 bits, respectively (one extra for the sign)
$endgroup$
– Morgan Rodgers
Jan 11 at 23:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Almost.
With $n$ bits, we can represent $2^n$ distinct values. For unsigned integer types, we can thus cover the range from $0$ to $2^n-1$. And with signed integer types (two's complement with symmetric range), we can cover $-2^{n-1}$ to $2^{n-1}-1$.
answered Jan 11 at 23:10
Hagen von EitzenHagen von Eitzen
279k23271502
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add a comment |
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$begingroup$
Almost.
With $n$ bits, we can represent $2^n$ distinct values. For unsigned integer types, we can thus cover the range from $0$ to $2^n-1$. And with signed integer types (two's complement with symmetric range), we can cover $-2^{n-1}$ to $2^{n-1}-1$.
answered Jan 11 at 23:10
Hagen von EitzenHagen von Eitzen
279k23271502
$endgroup$
add a comment |
$begingroup$
Almost.
With $n$ bits, we can represent $2^n$ distinct values. For unsigned integer types, we can thus cover the range from $0$ to $2^n-1$. And with signed integer types (two's complement with symmetric range), we can cover $-2^{n-1}$ to $2^{n-1}-1$.
answered Jan 11 at 23:10
Hagen von EitzenHagen von Eitzen
279k23271502
$endgroup$
add a comment |
$begingroup$
Almost.
With $n$ bits, we can represent $2^n$ distinct values. For unsigned integer types, we can thus cover the range from $0$ to $2^n-1$. And with signed integer types (two's complement with symmetric range), we can cover $-2^{n-1}$ to $2^{n-1}-1$.
answered Jan 11 at 23:10
Hagen von EitzenHagen von Eitzen
279k23271502
$endgroup$
Almost.
With $n$ bits, we can represent $2^n$ distinct values. For unsigned integer types, we can thus cover the range from $0$ to $2^n-1$. And with signed integer types (two's complement with symmetric range), we can cover $-2^{n-1}$ to $2^{n-1}-1$.
answered Jan 11 at 23:10
Hagen von EitzenHagen von Eitzen
279k23271502
answered Jan 11 at 23:10
Hagen von EitzenHagen von Eitzen
279k23271502
answered Jan 11 at 23:10
Hagen von EitzenHagen von Eitzen
279k23271502
answered Jan 11 at 23:10
Hagen von EitzenHagen von Eitzen
279k23271502
279k23271502
add a comment |
add a comment |
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1
$begingroup$
Those ranges would require 47 and 65 bits, respectively (one extra for the sign)
$endgroup$
– Morgan Rodgers
Jan 11 at 23:08