Why does command substitution not work in braces but replacing it with the actual number works?
I'm using the following command to try to max out the number of cores on my server:
for i in {1..`nproc --all`}; do while : ; do : ; done & done
For some reason, it only ever uses one core, even though my server has two. If I try to replace `nproc --all` with 2, it works fine. Why is this happening?
linux bash script shell-script bash-scripting
edited Dec 11 '18 at 21:05
Kamil Maciorowski
24.6k155277
asked Dec 11 '18 at 20:46
AndreasKralj
124
add a comment |
I'm using the following command to try to max out the number of cores on my server:
for i in {1..`nproc --all`}; do while : ; do : ; done & done
For some reason, it only ever uses one core, even though my server has two. If I try to replace `nproc --all` with 2, it works fine. Why is this happening?
linux bash script shell-script bash-scripting
edited Dec 11 '18 at 21:05
Kamil Maciorowski
24.6k155277
asked Dec 11 '18 at 20:46
AndreasKralj
124
add a comment |
I'm using the following command to try to max out the number of cores on my server:
for i in {1..`nproc --all`}; do while : ; do : ; done & done
For some reason, it only ever uses one core, even though my server has two. If I try to replace `nproc --all` with 2, it works fine. Why is this happening?
linux bash script shell-script bash-scripting
edited Dec 11 '18 at 21:05
Kamil Maciorowski
24.6k155277
asked Dec 11 '18 at 20:46
AndreasKralj
124
I'm using the following command to try to max out the number of cores on my server:
for i in {1..`nproc --all`}; do while : ; do : ; done & done
For some reason, it only ever uses one core, even though my server has two. If I try to replace `nproc --all` with 2, it works fine. Why is this happening?
linux bash script shell-script bash-scripting
linux bash script shell-script bash-scripting
edited Dec 11 '18 at 21:05
Kamil Maciorowski
24.6k155277
asked Dec 11 '18 at 20:46
AndreasKralj
124
edited Dec 11 '18 at 21:05
Kamil Maciorowski
24.6k155277
asked Dec 11 '18 at 20:46
AndreasKralj
124
edited Dec 11 '18 at 21:05
Kamil Maciorowski
24.6k155277
edited Dec 11 '18 at 21:05
Kamil Maciorowski
24.6k155277
edited Dec 11 '18 at 21:05
Kamil Maciorowski
24.6k155277
24.6k155277
asked Dec 11 '18 at 20:46
AndreasKralj
124
asked Dec 11 '18 at 20:46
AndreasKralj
124
asked Dec 11 '18 at 20:46
AndreasKralj
124
124
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Brace expansion happens before command substitution. This fragment
{1..`nproc --all`}
will not be expanded as brace at all. When the command substitution is expanded to 2, it becomes
{1..2}
but it's too late for brace expansion now. The same problem would be if you used a variable there.
This syntax works:
for ((i=1; i<=`nproc --all`; i++)); do echo "$i"; done
or
for i in $(seq 1 "$(nproc --all)"); do echo "$i"; done
Adjust the loop to your needs. Note in the last case I used $(…) instead of `…` because this form nests easily.
answered Dec 11 '18 at 20:53
Kamil Maciorowski
24.6k155277
That worked perfectly, thank you!
– AndreasKralj
Dec 11 '18 at 21:10
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Brace expansion happens before command substitution. This fragment
{1..`nproc --all`}
will not be expanded as brace at all. When the command substitution is expanded to 2, it becomes
{1..2}
but it's too late for brace expansion now. The same problem would be if you used a variable there.
This syntax works:
for ((i=1; i<=`nproc --all`; i++)); do echo "$i"; done
or
for i in $(seq 1 "$(nproc --all)"); do echo "$i"; done
Adjust the loop to your needs. Note in the last case I used $(…) instead of `…` because this form nests easily.
answered Dec 11 '18 at 20:53
Kamil Maciorowski
24.6k155277
That worked perfectly, thank you!
– AndreasKralj
Dec 11 '18 at 21:10
add a comment |
Brace expansion happens before command substitution. This fragment
{1..`nproc --all`}
will not be expanded as brace at all. When the command substitution is expanded to 2, it becomes
{1..2}
but it's too late for brace expansion now. The same problem would be if you used a variable there.
This syntax works:
for ((i=1; i<=`nproc --all`; i++)); do echo "$i"; done
or
for i in $(seq 1 "$(nproc --all)"); do echo "$i"; done
Adjust the loop to your needs. Note in the last case I used $(…) instead of `…` because this form nests easily.
answered Dec 11 '18 at 20:53
Kamil Maciorowski
24.6k155277
That worked perfectly, thank you!
– AndreasKralj
Dec 11 '18 at 21:10
add a comment |
Brace expansion happens before command substitution. This fragment
{1..`nproc --all`}
will not be expanded as brace at all. When the command substitution is expanded to 2, it becomes
{1..2}
but it's too late for brace expansion now. The same problem would be if you used a variable there.
This syntax works:
for ((i=1; i<=`nproc --all`; i++)); do echo "$i"; done
or
for i in $(seq 1 "$(nproc --all)"); do echo "$i"; done
Adjust the loop to your needs. Note in the last case I used $(…) instead of `…` because this form nests easily.
answered Dec 11 '18 at 20:53
Kamil Maciorowski
24.6k155277
Brace expansion happens before command substitution. This fragment
{1..`nproc --all`}
will not be expanded as brace at all. When the command substitution is expanded to 2, it becomes
{1..2}
but it's too late for brace expansion now. The same problem would be if you used a variable there.
This syntax works:
for ((i=1; i<=`nproc --all`; i++)); do echo "$i"; done
or
for i in $(seq 1 "$(nproc --all)"); do echo "$i"; done
Adjust the loop to your needs. Note in the last case I used $(…) instead of `…` because this form nests easily.
answered Dec 11 '18 at 20:53
Kamil Maciorowski
24.6k155277
edited Dec 11 '18 at 21:10
answered Dec 11 '18 at 20:53
Kamil Maciorowski
24.6k155277
answered Dec 11 '18 at 20:53
Kamil Maciorowski
24.6k155277
answered Dec 11 '18 at 20:53
Kamil Maciorowski
24.6k155277
24.6k155277
That worked perfectly, thank you!
– AndreasKralj
Dec 11 '18 at 21:10
add a comment |
That worked perfectly, thank you!
– AndreasKralj
Dec 11 '18 at 21:10
That worked perfectly, thank you!
– AndreasKralj
Dec 11 '18 at 21:10
That worked perfectly, thank you!
– AndreasKralj
Dec 11 '18 at 21:10
add a comment |
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