Revisions to Is there a reason to use "[ ! -f file ] || command" instead of "[ -f file ] && command"?

Fixed sheband, add missing quotes.

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edited Nov 22, 2023 at 12:20

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Yes, if you care about the exit code of the compound statement. Try the following:

#! /bin/sh -

! $1"$@" || true
echo $"$?"

$1"$@" && true
echo $"$?"

then run it with true and false as arguments.

./script true
0
0

./script false
0
1

This is because short-circuit evaluation of boolean expressions.

In OP's example, let's suppose the file doesn't exist:

In the first case, the first condition returns TRUE, no need to evaluate the second operation (TRUE OR x = TRUE). You get a TRUE for the compound statement.
In the second case, the first condition returns FALSE, no need to evaluate the second operation (FALSE AND x = FALSE). You get a FALSE for the compound statement.

Exit codes are very important. Check What does set -e mean in a bash script? to see possible implications of set -e, trap and set -o pipefail.

Yes, if you care about the exit code of the compound statement. Try the following:

#/bin/sh

! $1 || true
echo $?

$1 && true
echo $?

then run it with true and false as arguments.

./script true
0
0

./script false
0
1

This is because short-circuit evaluation of boolean expressions.

In OP's example, let's suppose the file doesn't exist:

In the first case, the first condition returns TRUE, no need to evaluate the second operation (TRUE OR x = TRUE). You get a TRUE for the compound statement.
In the second case, the first condition returns FALSE, no need to evaluate the second operation (FALSE AND x = FALSE). You get a FALSE for the compound statement.

Exit codes are very important. Check What does set -e mean in a bash script? to see possible implications of set -e, trap and set -o pipefail.

Yes, if you care about the exit code of the compound statement. Try the following:

#! /bin/sh -

! "$@" || true
echo "$?"

"$@" && true
echo "$?"

then run it with true and false as arguments.

./script true
0
0

./script false
0
1

This is because short-circuit evaluation of boolean expressions.

In OP's example, let's suppose the file doesn't exist:

In the first case, the first condition returns TRUE, no need to evaluate the second operation (TRUE OR x = TRUE). You get a TRUE for the compound statement.
In the second case, the first condition returns FALSE, no need to evaluate the second operation (FALSE AND x = FALSE). You get a FALSE for the compound statement.

Exit codes are very important. Check What does set -e mean in a bash script? to see possible implications of set -e, trap and set -o pipefail.

https link, clarify granmmar

Source Link

edited Nov 22, 2023 at 11:39

mmoya

6.3k
2
23
23

Yes, if you care about the exit code of the compound statement exit code of the compound statement. Try the following:

#/bin/sh

! $1 || true
echo $?

$1 && true
echo $?

then run it with true and false as arguments.

./script true
0
0

./script false
0
1

This is because short-circuit evaluation of boolean expressions.

In OP's example, let's suppose the file doesn't exist:

In the first case, the first condition returns TRUE, no need to evaluate the second operation (TRUE OR x = TRUE). You get a TRUE for the compound statement.
In the second case, the first condition returns FALSE, no need to evaluate the second operation (FALSE AND x = FALSE). You get a FALSE for the compound statement.

Exit codes are very important. Check this SO answerWhat does set -e mean in a bash script? to see possible implications of set -e, trap and set -o pipefail.

Yes, if you care about the exit code of the compound statement. Try the following:

#/bin/sh

! $1 || true
echo $?

$1 && true
echo $?

then run it with true and false as arguments.

./script true
0
0

./script false
0
1

This is because short-circuit evaluation of boolean expressions.

In OP's example, let's suppose the file doesn't exist:

In the first case, the first condition returns TRUE, no need to evaluate the second operation (TRUE OR x = TRUE). You get a TRUE for the compound statement.
In the second case, the first condition returns FALSE, no need to evaluate the second operation (FALSE AND x = FALSE). You get a FALSE for the compound statement.

Exit codes are very important. Check this SO answer to see possible implications of set -e, trap and set -o pipefail.

Yes, if you care about the exit code of the compound statement. Try the following:

#/bin/sh

! $1 || true
echo $?

$1 && true
echo $?

then run it with true and false as arguments.

./script true
0
0

./script false
0
1

This is because short-circuit evaluation of boolean expressions.

In OP's example, let's suppose the file doesn't exist:

In the first case, the first condition returns TRUE, no need to evaluate the second operation (TRUE OR x = TRUE). You get a TRUE for the compound statement.
In the second case, the first condition returns FALSE, no need to evaluate the second operation (FALSE AND x = FALSE). You get a FALSE for the compound statement.

Exit codes are very important. Check What does set -e mean in a bash script? to see possible implications of set -e, trap and set -o pipefail.

Add link to SO question about set -e

Source Link

edited May 17, 2016 at 15:58

mmoya

6.3k
2
23
23

Yes, if you care about the exit code of the compound statement. Try the following:

#/bin/sh

! $1 || true
echo $?

$1 && true
echo $?

then run it with true and false as arguments.

./script true
0
0

./script false
0
1

This is because short-circuit evaluation of boolean expressions.

In OP's example, let's suppose the file doesn't exist:

In the first case, the first condition returns TRUE, no need to evaluate the second operation (TRUE OR x = TRUE). You get a TRUE for the compound statement.
In the second case, the first condition returns FALSE, no need to evaluate the second operation (FALSE AND x = FALSE). You get a FALSE for the compound statement.

Exit codes are very important. Check this SO answer to see possible implications of set -e, trap and set -o pipefail.

Yes, if you care about the exit code of the compound statement. Try the following:

#/bin/sh

! $1 || true
echo $?

$1 && true
echo $?

then run it with true and false as arguments.

./script true
0
0

./script false
0
1

This is because short-circuit evaluation of boolean expressions.

In OP's example, let's suppose the file doesn't exist:

In the first case, the first condition returns TRUE, no need to evaluate the second operation (TRUE OR x = TRUE). You get a TRUE for the compound statement.
In the second case, the first condition returns FALSE, no need to evaluate the second operation (FALSE AND x = FALSE). You get a FALSE for the compound statement.

Yes, if you care about the exit code of the compound statement. Try the following:

#/bin/sh

! $1 || true
echo $?

$1 && true
echo $?

then run it with true and false as arguments.

./script true
0
0

./script false
0
1

This is because short-circuit evaluation of boolean expressions.

In OP's example, let's suppose the file doesn't exist:

In the first case, the first condition returns TRUE, no need to evaluate the second operation (TRUE OR x = TRUE). You get a TRUE for the compound statement.
In the second case, the first condition returns FALSE, no need to evaluate the second operation (FALSE AND x = FALSE). You get a FALSE for the compound statement.

Exit codes are very important. Check this SO answer to see possible implications of set -e, trap and set -o pipefail.

Source Link

answered May 17, 2016 at 15:03

mmoya

6.3k
2
23
23

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