Exit statuses of comparisons in test constructs
I was writing some "if then" statements and found what seemed to me an odd behavior. Upon investigation I realized that it boiled down to the exit code of the comparison I was making. I illustrate my findings in the following code snippet.
As you can see
rc=1
[ $rc -eq 0 ]
es_num=$?
[ $rc=0 ]
es_str=$?
echo "es_num is $es_num"
echo "es_str is $es_str"
Outputs
es_num is 1
es_str is 0
Is there any documentation, preferably from the POSIX standards, that talks about the difference in the exit statuses of -eq and = in a test construct?
What should I be aware of when writing conditional statements? What are some best practices regarding this?
Portable code is preferable to Bash code (which I'm using).
test control-flow
asked 8 hours ago
EleganceElegance
183
New contributor
Elegance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I was writing some "if then" statements and found what seemed to me an odd behavior. Upon investigation I realized that it boiled down to the exit code of the comparison I was making. I illustrate my findings in the following code snippet.
As you can see
rc=1
[ $rc -eq 0 ]
es_num=$?
[ $rc=0 ]
es_str=$?
echo "es_num is $es_num"
echo "es_str is $es_str"
Outputs
es_num is 1
es_str is 0
Is there any documentation, preferably from the POSIX standards, that talks about the difference in the exit statuses of -eq and = in a test construct?
What should I be aware of when writing conditional statements? What are some best practices regarding this?
Portable code is preferable to Bash code (which I'm using).
test control-flow
asked 8 hours ago
EleganceElegance
183
New contributor
Elegance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I didn't know about the space being necessary. What I meant with "odd" is "weird if you read as pseudo code, ignoring the quirks of the language". Thanks. @ilkkachu
– Elegance
8 hours ago
1
The only way I can get[ $rc=0 ]to fail withrc=1is to setIFSto 1 as well. That would cause both tests to error out and set$?to 2 (inbash).
– Kusalananda
8 hours ago
@ilkkachu There was a typo. Of course I can't ignore them, but if I knew them, I wouldn't have to ask. That's exactly the point of the question.
– Elegance
5 hours ago
@Elegance, ok, good, thanks. And yes, you're right, you wouldn't have to ask if you knew. It's just that even a typo like that can send the readers off in the wrong direction, looking for some really weird edge case that could explain the result. (The shell can be a bit quirky sometimes so there might have been a remote possibility of an edge case where both would return1...)
– ilkkachu
5 hours ago
@ilkkachu Thank you for the helpful feedback.
– Elegance
5 hours ago
add a comment |
I was writing some "if then" statements and found what seemed to me an odd behavior. Upon investigation I realized that it boiled down to the exit code of the comparison I was making. I illustrate my findings in the following code snippet.
As you can see
rc=1
[ $rc -eq 0 ]
es_num=$?
[ $rc=0 ]
es_str=$?
echo "es_num is $es_num"
echo "es_str is $es_str"
Outputs
es_num is 1
es_str is 0
Is there any documentation, preferably from the POSIX standards, that talks about the difference in the exit statuses of -eq and = in a test construct?
What should I be aware of when writing conditional statements? What are some best practices regarding this?
Portable code is preferable to Bash code (which I'm using).
test control-flow
asked 8 hours ago
EleganceElegance
183
New contributor
Elegance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I was writing some "if then" statements and found what seemed to me an odd behavior. Upon investigation I realized that it boiled down to the exit code of the comparison I was making. I illustrate my findings in the following code snippet.
As you can see
rc=1
[ $rc -eq 0 ]
es_num=$?
[ $rc=0 ]
es_str=$?
echo "es_num is $es_num"
echo "es_str is $es_str"
Outputs
es_num is 1
es_str is 0
Is there any documentation, preferably from the POSIX standards, that talks about the difference in the exit statuses of -eq and = in a test construct?
What should I be aware of when writing conditional statements? What are some best practices regarding this?
Portable code is preferable to Bash code (which I'm using).
test control-flow
test control-flow
asked 8 hours ago
EleganceElegance
183
New contributor
Elegance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
EleganceElegance
183
New contributor
Elegance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 hours ago
Elegance
asked 8 hours ago
EleganceElegance
183
New contributor
Elegance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
EleganceElegance
183
asked 8 hours ago
EleganceElegance
183
183
New contributor
Elegance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Elegance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Elegance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I didn't know about the space being necessary. What I meant with "odd" is "weird if you read as pseudo code, ignoring the quirks of the language". Thanks. @ilkkachu
– Elegance
8 hours ago
1
The only way I can get[ $rc=0 ]to fail withrc=1is to setIFSto 1 as well. That would cause both tests to error out and set$?to 2 (inbash).
– Kusalananda
8 hours ago
@ilkkachu There was a typo. Of course I can't ignore them, but if I knew them, I wouldn't have to ask. That's exactly the point of the question.
– Elegance
5 hours ago
@Elegance, ok, good, thanks. And yes, you're right, you wouldn't have to ask if you knew. It's just that even a typo like that can send the readers off in the wrong direction, looking for some really weird edge case that could explain the result. (The shell can be a bit quirky sometimes so there might have been a remote possibility of an edge case where both would return1...)
– ilkkachu
5 hours ago
@ilkkachu Thank you for the helpful feedback.
– Elegance
5 hours ago
add a comment |
I didn't know about the space being necessary. What I meant with "odd" is "weird if you read as pseudo code, ignoring the quirks of the language". Thanks. @ilkkachu
– Elegance
8 hours ago
1
The only way I can get[ $rc=0 ]to fail withrc=1is to setIFSto 1 as well. That would cause both tests to error out and set$?to 2 (inbash).
– Kusalananda
8 hours ago
@ilkkachu There was a typo. Of course I can't ignore them, but if I knew them, I wouldn't have to ask. That's exactly the point of the question.
– Elegance
5 hours ago
@Elegance, ok, good, thanks. And yes, you're right, you wouldn't have to ask if you knew. It's just that even a typo like that can send the readers off in the wrong direction, looking for some really weird edge case that could explain the result. (The shell can be a bit quirky sometimes so there might have been a remote possibility of an edge case where both would return1...)
– ilkkachu
5 hours ago
@ilkkachu Thank you for the helpful feedback.
– Elegance
5 hours ago
I didn't know about the space being necessary. What I meant with "odd" is "weird if you read as pseudo code, ignoring the quirks of the language". Thanks. @ilkkachu
– Elegance
8 hours ago
I didn't know about the space being necessary. What I meant with "odd" is "weird if you read as pseudo code, ignoring the quirks of the language". Thanks. @ilkkachu
– Elegance
8 hours ago
1
1
The only way I can get
[ $rc=0 ] to fail with rc=1 is to set IFS to 1 as well. That would cause both tests to error out and set $? to 2 (in bash).– Kusalananda
8 hours ago
The only way I can get
[ $rc=0 ] to fail with rc=1 is to set IFS to 1 as well. That would cause both tests to error out and set $? to 2 (in bash).– Kusalananda
8 hours ago
@ilkkachu There was a typo. Of course I can't ignore them, but if I knew them, I wouldn't have to ask. That's exactly the point of the question.
– Elegance
5 hours ago
@ilkkachu There was a typo. Of course I can't ignore them, but if I knew them, I wouldn't have to ask. That's exactly the point of the question.
– Elegance
5 hours ago
@Elegance, ok, good, thanks. And yes, you're right, you wouldn't have to ask if you knew. It's just that even a typo like that can send the readers off in the wrong direction, looking for some really weird edge case that could explain the result. (The shell can be a bit quirky sometimes so there might have been a remote possibility of an edge case where both would return
1...)– ilkkachu
5 hours ago
@Elegance, ok, good, thanks. And yes, you're right, you wouldn't have to ask if you knew. It's just that even a typo like that can send the readers off in the wrong direction, looking for some really weird edge case that could explain the result. (The shell can be a bit quirky sometimes so there might have been a remote possibility of an edge case where both would return
1...)– ilkkachu
5 hours ago
@ilkkachu Thank you for the helpful feedback.
– Elegance
5 hours ago
@ilkkachu Thank you for the helpful feedback.
– Elegance
5 hours ago
add a comment |
2 Answers
2
active
oldest
votes
-eq
True if the integers n1 and n2 are algebraically equal; otherwise, false.
test
=
True if the strings s1 and s2 are identical; otherwise, false.
test
So -eq compares integers and = compares strings (which will also work with some limited integer cases).
You do have a syntax issue though, it should be:
[ "$rc" = 0 ]
And not
[ $rc=0 ]
[ "$rc" = 0 ] should exit with 1 because rc does not equal 0
[ $rc=0 ] should actually exit with 0 because it's likely going to be treated as a string and the presence of a string within the [ test construct will evaluate to true
With the sh [ test there are a few differences:
# leading 0
$ [ 01 -eq 1 ]; echo $?
0
# adjacent whitespace
$ [ ' 1' -eq 1 ]; echo $?
0
# negative 0 vs positive 0
$ [ 0 -eq -0 ]; echo $?
0
However with the bash [[ test there are a large number of differences (Including the ones mentioned above):
# base 8
$ [[ 032 -eq 26 ]]; echo $?
0
# Arithmetic expressions
$ [[ 1*6+32/15*2-1 -eq 9 ]]; echo $?
0
# Base 64
$ [[ 64#Hello_world -eq -5506400892957379251 ]]; echo $?
0
answered 8 hours ago
Jesse_bJesse_b
13.2k23369
I'm just a bit confused about what is meant by numeric and string comparisons. Let's say I use=to compare numbers. When is it not going to not yield the same result as a numeric comparison?
– Elegance
8 hours ago
Any examples with strings that have nothing but digits, and don't have leading zeros?
– Elegance
8 hours ago
1
@Elegance The point is that with-eq, the left and right hand side are compared as integers, not as strings. The test would evaluate the strings as integers (probably by passing them throughstrtol()or some such C function) before carrying out the comparison.
– Kusalananda
7 hours ago
1
@Elegance In abashtest with[[ ... -eq ... ]]you could even have arithmetic calculations on either side, as in[[ 1+1 -eq 2 ]]. Obviously,[[ 1+1 == 2 ]]would not be true.
– Kusalananda
7 hours ago
@Jesse_b You shared the same link twice. I take it the first one is incorrect.
– Elegance
5 hours ago
|
show 1 more comment
For numeric comparisons you have to use -eq whereas = is for string comparisons (as from your variable naming you already seem to know).
One of the best introductions on the test aka [ command I know is The Unix Shell's Humble If
answered 8 hours ago
freiheitsnetzfreiheitsnetz
112
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
-eq
True if the integers n1 and n2 are algebraically equal; otherwise, false.
test
=
True if the strings s1 and s2 are identical; otherwise, false.
test
So -eq compares integers and = compares strings (which will also work with some limited integer cases).
You do have a syntax issue though, it should be:
[ "$rc" = 0 ]
And not
[ $rc=0 ]
[ "$rc" = 0 ] should exit with 1 because rc does not equal 0
[ $rc=0 ] should actually exit with 0 because it's likely going to be treated as a string and the presence of a string within the [ test construct will evaluate to true
With the sh [ test there are a few differences:
# leading 0
$ [ 01 -eq 1 ]; echo $?
0
# adjacent whitespace
$ [ ' 1' -eq 1 ]; echo $?
0
# negative 0 vs positive 0
$ [ 0 -eq -0 ]; echo $?
0
However with the bash [[ test there are a large number of differences (Including the ones mentioned above):
# base 8
$ [[ 032 -eq 26 ]]; echo $?
0
# Arithmetic expressions
$ [[ 1*6+32/15*2-1 -eq 9 ]]; echo $?
0
# Base 64
$ [[ 64#Hello_world -eq -5506400892957379251 ]]; echo $?
0
answered 8 hours ago
Jesse_bJesse_b
13.2k23369
I'm just a bit confused about what is meant by numeric and string comparisons. Let's say I use=to compare numbers. When is it not going to not yield the same result as a numeric comparison?
– Elegance
8 hours ago
Any examples with strings that have nothing but digits, and don't have leading zeros?
– Elegance
8 hours ago
1
@Elegance The point is that with-eq, the left and right hand side are compared as integers, not as strings. The test would evaluate the strings as integers (probably by passing them throughstrtol()or some such C function) before carrying out the comparison.
– Kusalananda
7 hours ago
1
@Elegance In abashtest with[[ ... -eq ... ]]you could even have arithmetic calculations on either side, as in[[ 1+1 -eq 2 ]]. Obviously,[[ 1+1 == 2 ]]would not be true.
– Kusalananda
7 hours ago
@Jesse_b You shared the same link twice. I take it the first one is incorrect.
– Elegance
5 hours ago
|
show 1 more comment
-eq
True if the integers n1 and n2 are algebraically equal; otherwise, false.
test
=
True if the strings s1 and s2 are identical; otherwise, false.
test
So -eq compares integers and = compares strings (which will also work with some limited integer cases).
You do have a syntax issue though, it should be:
[ "$rc" = 0 ]
And not
[ $rc=0 ]
[ "$rc" = 0 ] should exit with 1 because rc does not equal 0
[ $rc=0 ] should actually exit with 0 because it's likely going to be treated as a string and the presence of a string within the [ test construct will evaluate to true
With the sh [ test there are a few differences:
# leading 0
$ [ 01 -eq 1 ]; echo $?
0
# adjacent whitespace
$ [ ' 1' -eq 1 ]; echo $?
0
# negative 0 vs positive 0
$ [ 0 -eq -0 ]; echo $?
0
However with the bash [[ test there are a large number of differences (Including the ones mentioned above):
# base 8
$ [[ 032 -eq 26 ]]; echo $?
0
# Arithmetic expressions
$ [[ 1*6+32/15*2-1 -eq 9 ]]; echo $?
0
# Base 64
$ [[ 64#Hello_world -eq -5506400892957379251 ]]; echo $?
0
answered 8 hours ago
Jesse_bJesse_b
13.2k23369
I'm just a bit confused about what is meant by numeric and string comparisons. Let's say I use=to compare numbers. When is it not going to not yield the same result as a numeric comparison?
– Elegance
8 hours ago
Any examples with strings that have nothing but digits, and don't have leading zeros?
– Elegance
8 hours ago
1
@Elegance The point is that with-eq, the left and right hand side are compared as integers, not as strings. The test would evaluate the strings as integers (probably by passing them throughstrtol()or some such C function) before carrying out the comparison.
– Kusalananda
7 hours ago
1
@Elegance In abashtest with[[ ... -eq ... ]]you could even have arithmetic calculations on either side, as in[[ 1+1 -eq 2 ]]. Obviously,[[ 1+1 == 2 ]]would not be true.
– Kusalananda
7 hours ago
@Jesse_b You shared the same link twice. I take it the first one is incorrect.
– Elegance
5 hours ago
|
show 1 more comment
-eq
True if the integers n1 and n2 are algebraically equal; otherwise, false.
test
=
True if the strings s1 and s2 are identical; otherwise, false.
test
So -eq compares integers and = compares strings (which will also work with some limited integer cases).
You do have a syntax issue though, it should be:
[ "$rc" = 0 ]
And not
[ $rc=0 ]
[ "$rc" = 0 ] should exit with 1 because rc does not equal 0
[ $rc=0 ] should actually exit with 0 because it's likely going to be treated as a string and the presence of a string within the [ test construct will evaluate to true
With the sh [ test there are a few differences:
# leading 0
$ [ 01 -eq 1 ]; echo $?
0
# adjacent whitespace
$ [ ' 1' -eq 1 ]; echo $?
0
# negative 0 vs positive 0
$ [ 0 -eq -0 ]; echo $?
0
However with the bash [[ test there are a large number of differences (Including the ones mentioned above):
# base 8
$ [[ 032 -eq 26 ]]; echo $?
0
# Arithmetic expressions
$ [[ 1*6+32/15*2-1 -eq 9 ]]; echo $?
0
# Base 64
$ [[ 64#Hello_world -eq -5506400892957379251 ]]; echo $?
0
answered 8 hours ago
Jesse_bJesse_b
13.2k23369
-eq
True if the integers n1 and n2 are algebraically equal; otherwise, false.
test
=
True if the strings s1 and s2 are identical; otherwise, false.
test
So -eq compares integers and = compares strings (which will also work with some limited integer cases).
You do have a syntax issue though, it should be:
[ "$rc" = 0 ]
And not
[ $rc=0 ]
[ "$rc" = 0 ] should exit with 1 because rc does not equal 0
[ $rc=0 ] should actually exit with 0 because it's likely going to be treated as a string and the presence of a string within the [ test construct will evaluate to true
With the sh [ test there are a few differences:
# leading 0
$ [ 01 -eq 1 ]; echo $?
0
# adjacent whitespace
$ [ ' 1' -eq 1 ]; echo $?
0
# negative 0 vs positive 0
$ [ 0 -eq -0 ]; echo $?
0
However with the bash [[ test there are a large number of differences (Including the ones mentioned above):
# base 8
$ [[ 032 -eq 26 ]]; echo $?
0
# Arithmetic expressions
$ [[ 1*6+32/15*2-1 -eq 9 ]]; echo $?
0
# Base 64
$ [[ 64#Hello_world -eq -5506400892957379251 ]]; echo $?
0
answered 8 hours ago
Jesse_bJesse_b
13.2k23369
edited 6 hours ago
answered 8 hours ago
Jesse_bJesse_b
13.2k23369
answered 8 hours ago
Jesse_bJesse_b
13.2k23369
answered 8 hours ago
Jesse_bJesse_b
13.2k23369
13.2k23369
I'm just a bit confused about what is meant by numeric and string comparisons. Let's say I use=to compare numbers. When is it not going to not yield the same result as a numeric comparison?
– Elegance
8 hours ago
Any examples with strings that have nothing but digits, and don't have leading zeros?
– Elegance
8 hours ago
1
@Elegance The point is that with-eq, the left and right hand side are compared as integers, not as strings. The test would evaluate the strings as integers (probably by passing them throughstrtol()or some such C function) before carrying out the comparison.
– Kusalananda
7 hours ago
1
@Elegance In abashtest with[[ ... -eq ... ]]you could even have arithmetic calculations on either side, as in[[ 1+1 -eq 2 ]]. Obviously,[[ 1+1 == 2 ]]would not be true.
– Kusalananda
7 hours ago
@Jesse_b You shared the same link twice. I take it the first one is incorrect.
– Elegance
5 hours ago
|
show 1 more comment
I'm just a bit confused about what is meant by numeric and string comparisons. Let's say I use=to compare numbers. When is it not going to not yield the same result as a numeric comparison?
– Elegance
8 hours ago
Any examples with strings that have nothing but digits, and don't have leading zeros?
– Elegance
8 hours ago
1
@Elegance The point is that with-eq, the left and right hand side are compared as integers, not as strings. The test would evaluate the strings as integers (probably by passing them throughstrtol()or some such C function) before carrying out the comparison.
– Kusalananda
7 hours ago
1
@Elegance In abashtest with[[ ... -eq ... ]]you could even have arithmetic calculations on either side, as in[[ 1+1 -eq 2 ]]. Obviously,[[ 1+1 == 2 ]]would not be true.
– Kusalananda
7 hours ago
@Jesse_b You shared the same link twice. I take it the first one is incorrect.
– Elegance
5 hours ago
I'm just a bit confused about what is meant by numeric and string comparisons. Let's say I use
= to compare numbers. When is it not going to not yield the same result as a numeric comparison?– Elegance
8 hours ago
I'm just a bit confused about what is meant by numeric and string comparisons. Let's say I use
= to compare numbers. When is it not going to not yield the same result as a numeric comparison?– Elegance
8 hours ago
Any examples with strings that have nothing but digits, and don't have leading zeros?
– Elegance
8 hours ago
Any examples with strings that have nothing but digits, and don't have leading zeros?
– Elegance
8 hours ago
1
1
@Elegance The point is that with
-eq, the left and right hand side are compared as integers, not as strings. The test would evaluate the strings as integers (probably by passing them through strtol() or some such C function) before carrying out the comparison.– Kusalananda
7 hours ago
@Elegance The point is that with
-eq, the left and right hand side are compared as integers, not as strings. The test would evaluate the strings as integers (probably by passing them through strtol() or some such C function) before carrying out the comparison.– Kusalananda
7 hours ago
1
1
@Elegance In a
bash test with [[ ... -eq ... ]] you could even have arithmetic calculations on either side, as in [[ 1+1 -eq 2 ]]. Obviously, [[ 1+1 == 2 ]] would not be true.– Kusalananda
7 hours ago
@Elegance In a
bash test with [[ ... -eq ... ]] you could even have arithmetic calculations on either side, as in [[ 1+1 -eq 2 ]]. Obviously, [[ 1+1 == 2 ]] would not be true.– Kusalananda
7 hours ago
@Jesse_b You shared the same link twice. I take it the first one is incorrect.
– Elegance
5 hours ago
@Jesse_b You shared the same link twice. I take it the first one is incorrect.
– Elegance
5 hours ago
|
show 1 more comment
For numeric comparisons you have to use -eq whereas = is for string comparisons (as from your variable naming you already seem to know).
One of the best introductions on the test aka [ command I know is The Unix Shell's Humble If
answered 8 hours ago
freiheitsnetzfreiheitsnetz
112
add a comment |
For numeric comparisons you have to use -eq whereas = is for string comparisons (as from your variable naming you already seem to know).
One of the best introductions on the test aka [ command I know is The Unix Shell's Humble If
answered 8 hours ago
freiheitsnetzfreiheitsnetz
112
add a comment |
For numeric comparisons you have to use -eq whereas = is for string comparisons (as from your variable naming you already seem to know).
One of the best introductions on the test aka [ command I know is The Unix Shell's Humble If
answered 8 hours ago
freiheitsnetzfreiheitsnetz
112
For numeric comparisons you have to use -eq whereas = is for string comparisons (as from your variable naming you already seem to know).
One of the best introductions on the test aka [ command I know is The Unix Shell's Humble If
answered 8 hours ago
freiheitsnetzfreiheitsnetz
112
answered 8 hours ago
freiheitsnetzfreiheitsnetz
112
answered 8 hours ago
freiheitsnetzfreiheitsnetz
112
answered 8 hours ago
freiheitsnetzfreiheitsnetz
112
112
add a comment |
add a comment |
Elegance is a new contributor. Be nice, and check out our Code of Conduct.
Elegance is a new contributor. Be nice, and check out our Code of Conduct.
Elegance is a new contributor. Be nice, and check out our Code of Conduct.
Elegance is a new contributor. Be nice, and check out our Code of Conduct.
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I didn't know about the space being necessary. What I meant with "odd" is "weird if you read as pseudo code, ignoring the quirks of the language". Thanks. @ilkkachu
– Elegance
8 hours ago
1
The only way I can get
[ $rc=0 ]to fail withrc=1is to setIFSto 1 as well. That would cause both tests to error out and set$?to 2 (inbash).– Kusalananda
8 hours ago
@ilkkachu There was a typo. Of course I can't ignore them, but if I knew them, I wouldn't have to ask. That's exactly the point of the question.
– Elegance
5 hours ago
@Elegance, ok, good, thanks. And yes, you're right, you wouldn't have to ask if you knew. It's just that even a typo like that can send the readers off in the wrong direction, looking for some really weird edge case that could explain the result. (The shell can be a bit quirky sometimes so there might have been a remote possibility of an edge case where both would return
1...)– ilkkachu
5 hours ago
@ilkkachu Thank you for the helpful feedback.
– Elegance
5 hours ago