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Exit statuses of comparisons in test constructs

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Exit statuses of comparisons in test constructs

POSIX test and -adifference between non-builtin 'test' and '['unix test when to use eq vs = vs == in test commands?What does `[ EXPRESSION ], [ ] and [OPTION` mean in `man test`?-n Vs !(exclamation mark) behaves differently with test commandDifference in conditions /test( test -n $st ) != ( test -z $st ) right?How to test list of proxy servers?Bash test: what does “=~” do?How to correctly test file's extension in if statement?

3

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

share|improve this question

edited 1 hour ago

Elegance

asked 4 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
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  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
  3 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
  1 hour 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
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @ilkkachu Thank you for the helpful feedback.
  
  – Elegance
  1 hour ago
  

  
  
  
  

add a comment | 

3

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

share|improve this question

edited 1 hour ago

Elegance

asked 4 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
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  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
  3 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
  1 hour 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
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @ilkkachu Thank you for the helpful feedback.
  
  – Elegance
  1 hour 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

share|improve this question

edited 1 hour ago

Elegance

asked 4 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.

rc=1

[ $rc -eq 0 ]

es_num=$?

[ $rc=0 ]

es_str=$?

echo "es_num is $es_num"

echo "es_str is $es_str"

es_num is 1

es_str is 0

share|improve this question

edited 1 hour ago

Elegance

asked 4 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.

share|improve this question

edited 1 hour ago

Elegance

asked 4 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 4 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 4 hours ago

EleganceElegance

183

2 Answers
2

active

oldest

votes

3

-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

share|improve this answer

edited 2 hours ago

answered 4 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
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Any examples with strings that have nothing but digits, and don't have leading zeros?
  
  – Elegance
  3 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 through strtol() or some such C function) before carrying out the comparison.
  
  – Kusalananda
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  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
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Jesse_b You shared the same link twice. I take it the first one is incorrect.
  
  – Elegance
  1 hour ago
  

  
  
  
  

 | 
show 1 more comment

0

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

share|improve this answer

answered 4 hours ago

freiheitsnetzfreiheitsnetz

112

add a comment | 

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3

-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

share|improve this answer

edited 2 hours ago

answered 4 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
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Any examples with strings that have nothing but digits, and don't have leading zeros?
  
  – Elegance
  3 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 through strtol() or some such C function) before carrying out the comparison.
  
  – Kusalananda
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  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
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Jesse_b You shared the same link twice. I take it the first one is incorrect.
  
  – Elegance
  1 hour ago
  

  
  
  
  

 | 
show 1 more comment

3

-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

share|improve this answer

edited 2 hours ago

answered 4 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
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Any examples with strings that have nothing but digits, and don't have leading zeros?
  
  – Elegance
  3 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 through strtol() or some such C function) before carrying out the comparison.
  
  – Kusalananda
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  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
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Jesse_b You shared the same link twice. I take it the first one is incorrect.
  
  – Elegance
  1 hour 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

share|improve this answer

edited 2 hours ago

answered 4 hours ago

Jesse_bJesse_b

13.2k23369

[ "$rc" = 0 ]

[ $rc=0 ]

# leading 0

$ [ 01 -eq 1 ]; echo $?

0

# adjacent whitespace

$ [ ' 1' -eq 1 ]; echo $?

0

# negative 0 vs positive 0

$ [ 0 -eq -0 ]; echo $?

0

# 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

share|improve this answer

edited 2 hours ago

answered 4 hours ago

Jesse_bJesse_b

13.2k23369

answered 4 hours ago

Jesse_bJesse_b

13.2k23369

answered 4 hours ago

Jesse_bJesse_b

13.2k23369

0

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

share|improve this answer

answered 4 hours ago

freiheitsnetzfreiheitsnetz

112

add a comment | 

0

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

share|improve this answer

answered 4 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

share|improve this answer

answered 4 hours ago

freiheitsnetzfreiheitsnetz

112

share|improve this answer

answered 4 hours ago

freiheitsnetzfreiheitsnetz

112

answered 4 hours ago

freiheitsnetzfreiheitsnetz

112

answered 4 hours ago

freiheitsnetzfreiheitsnetz

112