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2

I have a file in the name of Build.number with the content value 012 which I need to increment by +1.
So, I tried this

BN=$($cat Build.number)
BN=$(($BN+1))
echo $BN >Build.number

but here I am getting the value 11 when I am expecting 013.
Can anyone help me?

bash shell-script numeric-data arithmetic

share|improve this question

edited 4 hours ago

Cbhihe

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user11668570user11668570

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  There are two good answers, both not only explaining the cause of the problem, but providing additional information that might be useful. Stephen's provides a reference and Stéphane's mentions other shells and provides a more efficient way of reading the data without using the cat program. The ideal answer would combine both. (The ideal answer would also use "%.3dn" format rather than the old leading 0 trick, which is ancient style, and misleading because in that case the leading 0 doesn't mean octal.)
  
  – Ray Butterworth
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Related: unix.stackexchange.com/a/36959/117549
  
  – Jeff Schaller♦
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Ray why is .3d better than 03d? Apart from the octal confusion...
  
  – Stephen Kitt
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @StephenKitt. For d, i, o, u, x, X, a, A, e, E, f, F, g, and G conversion specifiers, leading zeros (following any indication of sign or base) are used to pad to the field width rather than performing space padding, except when converting an infinity or NaN. If the 0 and - flags both appear, the 0 flag is ignored. For d, i, o, u, x, and X conversion specifiers, if a precision is specified, the 0 flag shall be ignored. [CX] [Option Start] If the 0 and <apostrophe> flags both appear, the grouping characters are inserted before zero padding. For other conversions, the behavior is undefined.
  
  – Ray Butterworth
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @StephenKitt, the .3d makes the intent much more obvious: An optional precision that gives the minimum number of digits to appear for the d, i, o, u, x, and X conversion specifiers. The 03d syntax dates back to the original 1970s version, and preserving backward compatibility has made its rules extremely complicated.
  
  – Ray Butterworth
  5 hours ago
  
  

  
  
  
  

 | 
show 2 more comments

2

I have a file in the name of Build.number with the content value 012 which I need to increment by +1.
So, I tried this

BN=$($cat Build.number)
BN=$(($BN+1))
echo $BN >Build.number

but here I am getting the value 11 when I am expecting 013.
Can anyone help me?

bash shell-script numeric-data arithmetic

share|improve this question

edited 4 hours ago

Cbhihe

48211 gold badge66 silver badges2020 bronze badges

asked 9 hours ago

user11668570user11668570

1911 bronze badge

New contributor

user11668570 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  There are two good answers, both not only explaining the cause of the problem, but providing additional information that might be useful. Stephen's provides a reference and Stéphane's mentions other shells and provides a more efficient way of reading the data without using the cat program. The ideal answer would combine both. (The ideal answer would also use "%.3dn" format rather than the old leading 0 trick, which is ancient style, and misleading because in that case the leading 0 doesn't mean octal.)
  
  – Ray Butterworth
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Related: unix.stackexchange.com/a/36959/117549
  
  – Jeff Schaller♦
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Ray why is .3d better than 03d? Apart from the octal confusion...
  
  – Stephen Kitt
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @StephenKitt. For d, i, o, u, x, X, a, A, e, E, f, F, g, and G conversion specifiers, leading zeros (following any indication of sign or base) are used to pad to the field width rather than performing space padding, except when converting an infinity or NaN. If the 0 and - flags both appear, the 0 flag is ignored. For d, i, o, u, x, and X conversion specifiers, if a precision is specified, the 0 flag shall be ignored. [CX] [Option Start] If the 0 and <apostrophe> flags both appear, the grouping characters are inserted before zero padding. For other conversions, the behavior is undefined.
  
  – Ray Butterworth
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @StephenKitt, the .3d makes the intent much more obvious: An optional precision that gives the minimum number of digits to appear for the d, i, o, u, x, and X conversion specifiers. The 03d syntax dates back to the original 1970s version, and preserving backward compatibility has made its rules extremely complicated.
  
  – Ray Butterworth
  5 hours ago
  
  

  
  
  
  

 | 
show 2 more comments

I have a file in the name of Build.number with the content value 012 which I need to increment by +1.
So, I tried this

BN=$($cat Build.number)
BN=$(($BN+1))
echo $BN >Build.number

but here I am getting the value 11 when I am expecting 013.
Can anyone help me?

bash shell-script numeric-data arithmetic

share|improve this question

edited 4 hours ago

Cbhihe

48211 gold badge66 silver badges2020 bronze badges

asked 9 hours ago

user11668570user11668570

1911 bronze badge

New contributor

user11668570 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

BN=$($cat Build.number)
BN=$(($BN+1))
echo $BN >Build.number

share|improve this question

edited 4 hours ago

Cbhihe

48211 gold badge66 silver badges2020 bronze badges

asked 9 hours ago

user11668570user11668570

1911 bronze badge

New contributor

user11668570 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 4 hours ago

Cbhihe

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asked 9 hours ago

user11668570user11668570

1911 bronze badge

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user11668570 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 4 hours ago

Cbhihe

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edited 4 hours ago

Cbhihe

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asked 9 hours ago

user11668570user11668570

1911 bronze badge

New contributor

user11668570 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 9 hours ago

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2 Answers
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11

The leading 0 causes Bash to interpret the value as an octal value; 012 octal is 10 decimal, so you get 11.

To force the use of decimal, add 10#:

BN=10#$(cat Build.number)
echo $((++BN)) > Build.number

To print the number using at least three digits, use printf:

printf "%.3dn" $((++BN)) > Build.number

share|improve this answer

edited 53 mins ago

answered 9 hours ago

Stephen KittStephen Kitt

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 | 

6

bash treats constants that start with 0 as octal numbers in its arithmetic expressions, so 011 is actually 9.

That's actually a POSIX requirement.

Some other shells like mksh or zsh ignore it (unless in POSIX compliant mode) as it gets in the way far more often than it is useful.

With ksh93, BN=011; echo "$(($BN))" outputs 9, but echo "$((BN))" outputs 11.

In bash, you can use BN=$((10#$(<Build.number))), which should work as long as the number doesn't start with - or +.

share|improve this answer

answered 9 hours ago

Stéphane ChazelasStéphane Chazelas

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11

The leading 0 causes Bash to interpret the value as an octal value; 012 octal is 10 decimal, so you get 11.

To force the use of decimal, add 10#:

BN=10#$(cat Build.number)
echo $((++BN)) > Build.number

To print the number using at least three digits, use printf:

printf "%.3dn" $((++BN)) > Build.number

share|improve this answer

edited 53 mins ago

answered 9 hours ago

Stephen KittStephen Kitt

208k2727 gold badges493493 silver badges559559 bronze badges

add a comment
 | 

11

The leading 0 causes Bash to interpret the value as an octal value; 012 octal is 10 decimal, so you get 11.

To force the use of decimal, add 10#:

BN=10#$(cat Build.number)
echo $((++BN)) > Build.number

To print the number using at least three digits, use printf:

printf "%.3dn" $((++BN)) > Build.number

share|improve this answer

edited 53 mins ago

answered 9 hours ago

Stephen KittStephen Kitt

208k2727 gold badges493493 silver badges559559 bronze badges

add a comment
 | 

The leading 0 causes Bash to interpret the value as an octal value; 012 octal is 10 decimal, so you get 11.

To force the use of decimal, add 10#:

BN=10#$(cat Build.number)
echo $((++BN)) > Build.number

To print the number using at least three digits, use printf:

printf "%.3dn" $((++BN)) > Build.number

share|improve this answer

edited 53 mins ago

answered 9 hours ago

Stephen KittStephen Kitt

208k2727 gold badges493493 silver badges559559 bronze badges

BN=10#$(cat Build.number)
echo $((++BN)) > Build.number

printf "%.3dn" $((++BN)) > Build.number

share|improve this answer

edited 53 mins ago

answered 9 hours ago

Stephen KittStephen Kitt

208k2727 gold badges493493 silver badges559559 bronze badges

answered 9 hours ago

Stephen KittStephen Kitt

208k2727 gold badges493493 silver badges559559 bronze badges

answered 9 hours ago

Stephen KittStephen Kitt

208k2727 gold badges493493 silver badges559559 bronze badges

6

bash treats constants that start with 0 as octal numbers in its arithmetic expressions, so 011 is actually 9.

That's actually a POSIX requirement.

Some other shells like mksh or zsh ignore it (unless in POSIX compliant mode) as it gets in the way far more often than it is useful.

With ksh93, BN=011; echo "$(($BN))" outputs 9, but echo "$((BN))" outputs 11.

In bash, you can use BN=$((10#$(<Build.number))), which should work as long as the number doesn't start with - or +.

share|improve this answer

answered 9 hours ago

Stéphane ChazelasStéphane Chazelas

337k5858 gold badges661661 silver badges10381038 bronze badges

add a comment
 | 

6

bash treats constants that start with 0 as octal numbers in its arithmetic expressions, so 011 is actually 9.

That's actually a POSIX requirement.

Some other shells like mksh or zsh ignore it (unless in POSIX compliant mode) as it gets in the way far more often than it is useful.

With ksh93, BN=011; echo "$(($BN))" outputs 9, but echo "$((BN))" outputs 11.

In bash, you can use BN=$((10#$(<Build.number))), which should work as long as the number doesn't start with - or +.

share|improve this answer

answered 9 hours ago

Stéphane ChazelasStéphane Chazelas

337k5858 gold badges661661 silver badges10381038 bronze badges

add a comment
 | 

bash treats constants that start with 0 as octal numbers in its arithmetic expressions, so 011 is actually 9.

That's actually a POSIX requirement.

Some other shells like mksh or zsh ignore it (unless in POSIX compliant mode) as it gets in the way far more often than it is useful.

With ksh93, BN=011; echo "$(($BN))" outputs 9, but echo "$((BN))" outputs 11.

In bash, you can use BN=$((10#$(<Build.number))), which should work as long as the number doesn't start with - or +.

share|improve this answer

answered 9 hours ago

Stéphane ChazelasStéphane Chazelas

337k5858 gold badges661661 silver badges10381038 bronze badges

share|improve this answer

answered 9 hours ago

Stéphane ChazelasStéphane Chazelas

337k5858 gold badges661661 silver badges10381038 bronze badges

answered 9 hours ago

Stéphane ChazelasStéphane Chazelas

337k5858 gold badges661661 silver badges10381038 bronze badges

answered 9 hours ago

Stéphane ChazelasStéphane Chazelas

337k5858 gold badges661661 silver badges10381038 bronze badges