why integer result in 0 when I change declared integer to string in Bash script

Question

Code:

#!/bin/bash

declare -i number
# The script will treat subsequent occurrences of "number" as an integer.       

number=3
echo "Number = $number"     # Number = 3

number=three
echo "Number = $number"     # Number = 0
# Tries to evaluate the string "three" as an integer.

I cannot figure out why number changed when I assign a string "three" to number. I think number should stay the same. That really surprised me.

This is just not supported. The assignment operator does not translate english words into their numerical meaning. If it isn't a number (in digits) then 0 will result. What are you trying to do, in a wider perspective? — geert3
– geert3, Commented Dec 1, 2015 at 8:24
Why would it stay the same, you just assigned it a new value... — 123
– 123, Commented Dec 1, 2015 at 8:26
But assign string to a declared integer is illegal, so I think number should stay same just like it failed. Set number to 0 destroy my original data. It is just a example from a bash script book. — Feng Yu
– Feng Yu, Commented Dec 1, 2015 at 8:34
no assigning a string is not illegal, it is "interpreted" as a number. e.g. "5+5" is interpreted as 10. You string is interpreted as 0 as a last resort. — geert3
– geert3, Commented Dec 1, 2015 at 8:36
@geert3 Not quite a last resort; strings are (recursively) evaluated as parameter expansions in an arithmetic context; undefined parameters default to 0. — chepner
– chepner, Commented Dec 1, 2015 at 13:22

user513951 · Accepted Answer · 2015-12-01 09:47:12Z

2

From the declare section of man bash:

-i The variable is treated as an integer; arithmetic evaluation (see ARITHMETIC EVALUATION) is performed when the variable is assigned a value.

From the ARITHMETIC EVALUATION section of man bash:

The value of a variable is evaluated as an arithmetic expression when...a variable which has been given the integer attribute using declare -i is assigned a value. A null value evaluates to 0.

Together, these clearly state that the behavior you're seeing is the expected behavior. When the characters t h r e e are evaluated arithmetically, the resulting null value is evaluated as 0, which is then assigned to the variable number.

All assignments in bash are interpreted first as strings. number=10 interprets the 1 0 as a string first, recognizes it as a valid integer, and leaves it as-is. number=three is just as syntactically and semantically valid as number=10, which is why your script continues without any error after assigning the evaluated value of 0 to number.

edited Dec 1, 2015 at 9:47

answered Dec 1, 2015 at 8:32

user513951

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2 Comments

123 Over a year ago

then casts it to an integer, it doesn't, it's always a string.

user513951 Over a year ago

@123 Adjusted to recognizes it as a valid integer.

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