is declare -A explicit declaration mandatory in associative arrays in Bash?

Question

I have defined a bash associative array without the explicit declare -A command. But I am not sure if it really is the associative array since I read somewhere without the declare -A, the array is treated as indexed array by Bash.

I tried the below code:

players=([tennis]=fedExpress [cricket]=buzz [football]=cr7) echo "${players[tennis]}"

the o/p is: cr7 but should be fedExpress

but in geeks for geeks article, they declared an indexed array as follows:

ARRAYNAME=([1]=10 [2]=20 [3]=30)

so, w/o the declare -A, is the player's array also treated as an indexed array? or associative array? please clarify this. I am writing an article so I have to be corroborative in what I write.

KamilCuk · Accepted Answer · 2023-10-11 06:26:17Z

is declare -A explicit declaration mandatory in associative arrays in Bash?

Yes.

w/o the declare -A, is the player's array also treated as an indexed array? or associative array?

Indexed. You do not need us for this - use declare -p to inspect a variable with all the flags.

The string inside [this] is treated as an arithmetic expression (unless it is an associative array). Inside arithmetic expression undefined variables are equal to 0, i.e. echo $((something)) outputs 0, see documentation. The following with undefined variables tennis cricket and football:

 players=([tennis]=fedExpress [cricket]=buzz [football]=cr7)

Is just equal to:

players=([0]=fedExpress [0]=buzz [0]=cr7)

However, consider:

$ tennis=123 $ cricket=123*2 $ players=([tennis]=fedExpress [cricket]=buzz [football]=cr7) $ declare -p players declare -a players=([0]="cr7" [123]="fedExpress" [246]="buzz")

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