CODEWITHSUNDEEP
shellbatch-file
JiangFeng
JiangFeng

Reputation: 355

why set used in array cannot use echo?

I'm learning batch script in tutorialspoint now, here's a very simple script:

@echo off

set a[0]=0
set a[1]=1
set a[2]=2
set a[3]=3
set a[4]=4
set a[5]=5

for /l %%n in (0,1,5) do (echo %a[%%n]%)

why result is "ECHO is off"

if I write like for /l %%n in (0,1,5) do (echo a[%%n]) I can get 

a[0]
a[1]
a[2]
a[3]
a[4]
a[5]

so why cannot I use echo to get the value of array?

Upvotes: 1

Views: 93

Answers (2)

aschipfl
aschipfl

Reputation: 34989

Here is an alternative method without having to enable delayed expansion:

for /l %%n in (0,1,5) do (call echo %%a[%%n]%%)

The body of the loop call echo %%a[%%n]%% is parsed two times due to call; during the first time it becomes call echo %a[0]%, supposing the current value of %%n is 0, because %% becomes %, and during the second time, %a[0]% is replaced by the respective value of the variable.

In your approach, echo %a[%%n]%, the command line interpreter sees two variables to expand, namely %a[% and %n]%, which are both undefined and become therefore replaced by nothing.

Upvotes: 1

jeb
jeb

Reputation: 82438

As you shouldn't use percent expansion in a block (here it's a FOR-block), as percents are expanded while the block is parsed.

Use delayed expansion instead.

setlocal EnableDelayedExpansion
for /l %%n in (0,1,5) do (echo !a[%%n]!)

Upvotes: 1

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