The echo behavior seems incorrect with/without -e option.

Could any one tell me why the below echo command seems to behave inversely? Thanks.

Quote:

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[justdemon ~]$ t="1\n"; t=$(echo -e $t); echo -e $t
1
[justdemon ~]$ t="1\n"; t=$(echo $t); echo -e $t
1

[justdemon ~]$

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what do you mean by inversely??

what answer are you expecting?

let me explain;

take this one first:
t="1\n"; t=$(echo -e $t); echo -e $t

split it away;
$ t="1\n" (it will assign the value to "t")
$ t=$(echo -e $t)

in this you have mentioned -e flag so it will enable interpretation of the backslash-escaped characters, in our case it is "\n"(new line character).
so t will be assigned "1" after this step.
at last, $ echo -e $t
it will print the value of t (this time there is no backslash-escaped characters ).

Now consider second one,

t="1\n"; t=$(echo $t); echo -e $t

$ t="1\n" (it will assign the value to "t")
$ t=$(echo $t)
here you have not mentioned -e flag so echo will not enable interpretation of the backslash-escaped characters, and t will be assigned "1\n" only.

at last, $ echo -e $t
it will print the value of t (this time it will enable interpretation of the backslash-escaped characters, so it will print "1" + new line).

Hope you have got you answer.

Dear patelhemal2210,

Thanks very much for your detailed explanation.
It looks like I got incorrect idea of -e option.

Best Regards,
justdemon

you are always welcome.:-)

Quote:

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Originally Posted by justdemon

Could any one tell me why the below echo command seems to behave inversely? Thanks.

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Bash is doing what it's supposed to do.

When you execute the first $(echo -e $t), bash is already interpreting the escape character, and so it assigns 1 to your variable.

try this:

Code:

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user@host$ A="
>
> 1"; echo -e $A
1
user@host$

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I found that I still have question on this.

Even I use the following command, I can't put the "newline" character into the variable.

Quote:

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[justdemon ~]$ t="1\n"; t="$(echo -e $t)"; echo "$t"
1
[justdemon ~]$

---

How to do to assign a newline included in the original variable into a new variable by the echo command?

Thanks.

command you are looking for might somthing look like this:-

Code:

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[justdemon ~]$ t="1\n"; t="$(echo $t)"; echo -e "$t"

---

If you still have a problem with this then provide us proper information what you exactly want to do? if possible provide example for the same.

If t="1"$'\n', how to copy its content to other variables?

Here is the example without echo command.

Code:

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[justdemon ~]$ t="1
> "; s="$t"; echo "$s"
1

[justdemon ~]$

---

Thanks.

Quote:

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Originally Posted by justdemon

Code:

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[justdemon ~]$ t="1
> "; s="$t"; echo "$s"
1

[justdemon ~]$

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.

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Yeap.. this is working.

you could copy the content of "t" into "s" variable.:)

what else do you want? :twisted:

what exactly do you want do?