3 added 1 character in body
source | link

No, it shouldn't, because of the way bash operate the command.

When you type echo one|name, bash parse the command, treats | as a pipe token, so | perform pipeline.

When you type echo $variable, because token parsing occur before variable expansion, bash parsing the command into two partparts, echo and $variable. After that, it performs variable expansion, expand $variable to one|name. In this case, one|name is a string, | is a part of string and can not be treated as a pipe token (of course, the token recognition phrase was done). The only thing it can be special if IFS variable contains |, bash will use | as delimiter to perform field spliting:

$ variable='one|name'
$ IFS='|'
$ echo $variable
one name

No, it shouldn't, because of the way bash operate the command.

When you type echo one|name, bash parse the command, treats | as a pipe token, so | perform pipeline.

When you type echo $variable, because token parsing occur before variable expansion, bash parsing the command into two part, echo and $variable. After that, it performs variable expansion, expand $variable to one|name. In this case, one|name is a string, | is a part of string and can not be treated as a pipe token (of course, the token recognition phrase was done). The only thing it can be special if IFS variable contains |, bash will use | as delimiter to perform field spliting:

$ variable='one|name'
$ IFS='|'
$ echo $variable
one name

No, it shouldn't, because of the way bash operate the command.

When you type echo one|name, bash parse the command, treats | as a pipe token, so | perform pipeline.

When you type echo $variable, because token parsing occur before variable expansion, bash parsing the command into two parts, echo and $variable. After that, it performs variable expansion, expand $variable to one|name. In this case, one|name is a string, | is a part of string and can not be treated as a pipe token (of course, the token recognition phrase was done). The only thing it can be special if IFS variable contains |, bash will use | as delimiter to perform field spliting:

$ variable='one|name'
$ IFS='|'
$ echo $variable
one name
2 deleted 2 characters in body
source | link

No, it shouldn't, because of the way bash operate the command.

When you type echo one|name, bash parsingparse the command, treats | as a pipe token, so | perform pipeline.

When you type echo $variable, because token parsing occur before variable expansion, bash parsing the command into two part, echo and $variable. After that, it performs variable expansion, expand $variable to one|name. In this case, one|name is a string, | is a part of string and can not be treated as a pipe token (of course, the token recognition phrase was done). The only thing it can be special if IFS variable contains |, bash will use | as delimiter to perform field spliting:

$ variable='one|name'
$ IFS='|'
$ echo $variable
one name

No, it shouldn't.

When you type echo one|name, bash parsing the command, treats | as a token, so | perform pipeline.

When you type echo $variable, because token parsing occur before variable expansion, bash parsing the command into two part, echo and $variable. After that, it performs variable expansion, expand $variable to one|name. In this case, one|name is a string, | is a part of string and can not be treated as a pipe. The only thing it can be special if IFS variable contains |, bash will use | as delimiter to perform field spliting:

$ variable='one|name'
$ IFS='|'
$ echo $variable
one name

No, it shouldn't, because of the way bash operate the command.

When you type echo one|name, bash parse the command, treats | as a pipe token, so | perform pipeline.

When you type echo $variable, because token parsing occur before variable expansion, bash parsing the command into two part, echo and $variable. After that, it performs variable expansion, expand $variable to one|name. In this case, one|name is a string, | is a part of string and can not be treated as a pipe token (of course, the token recognition phrase was done). The only thing it can be special if IFS variable contains |, bash will use | as delimiter to perform field spliting:

$ variable='one|name'
$ IFS='|'
$ echo $variable
one name
1
source | link

No, it shouldn't.

When you type echo one|name, bash parsing the command, treats | as a token, so | perform pipeline.

When you type echo $variable, because token parsing occur before variable expansion, bash parsing the command into two part, echo and $variable. After that, it performs variable expansion, expand $variable to one|name. In this case, one|name is a string, | is a part of string and can not be treated as a pipe. The only thing it can be special if IFS variable contains |, bash will use | as delimiter to perform field spliting:

$ variable='one|name'
$ IFS='|'
$ echo $variable
one name