[Solved] [Bash] How bash treats variables

[LinuxOS]

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$ foo=bar
$ echo $foo
bar
$ foo=baz echo $foo
bar
$ foo=baz sh -c 'echo $foo'
baz

Hello!
Can anybody explain to me why foo=baz echo $foo results in bar, but then it works properly when I use sh -c?
Thanks!

[steve_v]

This is an ancient question, and is answered multiple times elsewhere. It also smells suspiciously like homework...
Short version: variables are evaluated (i.e. env for the child process is constructed) before the command-line is executed, unless they are protected by single quotes. This applies also to builtins like echo.

it works properly when I use sh -c?

There's nothing "improper" about the first result, it's expected (though arguably somewhat confusing) behaviour. You'll find shells other than bash do this as well.