mjhjmtu

Why the following command prints Smith but not Smith?

echo Smith

The backslash is an escape character in the shell syntax that:

shall preserve the literal value of the following character, with the exception of a <newline>. ... The <backslash> ... shall be removed

So S means the same thing as S, because S is not a newline character and also not a shell special character that could be escaped ($, ", ', {, [, `, , |, &, ;, <, >, (, ), ?, *, [, #, ~, =, %, , or tab). To include a literal backslash in the argument given to echo, escape it in turn with or any other quoting operator:

echo \Smith

or

echo 'Smith'

or

echo "Smith"

( still retains a special meaning within double quotes, but not when followed by S, only when followed by , `, $, " or newline).

Now, many echo implementations also treat specially when found in their argument (though generally not when followed by S, though there's no guarantee), so you may want to use printf instead:

printf '%sn' 'Smith'

note: this is the specified behavior for your example unquoted case, but it can change if it is otherwise quoted or read in from a file

S escapes the S, which is not a special character, thus gives S. You need to double the backslash to print it: echo \Smith

EDIT: But more generally, it's better to use printf. See the difference between echo x\by, which outputs "y" (the "x" gets overwritten by the backspace b) with some versions of echo (dash, zsh), and printf "%sn" x\by, which outputs "xby".