What's the difference between $@ and $*

They aren't the same. $* is a single string, whereas $@ is an actual array. To see the difference, execute the following script like so:

 > ./test.sh one two "three four"

The script:

#!/bin/bash

echo "Using \"\$*\":"
for a in "$*"; do
    echo $a;
done

echo -e "\nUsing \$*:"
for a in $*; do
    echo $a;
done

echo -e "\nUsing \"\$@\":"
for a in "$@"; do
    echo $a;
done

echo -e "\nUsing \$@:"
for a in $@; do
    echo $a;
done              

The explanation and the results for the four cases are below.

In the first case, the parameters are regarded as one long quoted string:

Using "$*":
one two three four

Case 2 (unquoted) - the string is broken into words by the for loop:

Using $*:
one
two
three
four

Case 3 - it treats each element of $@ as a quoted string:

Using "$@":
one
two
three four

Last case - it treats each element as an unquoted string, so the last one is again split by what amounts to for three four:

Using $@:
one
two
three
four

The difference comes in how they are expanded.

$* expands to a single argument with all the elements delimited by spaces (actually the first character of $IFS).
$@ expands to multiple arguments.

For example

#!/bin/bash
echo "With *:"
for arg in "$*"; do echo "<$arg>"; done
echo
echo "With @:"
for arg in "$@"; do echo "<$arg>"; done

 

$ /tmp/test.sh 1  2 "3  4"
With *:
<1 2 3  4>

With @:
<1>
<2>
<3  4>

You can review Bash Beginners Guide for more information. These to do pretty much the same thing with the difference of how it's separated:

$* - Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, it expands to a single word with the value of each parameter separated by the first character of the IFS special variable.

$@ - Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, each parameter expands to a separate word.

But unless you set IFS to value other then default it could look identical.