javascript - Why does decodeURI decode more characters than it should? -

i reading decodeuri (mdn, es6 spec) , caught eye:

escape sequences not have been introduced encodeuri not replaced.

so, should decode characters encodeuri encodes.

// none of these should escaped `encodeuri`.  const unescaped = "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz0123456789-_.!~*'();/?:@&=+$,#";    const data = [...unescaped].map(char => ({    "char": char,    "encodeuri(char)": encodeuri(char),    "encodepercent(char)": encodepercent(char),    "decodeuri(encodepercent(char))": decodeuri(encodepercent(char))  }));    console.table( data );  console.log( "check browser's console." );    function encodepercent(string) {    return string.replace(/./g, char => "%" + char.charcodeat(0).tostring(16));  }

why true ; / ? : @ & = + $ , #?

the specification states following step:

3. let unescapeduriset string containing 1 instance of each code unit valid in urireserved , uriunescaped plus "#"

let’s take @ urireserved, , voilà:

urireserved ::: 1 of

; / ? : @ & = + $ , 

the following step then:

4. return encode(uristring, unescapeduriset).

encode in encodes string except characters in unescapeduriset, include ; / ? : @ & = + $ ,.

this means encodeuri can never introduce escape sequences in urireserved , uriunescaped.

interestingly enough, decodeuri defined this:

3. let reserveduriset string containing 1 instance of each code unit valid in urireserved plus "#".

4. return decode(uristring, reserveduriset).

decode works encode , decodes except characters in reserveduriset. obviously, characters of urireserved excluded being decoded. , happen ; / ? : @ & = + $ ,!

the questions remains why standard specifies this. if had included uriunescaped in reserveduriset behaviour introduction states. mistake?