450338 - Last parenthesis of conditional expression not stripped when it is in if statement

Status: RESOLVED WORKSFORME

(Core :: JavaScript Engine, defect)

Description

[Biju]

Something nice to have....
http://groups.google.com/group/mozilla.dev.tech.js-engine/browse_thread/thread/46237fccd2df7b21#

function a() {
    x = (a == b) && (c == d) ;
    if ((a == b) && (c == d) ) {
        p();
    }

}

alert(a);

//  gives ==>

function a() {
    x = a == b && c == d;
    if (a == b && (c == d)) {
        p();
    }

}



Here all parenthesis of first expression in the function was striped
off
      "x = (a == b) && (c == d) ;"
  ==> "x = a == b && c == d;"

But for the next statement with similar conditional expression the
last parenthesis is not removed.
ie,  "if ((a == b) && (c == d) ) {" 
==>  "if (a == b && (c == d)) {"

now if we add an assignment operator in conditional expression, 
we will see the problem is gone.
ie,  " if(x=(a == b) && (c == d) ) {" 
 ==> "if (x = a == b && c == d) {"

Comment 1

[Biju]

 // other cases

function a() {
    x = (a == b) && (c == d) && (e == f) && (g == h) ;
    if((a == b) && (c == d) && (e == f) && (g == h) );
    if(x=(a == b) && (c == d) && (e == f) && (g == h) ) ;
    while((a == b) && (c == d) && (e == f) && (g == h) ) ;
    while(x=(a == b) && (c == d) && (e == f) && (g == h) ) ;
    for((a == b) && (c == d) && (e == f) && (g == h) ;;) ;
    for(;(a == b) && (c == d) && (e == f) && (g == h) ;) ;
    for(;x=(a == b) && (c == d) && (e == f) && (g == h) ;) ;
    for(;;(a == b) && (c == d) && (e == f) && (g == h) ) ;
    do{}while((a == b) && (c == d) && (e == f) && (g == h) ) ;
    do{}while(x=(a == b) && (c == d) && (e == f) && (g == h) ) ;
}

// gives ==>

function a() {
    x = a == b && c == d && e == f && g == h;
    if (a == b && c == d && e == f && (g == h)) {
    }
    if (x = a == b && c == d && e == f && g == h) {
    }
    while (a == b && c == d && e == f && (g == h)) {
    }
    while (x = a == b && c == d && e == f && g == h) {
    }
    for (a == b && c == d && e == f && g == h;;) {
    }
    for (; a == b && c == d && e == f && (g == h);) {
    }
    for (; x = a == b && c == d && e == f && g == h;) {
    }
    for (;; a == b && c == d && e == f && g == h) {
    }
    do {
    } while (a == b && c == d && e == f && (g == h));
    do {
    } while (x = a == b && c == d && e == f && g == h); 
}

Comment 2

[Biju]

Thanks this is cool 

function a() {
    y=(1==(3&4))?((a)+(b)):(a-c);
    x = ((a == b) && ((c|(c+2)) == (d*d)))  && ((c == d)  && (c == d)) ;
    if (((a == b) && ((c|(c+2)) == (d*d)))  && ((c == d)  && (c == d)) ) {
        p();
    }

}

alert(a);

// strips down all unnecessary parenthesis 
// and gives ==>

function a() {
    y = 1 == (3 & 4) ? a + b : a - c;
    x = a == b && (c | c + 2) == d * d && c == d && c == d;
    if (a == b && (c | c + 2) == d * d && c == d && c == d) {
        p();
    }
}

Comment 3

[Brendan Eich [:brendan]]

Thanks to jorendorff for pn_parens.

/be