Does && has higher precedence than || in if?

Recently I used:

if p==0&&q==0&&r==0||p==0&&q==0||q==0&&r==0||p==0&&r==0

if p>=0&&q>=0&&r>=0||p<=0&&q<=0&&r<=0

if a>0&&b>0&&c>0||d>0&&e>0&&f>0

if g>0&&h>0

whereas I actually mean:

if (p==0&&q==0&&r==0)||(p==0&&q==0)||(q==0&&r==0)||(p==0&&r==0)

if (p>=0&&q>=0&&r>=0)||(p<=0&&q<=0&&r<=0)

if (a>0&&b>0&&c>0)||(d>0&&e>0&&f>0)

if (g>0&&h>0)

The problem is I have done months of calculations without brackets, thingking && has higher precedence than . Now I just reflected: is it true? Was my code right?

Please tell me if my code was correct without brackets or not before I publish my calculation results. This is a serious matter to me if not; I'm really worried.

P.S. I'm aware of short circuits.

[Stephen23]

Why not just read the documentation?:

http://www.mathworks.com/help/matlab/matlab_prog/operator-precedence.html

[Guillaume]

It is also trivial to check for yourself:

0 && 0 || 1

If it returns 1, it means it's been evaluated as

(0 && 0) || 1

If it returns 0, it's been evaluated as

0 && (0 || 1)

[the cyclist]

This test does not distinguish between the following two rules:

• && has precedence over ||
• && and || have equal precedence, with left-to-right evaluation

Answer 1

[the cyclist]

According to this page about order precedence, && does indeed have precedence over ||.