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

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Iila
Iila 2016-1-15
评论： the cyclist 2016-1-15
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.
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Guillaume
Guillaume 2016-1-15
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
the cyclist 2016-1-15
This test does not distinguish between the following two rules:
  • && has precedence over ||
  • && and || have equal precedence, with left-to-right evaluation

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the cyclist
the cyclist 2016-1-15
According to this page about order precedence, && does indeed have precedence over ||.

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