Saurabh Jain asked:
are you sure x&y == x&&y???
Am I missing something?
Pedro's reply:
Shyan Lam, thinking that I mistyped, meant that:
if ((x && y) == true)
may be shortened to:
if (x && y)
But my bug was to think that:
if (x&y)
is equivalent to:
if ((x&y)==true)
My opinion is that the ISO committee made a bad decision when it defined the
boolean type at 6.3.1.2 but defined the macro true to be the integer 1 at
7.16ยง3 . The macro true should be of type boolean, not an integer, and to
compare between an integer and a boolean without casting should generate a
warning.
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