Given by a colleague as a puzzle, I cannot figure out how this C program actually compiles and runs. What is this >>>=
operator and the strange 1P1>
The line:
while( a[ 0xFULL?'\0':-1:>>>=a<:!!0X.1P1 ] )
contains the digraphs :>
and <:
, which translate to ]
and [
respectively, so it's equivalent to:
while( a[ 0xFULL?'\0':-1 ] >>= a[ !!0X.1P1 ] )
The literal 0xFULL
is the same as 0xF
(which is hex for 15
); the ULL
just specifies that it's an unsigned long long literal. In any case, as a boolean it's true, so 0xFULL ? '\0' : -1
evaluates to '\0'
, which is a character literal whose numerical value is simply 0
.
Meanwhile, 0X.1P1
is a hexadecimal floating point literal equal to 2/16 = 0.125. In any case, being non-zero, it's also true as a boolean, so negating it twice with !!
again produces 1
. Thus, the whole thing simplifies down to:
while( a[0] >>= a[1] )
The operator >>=
is a compound assignment that bit-shifts its left operand right by the number of bits given by the right operand, and returns the result. In this case, the right operand a[1]
always has the value 1
, so it's equivalent to:
while( a[0] >>= 1 )
or, equivalently:
while( a[0] /= 2 )
The initial value of a[0]
is 10. After shifting right once, it become 5, then (rounding down) 2, then 1 and finally 0, at which point the loop ends. Thus, the loop body gets executed three times.