shell bit shifting producing different numbers

Question

in my local machines bash when I run:

echo $((192 << 24))
3221225472

but on my embedded targets busy box SHELL I get something else:

echo $((192 << 24))
-1073741824

it works when I left shift by a smaller number though. The embedded device is 64bit, where my local host is 32 bit.

Just to be clear, on the 32bit machine, the value is positive, on the 64 bit machine it's negative.

EDIT: This is on the embedded device which is 64 bit machine with SHELL. It doesn't happen when left shifting by 23.

echo $((192 << 23))
1610612736
echo $((192 << 24))
-1073741824

On the local host, which is a 32 machine with BASH:

echo $((192 << 55))
6917529027641081856
echo $((192 << 56))
-4611686018427387904

Answer 1

The binary representation of 192 is 11000000. When you shift it left 24 places, the only two bits which are set are the two most significant bits - the representation is 11000000 00000000 00000000 00000000. When a 32 bit system sees the most significant bit set, it interprets it as a negative number in "two's complement" format. For a 64 bit system, the most significant bit is still zero, so it is interpreted as a positive number.

This is simply an integer overflow on the 32 bit machine. You could expect the same behavior in C or any other language when using 32 vs 64 bit signed integer types.

Answer 2

POSIX (here) says "Only signed long integer arithmetic is required", and in C a long is at least 32 bits; that being said, some shells explicitly choose a fixed width, eg mksh uses 32 bits arithmetic, and peeking at the busybox' source (math.h) it seems like they only use 64 bits is ENABLE_SH_MATH_SUPPORT_64 is #define'd, regardless of whether the underlying system is 32/64 bits. If anyone knows better, speak up!