c - bitwise: different behaviour with negative input value -

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i'm on hardware/software interface course hosted on coursera, i'm sure you've seen fair few of these questions..

one of assignments determine whether 2's complement integer x can fit n-bits, using subset of operators, etc. i'm doing well, came across 2 separate approaches, , hit whiteboard them both understand them fully. came confusion.

the function returns 1 if possible fit, 0 if not.

method 1 (correct output)

int foobar(int x, int n) {     return !(((x << (32-n)) >> (32-n)) ^x); }

executed positive integer

foobar(5,3)  == 0

correctly calculates 5 cannot represented 2's complement 3 bit integer.

executed negative integer

foobar(-4,3) == 1

correctly calculates -4 can represented 2's complement 3 bit integer.

method 1 analysis

x=5, n=3  00000000000000000000000000011101    32 - n == 29 10100000000000000000000000000000    x<<29 00000000000000000000000000000101    >> 29   00000000000000000000000000000101    5              xor 00000000000000000000000000000101    5 --------------------------------     00000000000000000000000000000000    0  00000000000000000000000000000001   !0 == answer, 1.

as can see, returns 0, in, no 5 may not represented 2's complement integer within 3 bits.

method 2 (incorrect output)

int foobar(int x, int n) {     return !((x & ~(1 << (32-n))) ^x); }

executed positive integer

foobar(5,3)  == 1

false positive.

executed negative integer

foobar(-4,3) == 0

false negative.

method 2 analysis

x=5, n=3  00000000000000000000000000011101    32 - n == 29  11011111111111111111111111111111    ~(1<<29)               , 00000000000000000000000000000101    5 --------------------------------     00000000000000000000000000000101    5   00000000000000000000000000000101    5               xor 00000000000000000000000000000101    5 -------------------------------- 00000000000000000000000000000000    0  00000000000000000000000000000001   !0 == answer, 1.

i compiling with:

gcc version 4.7.2 (debian 4.7.2-5)

question

i'm @ loss explain difference in output, when analysis shows identical @ bit level, hints/tips can shed light on myself highly appreciated.

thank time!

sc.

in method 1, write

10100000000000000000000000000000    x<<29 00000000000000000000000000000101    >> 29

but not correct (note analysis mean foobar(5,3) == 1).

first, result of 5 << 29 causes overflow signed 32-bit (or smaller) ints, undefined behaviour.

next, result, if shift creates indicated bit pattern (as does), negative.

right-shifting negative integers implementation-defined, common arithmetic right shift, sign-extend, , here result in

11111111111111111111111111111101    >> 29

which, when xor-ed 5 gives non-zero result (and applying ! produces 0).

method 2 doesn't work @ all, because, leaving aside undefined behaviour inputs, check whether (32-n)-th bit set.


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