Shift operator operations

int i=1;
i <<= 31;
i >>= 31;
i >>= 1;
can anyone explain how the result i value=-1
thanks

The high order bit of an integer is the sign. The ">>" operator maintains the sign. See the results of your shifts below which results in a value of -1. Use ">>>" if you do not want to propagate the sign.
i = 1      00000000000000000000000000000001
i <<= 31   10000000000000000000000000000000
i>>= 31    11111111111111111111111111111111
i>>=1      11111111111111111111111111111111

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    <input message="client:myServiceRegisterMessage"/>
    </operation>
    </portType>
    <portType name="myServiceCallback">
    <operation name="onResult">
    <input message="client:myServiceResponseMessage"/>
    </operation>
    <operation name="onComplete">
    <input message="client:myServiceCompleteMessage"/>
    </operation>
    <operation name="onError">
    <input message="client:myServiceErrorMessage"/>
    </operation>
    </portType>
    <plnk:partnerLinkType name="myService">
    <plnk:role name="myServiceProvider">
    <plnk:portType name="client:myService"/>
    </plnk:role>
    <plnk:role name="myServiceRequester">
    <plnk:portType name="client:myServiceCallback"/>
    </plnk:role>
    </plnk:partnerLinkType>
    Hope this helps. Otherwise send me an email and I'll send you my (working) example.
    Mike.van.Alst#AT#it-eye.nl (replace with @)

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