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left shift count >= width of type

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Hi,
here's a simple program snippet that I'm dealing with and worried about
potential pitfalls:

#define BITS_PER_UNIT 32
[...]
int foo, bar;

bar = 0xAAAA5555;
foo = bar & ~ ((int) 1 << BITS_PER_UNIT);
[...]

Strictly speaking the shift operations should report a warning (left
shift count >= width of type), since 1 is of type int (correct me if I'm
wrong).
Since this snippet is part of a very old implementation of g21k compiler
for ADSP21xxx, I'm wondering why do we need to shift at all, isn't the
above foo assignment equal to the following:

foo = bar;

Hope someone can clarify.

Al

-- 
A: Because it fouls the order in which people normally read text.
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left shift count >= width of type Alessandro Basili <alessandro.basili@cern.ch> - 2011-11-02 10:55 +0100
  Re: left shift count >= width of type James Kuyper <jameskuyper@verizon.net> - 2011-11-02 07:34 -0400
    Re: left shift count >= width of type Keith Thompson <kst-u@mib.org> - 2011-11-02 13:47 -0700
    Re: left shift count >= width of type Alessandro Basili <alessandro.basili@cern.ch> - 2011-11-04 17:46 +0100
      Re: left shift count >= width of type James Kuyper <jameskuyper@verizon.net> - 2011-11-04 13:12 -0400
        Re: left shift count >= width of type Alessandro Basili <alessandro.basili@cern.ch> - 2011-11-05 19:17 +0100
  Re: left shift count >= width of type Thad Smith <ThadSmith@acm.org> - 2011-11-02 20:22 -0600

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