@Wujin: Okay. But b is an integer and it is being printed as a long
long. Since there are no implicit type conversions in procedure calls
such as the printf statement, b is taken as a long long, not of an
int. So b and the next 32-bits are printed.
Dave
On Sep 19, 8:18 pm, wujin chen <wujinchen...@gmail.com> wrote:
> @Dave printf("a=%x, b=%llx",a,b,c); i think c will be ignored~~ , and the
> output is a=9,b=10
>
> 2011/9/19 Dave <dave_and_da...@juno.com>
>
>
>
> > @Wujin: What do you expect the output to be? How does it differ from
> > what you actually get?
>
> > Dave
>
> > On Sep 18, 8:47 am, wujin chen <wujinchen...@gmail.com> wrote:
> > > usigned long long x = 0x12345678;
> > > int a = 0x09;
> > > int b = 0x10;
> > > printf("a=%x, b=%llx",a,b,c);
>
> > > the result is: a=9,b=1234567800000010
>
> > > i wonder why~~
>
> > > can anyone explain it?
> > > thanks.
>
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