@Wujin: Okay. b is but an integer, yet 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. The format that these are
printed in is a figment of the little-endianness of your computer. and
the order in which the compiler happens to allocate memory for a, b,
and c, and perhaps for the constants used to initialize them.
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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