@above 2 ... nice explanation both of you
On Mon, Jun 21, 2010 at 10:31 AM, manisha nandal wrote:
> int main()
> {
> int a=7,b=2;
> char* p=(char*)a;
>
> printf("%p",p); //prints 00 00 00 07
> printf("\n%p",p+2); //prints 00 00 00 09
>
> int s=(int)&p[b];//converted to &( *(p
int main()
{
int a=7,b=2;
char* p=(char*)a;
printf("%p",p); //prints 00 00 00 07
printf("\n%p",p+2); //prints 00 00 00 09
int s=(int)&p[b];//converted to &( *(p+b)) i.e p+b same as above
printf("\n%d",s); // 9
}
hope its clear now
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p=(char *)a;
this statement assigns value of 7 to p after type converting to char*,
so as p is a pointer it points to memory location 7
p[b] will give to the value at location 9 same as *(p+b) = *(7 + 2)
just like array
&p[b] will give the address of that location i.e. 9 which is char*
so we type
This can be answer for addition for the question 2 numbers without
using + operator.
Interesting fact is that it works for -ve numbers also.actually it get
converted to unsigned and type converted back again to -ve number.
#include
int main()
{
int a=-7,b=-2,s;
char *p;
p=(char *)a;
s= (int)&p[b]
@ above all
i dint get how the value of a and b gets added in s... after p=(char *)a...
p has 7 in it... but how the value gets added
thru s=(int)&p[b] p[b] seems as if p is an array.. m confused in
this line .. please explain in a bit detail.. :O
On Sun, Jun 20, 2010 at 4:44 PM, anan
thanks
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@anand:he takes the value at particular byte by using char*...same u do for
chcking endianess rite...same concept
On Sun, Jun 20, 2010 at 3:39 PM, sharad kumar wrote:
> @anand:its similar to reference an array element using pointer...*(p+i)
> which is imilar to &p[i] when u put *before & the valu
@anand:its similar to reference an array element using pointer...*(p+i)
which is imilar to &p[i] when u put *before & the value gets dereferenced
and hence p+i takes place same concept
On Sun, Jun 20, 2010 at 3:27 PM, anand wrote:
> it caculate the sum of two numbers.
>
> #include
> int main()
>
it caculate the sum of two numbers.
#include
int main()
{
int a=7,b=2,s;
char *p;
p=(char *)a;
s= (int)&p[b]; //adding a & b//HOW
printf("sum=%d\n",s);
return 0;
}
can any one explain the code..??
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