Nice solution. Similarly you could #define:
void change()
{
#define change() i=10
}
On Feb 16, 12:55 pm, ankit sablok <ankit4...@gmail.com> wrote:
> nice solution
>
> On Feb 15, 11:22 pm, jagannath prasad das <jpdasi...@gmail.com> wrote:
>
>
>
>
>
>
>
> > void change()
> > {
> > #define i i=10,n}
>
> > this will do..
>
> > On Tue, Feb 15, 2011 at 11:33 PM, Rel Guzman Apaza <rgap...@gmail.com>wrote:
>
> > > Nothing... 10 in base 5 = 5 in base 10.
>
> > > void change(){
> > > printf(""); //...?
> > > }
>
> > > 2011/2/15 Don <dondod...@gmail.com>
>
> > > A semicolon is valid in the middle of a line in C or C++.
>
> > >> For instance, no one says that
>
> > >> for(i = 0; i < 10; ++i)
>
> > >> is three lines of code.
>
> > >> Don
>
> > >> On Feb 15, 11:31 am, jalaj jaiswal <jalaj.jaiswa...@gmail.com> wrote:
> > >> > after termination of semicolon , that will be considered a separate
> > >> > line
> > >> i
> > >> > guess
>
> > >> > On Tue, Feb 15, 2011 at 10:59 PM, Don <dondod...@gmail.com> wrote:
> > >> > > void change()
> > >> > > {
> > >> > > printf("10"); while(1) {}
> > >> > > }
>
> > >> > > On Feb 15, 10:17 am, Balaji S <balaji.ceg...@gmail.com> wrote:
> > >> > > > Insert only one line in the function change() so that the output of
> > >> the
> > >> > > > program is 10.
> > >> > > > You are not allowed to use exit(). You are not allowed to edit the
> > >> > > function
> > >> > > > main() or to
> > >> > > > pass the parameter to change()
>
> > >> > > > void change()
> > >> > > > {
> > >> > > > // Code here}
>
> > >> > > > int main()
> > >> > > > {
> > >> > > > int i=5;
> > >> > > > change();
> > >> > > > printf(“%d” ,i);
> > >> > > > return 0;
>
> > >> > > > }
>
> > >> > > --
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>
> > >> > --
> > >> > With Regards,
> > >> > *Jalaj Jaiswal* (+919019947895)
> > >> > Software developer, Cisco Systems
> > >> > B.Tech IIIT ALLAHABAD
>
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