print_array( cout, ai, 5);
in this code, does ai get changed to p1 in all cases
Robert Ryan <[EMAIL PROTECTED]> wrote: original
code to be used as prototype
void print_array(ostream& os, int* ai, int n)
{
for(int i=0; i<n; ++i)
os << ai[i];
} const int max = 5;
int i, n = 0;
int f(int n)
{
int *a=new int[5];
int ai[5];
int* p = &ai[5];
ai[0]=1, ai[1]=2, ai[2]=3, ai[3]=4, ai[4]=5;
print_array(cout, ai, 5);
return 0;
}
int main( )
{
f(0);
}
Robert Ryan <[EMAIL PROTECTED]> wrote: the
problem states: allocate an int in C++, initialize it to 7 and assign it to a
variable p1
print out the value of p1 and of the ints it
points to
Jacob Lund Fisker <[EMAIL PROTECTED]> wrote:
Not sure what you intend to do.
"new" allocates
"=" assigns
"int,double,float, ..." initializes.
So for instance ... int *i = new int(5);
new allocates space for an int on the heap, assigns the value 5 to that
space and returns the address which is assigned to a pointer called i
that is initialized to that address. Phew!
On Nov 13, 2007, at 6:41 PM, Robert Ryan wrote:
> allocate an int: int i =0;
> initialize an int: int i = 5;
> assign it to variable p1: p1 = 5;
>
> or is it this:
>
> int i =new int[ ];
> int p1[5];
> int* p1 = &p1[5];
>
> is this right
> bob ryan
>
>
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