Python is just as good as C++ for real apps

[Grant Edwards]

In article <d1645ug80n2mvcvtdkctujkruepotv5p08 at 4ax.com>, Courageous wrote:

>>IOW, *p behaves like any other int.
> 
> NO. It doesn't. For one thing, adding one to it actually adds
> 4 to it.

 (*p) + 1  adds one to it.

> More generally, what happens can be shown by the
> example program below. Pointers are _different_. 

I know pointers are different.  I never said they weren't.

I'm saying that when you declare

int  *p;

*p is an int.  
p is a pointer to an int.  

I never propsed that you think of p as an int.  I said you
should think of *p as an int.

> They should be cognited as their own complete type. They even
> effectively have operator overload. To wit:
> 
> #include <stdio.h>
> 
> void main()
> {
>     char    c=0,   *cp=0;
>     short   s=0,   *sp=0;
>     int     i=0,   *ip=0;
>     long    l=0,   *lp=0;
>     float   f=0.0, *fp=0;
> 
>     fprintf(stderr, "c = %d (size %d), cp =%d (size %d)\n",
> 		(int)c+1, sizeof(c), (int)(cp+=1), sizeof(cp));

What I'm talking about is that *cp is a char.  sizeof *cp is 1;

>     fprintf(stderr, "s = %d (size %d), sp =%d (size %d)\n",
> 		(int)s+1, sizeof(s), (int)(sp+=1), sizeof(sp));
>     fprintf(stderr, "i = %d (size %d), ip =%d (size %d)\n",
> 		(int)i+1, sizeof(i), (int)(ip+=1), sizeof(ip));
>     fprintf(stderr, "l = %d (size %d), lp =%d (size %d)\n",
> 		(int)l+1, sizeof(l), (int)(lp+=1), sizeof(lp));
>     fprintf(stderr, "f = %d (size %d), fp =%d (size %d)\n",
> 		(int)f+1, sizeof(f), (int)(fp+=1), sizeof(fp));
> }
> 
> % gcc test.C
> % a.exe
> c = 1 (size 1), cp =1 (size 4)
> s = 1 (size 2), sp =2 (size 4)
> i = 1 (size 4), ip =4 (size 4)
> l = 1 (size 4), lp =4 (size 4)
> f = 1 (size 4), fp =4 (size 4)
> 
> --------------

It appears we're talking past each other.


-- 
Grant Edwards                   grante             Yow!  It's OKAY --- I'm an
                                  at               INTELLECTUAL, too.
                               visi.com