Re: [ECOS] Deciphering ISO C (Chap 6.3.2.3 - Pointers)

[Andrew Lunn <andrew@lunn.ch>]

> 	cyg_uint16* res_16;
> 	cyg_uint16* p_16;
> 	cyg_uint16 u_16;
> 
> 	p_16 = 0x0u;
> 	u_16 = 0x0u;'
> 	res_16 = (p_16 + 0x555u);
> 
> The above code yields res_16 == 0xaaa
> I was sort of hoping for 0x555 instead...


> 
> Altering the addition line to this:
> 	res_16 = (cyg_uint16*)(u_16 + 0x555);
> corrects the problem... 
> 
> why?

Pointer arithmetic. Take the following example bits of code....(which probably has many syntax errors etc)

char foo[4]="bar";
char foo1[4];

int  a[]={0,1,2,3};
int  a1[3];

for (i = 0; i < 3; i++) 
{
  char * pfoo = &foo;
  char * pa = & foo1;
  
  *(pfoo+i) = pfoo[i];
  *(pa+i) = a[i];
}

Its a contrived way of copying foo into foo1 and a into a1.

You want *(pfoo+i) to be byte by byte since you are copying characters, but 
you want *(pa+i) to be word by word since its copying ints.

In 
 	res_16 = (p_16 + 0x555u);

p_16 is of type cyg_uint16 *, so you are adding 0x555 16bit words, ie 0xaaa.

With 
        res_16 = (cyg_uint16*)(u_16 + 0x555);

u_16 is a plain cyg_uint16, so its not pointer arithmetic, its normal 
arithmetic which you then cast to a pointer afterwards.

     Andrew

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