Pointer/struct syntax 1

[Guest42]

Quick syntax question

If I have some field **go
and I say *go = malloc(sizeof(whatever))
If go is a field in struct v, how would I achieve the same thing? For example, "v->go = malloc" would be the same thing as go = malloc, but I need *go = malloc.

 

[xwb]

*(v->go) = malloc(sizeof(whatever))

 

[DaveSinkula]

http://faq.cprogramming.com/cgi-bin/smartfaq.cgi?answer=1047673478&id=1043284351

p = malloc ( n * [blue]sizeof *p[/blue] );

 

[cpjust]

Whoa! What the hell??
I tried this code, which I assumed would blow up since I'm de-referencing a NULL pointer, but it worked normally. Why isn't it blowing up?

#include <stdlib.h>

int main( void )
{
   char* p = NULL;
   int size = sizeof( *p );  /* Why no boom here? */

   p = malloc( 5 * size );
   return 0;
}

 

[ArkM]

Because of sizeof operands never elaborated, only the expression type determined. So it's a correct construct (no pointer dereferencing).

 

[dEVooXiAm]

In addition to ArkM's answer, I would like to say that sizeof is calculated at compile time, so your linked application would appear like

int size = 1;

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When you do it, do it right.