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Nov 3rd, 2009

32 bits/64 bits case

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Hi All,
I have the application which will be support 32 bits and 64 bits
and below are the statement to handle 32bits/64bits data
and new_node->data
is void * new_node->data
for ( index = 0 ; index < items ;index++ ) {
if (data_size == SIZE_32 ) {
* ( ((int32_t*)(new_node->data)) + index) = (int32_t) index*2 ;
* ( ((int32_t*)(new_node-> data)) + index) = (int32_t) index*3
}
else {
* ( ((int64_t*)(new_node-> data)) + index) = (int64_t) index*4;
* ( ((int64_t*)(new_node-> data)) + index) = (int64_t) index*5;
}
}
Is there any better to handle this without the if/else statement (if ((data_size == SIZE_32 ) {
Thanks
J

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jodie121997

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Nov 3rd, 2009

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Re: 32 bits/64 bits case

I would use a typedef

C Syntax (Toggle Plain Text)
#ifdef SIZE_32
   typedef __int32 inttype;
#else
   typedef __int64 inttype;
#endif
 
int main()
{
    inttype a = 0;
}
#ifdef SIZE_32
   typedef __int32 inttype;
#else
   typedef __int64 inttype;
#endif

int main()
{
    inttype a = 0;
}

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Nov 3rd, 2009

Re: 32 bits/64 bits case

(Toggle Plain Text)

  if (data_size == SIZE_32 ) {
      * ( ((int32_t*)(new_node->data)) + index) = (int32_t) index*2 ; 
      * ( ((int32_t*)(new_node-> data)) + index) = (int32_t) index*3;
    }
   else {
       * ( ((int64_t*)(new_node-> data)) + index) = (int64_t) index*4; 
      * ( ((int64_t*)(new_node-> data)) + index) = (int64_t) index*5;
}

What's the purpose of this assignment :
* ( ((int32_t*)(new_node->data)) + index) = (int32_t) index*2 ; if in the next expression it gets changed right away?
* ( ((int32_t*)(new_node-> data)) + index) = (int32_t) index*3; Same in the else

Aia

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Nearly a Posting Maven

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Nov 3rd, 2009

Re: 32 bits/64 bits case

The purpose of these statement to cast the void point to either 32 or 64bits data type.
Thanks
J

jodie121997

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Nov 3rd, 2009

Re: 32 bits/64 bits case

Click to Expand / Collapse Quote originally posted by jodie121997 ...

The purpose of these statement to cast the void point to either 32 or 64bits data type.
Thanks
J

It was intended to be a hint, rather that a question to answer.

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