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Re: amd64: why is sizeof(int) =4? why not =8?

On Wednesday 05 April 2006 15:16, Mike McCarty wrote:

> This program has defects. The type of the results of
> the sizeof operator is an unsigned integer of unspecified
> size. The only portable way to do this is...
>      printf("Size of int is %lu\n",(unsigned long)sizeof(int));

Why do I have cast something from size_t? I see the construct I used in 
"C a reference Manual" by Harbisson and Steele p148.

I am not a c expert - and I dont understand why you do the cast.

Do you have an example of a compiler where the length of sizeof(int) is 
greater than size_t ? 

> > and the output is  no different from on my 32 bit system.
> As expected. Once the compiler is run, and the program
> created, the instructions are whatever they are. There is
> no run time size detection and re-writing of the program.
> C is not an interpreted language.
> You need to get a 64 bit compiler. The programs it creates will
> not run on a 32 bit machine, and likely not with a 32 bit OS
> even on your 64 bit machine.

I was (of course) using  gcc - a 64 bit compiler on the debian 64 bit system.

I compiled the same program on each system - one a 32 bit and one a 64   bit 
compiler/ kernel etc. 

The problem that I needed help with was -
as was pointed out by Hendrik and Ron 
is that 

1) we need to declare using int64 to get 64 bit integers and 
2) we always now have 64 bit addresses 
which i can test for with int * or int64*.



> Mike
> --
> p="p=%c%s%c;main(){printf(p,34,p,34);}";main(){printf(p,34,p,34);}
> This message made from 100% recycled bits.
> You have found the bank of Larn.
> I can explain it for you, but I can't understand it for you.
> I speak only for myself, and I am unanimous in that!

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