On Wed, Sep 19, 2012 at 3:59 PM, Randi Botse <nightdecoder@xxxxxxxxx> wrote:
> Hi Phil, Jon
>
> Thanks, now I'm clear with this, assignment doesn't care with type modifier.
>
> Code such as
>
> unsigned int j = 0xffeeddcc;
> int i = j;
>
> Both has the same value depending on how them interpreted (is this
> assumption correct?)
>
According to C99, when applying integer conversion, "if the new type
is signed and the value cannot be represented in it, either the result
is implementation-defined or an implementation-defined signal is
raised". But most implementation keeps the same memory representation.
> Because,
>
> printf("%u", i) will be different to printf("%i", i)
> - but -
> printf("%u", i) wlll be same as printf("%u", j)
>
>
> Actually why asking this because I often see a pointer to a char* cast
>
> Let me show you with this example.
> Consider some structures...
>
> struct a_data {
> unsigned char f1[4];
> unsigned char f2[6];
> unsigned short f3[2];
> };
>
> and another struct named b_data, c_data, etc.
>
> Then there is a general function to process all type of structure,
> maybe something like this:
>
> int process_data(char *buffer, size_t len);
>
> Then if we cast for example a pointer to a_data struct to a char* as follow:
>
> struct a_data a;
> process_data((char*) &a, sizeof(a));
>
> I though since it was cast to char*, the cast is "problem" because
> every signed char buffer will have a range CHAR_MIN to CHAR_MAX,
> therefore value of CHAR_MAX to UCHAR_MAX will broken (signed char
> overflow)
>
Actually, whether char is signed or unsigned is
implementation-defined, though, normally, it is signed. SCHAR_MAX+1 ~
UCHAR_MAX can be mapped to SCHAR_MIN ~ -1.
For a pointer that denotes a memory region, what type it points to
doesn't cause much problem as long as you don't simply dereference it.
In such cases, void * might be less confusing.
Regards,
Leon
> I think process_data() should be declared with
>
> int process_data(unsigned char *buffer, size_t len)
>
> this declaration in seem correct and work for me.
>
> However, now I'm conceptually understand why this works.
>
> Thanks.
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