Doubt cast with pointer

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 while( ++idx <= fp_size)
 {  
     byte current = buff[idx];
     int integer = 0 ; 
     short shortint = 0 ; 

     if(idx < fp_size - 3)
         integer = *((int *))&buff[idx];
 }

What kind of cast is this *((int *))? variable integer is the type int is buff unsigned char.

c variables

1 answer

6


The cast is just the (int *). The rest is reference and reference.

What happens on that line is this:

First, the address of an element of the vector is taken buff: &buff[idx]. If buff is the type unsigned char, will be returned a unsigned char *.

Once this is done, a cast of unsigned char* for int*: (int *).

Finally, a melt is made in *, returning a int.

The variable int will contain the following values in your bytes:

buff[idx]
buff[idx + 1]
buff[idx + 2]
buff[idx + 3]

That if the endianness of your machine for little endiann. If it is big endiann, will be the reverse order.

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