Subj : Re: Math error ??? doen't make sense!
To   : borland.public.cpp.borlandcpp
From : Stan DeGroff
Date : Fri Jul 02 2004 09:31 pm

Lets change the program to find out more...
------------------------------
#include<stdio.h>
#include<math.h>
main()
{
int x,y;
long int z;
x=7000;
y=12000;
z=x*y;
printf("x = %d, y = %d\n",x,y);
printf("sizes of x, y, z = %d, %d, %d\n",sizeof(x),sizeof(y),sizeof(z));
printf("x * y == (unsigned)%lu\n",z);
printf("x * y == (signed) %ld\n",z);
printf("x = 0x%08X\n",x);
printf("y = 0x%08X\n",y);
printf("z = 0x%08X\n",z);
return 0;
}
-------------------------------
Do the math on a hex calculator:

(1) if you set up the ide for a 16 bit task,
the above will report sizeof for x=2, y=2, z=4 bytes

(2) if you set format for Unsigned (even though z is signed
it will return 4924950144.

(3) if you set format as signed, it will return -17152
thats only changing the printf format.

NOW!

lets use a hex calculator.

x = 7000 = 0x1B58
y = 12000 = 0x2EE0
z = x * y = 0x0501 BD00 = 84,000,000 decimal

fact, always true:
a 16 bit x 16 bit will always fit in 32 bits. z size is 32 bits

x & y are initialized as 16 bit integers.
z is set up as a 32 bit integer.

There ain't no overflow. as shown in the z result above.

----------------------------------
If we set up the task to be a 32 bit task.
Sizeof x, y, & z are all 32 bits (4 byte)
correct answer is produced.

 OR

if we set
int x,y

 to

long int x,y
correct answer is produced.

conclusion :

Multiply in Borland product must have both inputs to a multiply
and the output all the same size variables.

interesting, since most assembly instructions do things like 16 x 16 = 32
and same for most PLC's (Programmable Logic COntrollers).

Is this a deviation from the ANSI C spec? ie: the "Sams Learning C in 24
hrs"
book uses 16 x 16 for a 32 bit result as I first brought it to the table.
(See my first post, as it came from the book.)

Thanks for your input.

Stan

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