C++ Pointer Dereference Multiplication - c++

Question

I am currently programming an arduino and am using C++ objects to do so. I've run into a weird issue when I try to multiply the values that are being pointed at. Referring to the code below, when I run the program, var3 and var4 end up having two different values. Why is this? They are essentially multiplying the same values (or so I believe). Any help?
long var1 = info->accelXYZ[0];
long var2 = info->taughtAccelXYZ[0];
long var3 = var1*var2;
long var4 = info->accelXYZ[0]*info->taughtAccelXYZ[0];

Answer 1

It's possible you're overflowing in one of the situations.
The multiplication of var1 and var2 (both long) gives a long which is then loaded into var3.
If both info->accelXYZ[0] and info->taughtAccelXYZ[0] are int (for example), the result of the multiplication will be int which is then loaded into a long.
The intermediate int form may be overflowing, something you can see in the following snippet:
#include <stdio.h>
#include <limits.h>
int main(void) {
printf("int has %d bytes\n",sizeof(int));
printf("long has %d bytes\n",sizeof(long));
int a = INT_MAX;
int b = 2;
long var1 = a;
long var2 = b;
long var3 = a * b;
long var4 = var1 * var2;
printf ("var3=%ld\n", var3);
printf ("var4=%ld\n", var4);
return 0;
}
which outputs:
int has 4 bytes
long has 8 bytes
var3=-2
var4=4294967294

Answer 2

One reason why var3 may end up with a different value than var4 is integer overflow. This happens when both multiplicands fit in an int, but the product doesn't.
Since ints and longs have different sizes on Arduino Uno*, the computation of var3 is different from computation of var4.
When you compute var3, the multiplication is done in longs on the initial values that fit in an int, so the result of multiplication is not translated. When you compute var4, the computation is done in ints, and then promoted to long. However, by then the result is already truncated, which results in the discrepancy that you are observing.
To make var4 the same correct value as var3, add a cast to long to one of the multiplicands, like this:
long var4 = (info->accelXYZ[0])*((long)info->taughtAccelXYZ[0]);
* int has 16 bits, while long has 32 bits.