C ++指针解引用乘法

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 int s and long s 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 long s 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 int s, 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.