How do I detect unsigned integer multiply overflow?


I was writing a program in C++ to find all solutions of ab = c, where a, b and c together use all the digits 0-9 exactly once. The program looped over values of a and b, and it ran a digit-counting routine each time on a, b and ab to check if the digits condition was satisfied.

However, spurious solutions can be generated when ab overflows the integer limit. I ended up checking for this using code like:

unsigned long b, c, c_test;
c_test=c*b;         // Possible overflow
if (c_test/b != c) {/* There has been an overflow*/}
else c=c_test;      // No overflow

Is there a better way of testing for overflow? I know that some chips have an internal flag that is set when overflow occurs, but I've never seen it accessed through C or C++.

Beware that signed int overflow is undefined behaviour in C and C++, and thus you have to detect it without actually causing it. For signed int overflow before addition, see Detecting signed overflow in C/C++.

4/2/2019 10:48:09 PM

I see you're using unsigned integers. By definition, in C (don't know about C++), unsigned arithmetic does not overflow ... so, at least for C, your point is moot :)

With signed integers, once there has been overflow, Undefined Behaviour has occurred and your program can do anything (for example: render tests inconclusive). 

#include <limits.h>
int a = <something>;
int x = <something>;
a += x;              /* UB */
if (a < 0) {         /* unreliable test */
  /* ... */

To create a conforming program you need to test for overflow before generating said overflow. The method can be used with unsigned integers too

// for addition
#include <limits.h>
int a = <something>;
int x = <something>;
if ((x > 0) && (a > INT_MAX - x)) /* `a + x` would overflow */;
if ((x < 0) && (a < INT_MIN - x)) /* `a + x` would underflow */;

// for subtraction
#include <limits.h>
int a = <something>;
int x = <something>;
if ((x < 0) && (a > INT_MAX + x)) /* `a - x` would overflow */;
if ((x > 0) && (a < INT_MIN + x)) /* `a - x` would underflow */;

// for multiplication
#include <limits.h>
int a = <something>;
int x = <something>;
// there may be need to check for -1 for two's complement machines
// if one number is -1 and another is INT_MIN multiplying them we get abs(INT_MIN) which is 1 higher than INT_MAX
if ((a == -1) && (x == INT_MIN)) /* `a * x` can overflow */
if ((x == -1) && (a == INT_MIN)) /* `a * x` (or `a / x`) can overflow */
// general case
if (a > INT_MAX / x) /* `a * x` would overflow */;
if ((a < INT_MIN / x)) /* `a * x` would underflow */;

for division (except for the INT_MIN and -1 special case) there is no possibility of going over INT_MIN or INT_MAX.

5/7/2019 6:55:10 AM

Licensed under: CC-BY-SA with attribution
Not affiliated with: Stack Overflow