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To detect int overflow/underflow in C, I use the below code.

What might be simpler and portable code?
(That is: fewer tests)

Assume 2's complement and don't use wider integers.

int a,b,sum;
sum = a + b;
// out-of-range only possible when the signs are the same.
if ((a < 0) == (b < 0)) {
  if (a < 0) {
    // Underflow here means the result is excessively negative.
    if (sum > b) UnderflowDetected();
  else {
    if (sum < b) OverflowDetected();  

Answer: Combining @Gareth Rees answer with some other ideas resulted in:

#include <limits.h>
int safe_add(int a, int b) {
  if (a >= 0) {
    if (b > INT_MAX - a) {
      ; /* handle overflow */
  } else {
    if (b < INT_MIN - a) {
      ; /* handle underflow */
  return a + b;

Note: Solution does not require 2's complement.
Note: This post does not address unsigned overflow.

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I think you're misunderstanding underflow ... or am I? Let's say, as an example, the smallest number a float can represent is 0.001. 1.0 / 10000 would result in a value of 0.0 because the actual value is too small. –  Fiddling Bits Dec 12 '13 at 0:44
@BitFiddlingCodeMonkey this is on integers - there are various wrap-around cases where the result of an addition does not fit in the same size integer. Sometimes it's called underflow when it's the sum of two negative numbers that doesn't fit. –  Michael Urman Dec 12 '13 at 1:01
If you just change from using int to using unsigned int, or better still, uint32_t and size_t, you'll be able to do those checks after the operation. For signed ints, overflow and underflow can't be detected after-the-fact because of undefined behaviour. And be warned: undefined behaviour can exhibit itself as anything from the program appearing to work properly right through to malware being installed on your machine and being used to steal your credit card information. –  Matt Dec 12 '13 at 12:10
@Matt Are you suggesting a method that would detect the out-of-range sum of 2 ints by employing unsigned conversion? OTOH if you are talking about overflow detection for 2 unsigned, that is a different question. if (sum < a) OverflowDetected(); seems to work for that. –  chux Dec 12 '13 at 15:46
@chux: No - I was merely pointing out that the method employed here (i.e. do an add and then see if an overflow occurred) is valid only on unsigned integers. For signed integers it is never valid because overflow of signed integers is inherently undefined in the language. –  Matt Dec 12 '13 at 15:52
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1 Answer

up vote 10 down vote accepted

Your code doesn't work: if the addition overflows then there is undefined behaviour here:

sum = a + b;

so your test is too late. You have to test for possible overflow before you do the addition. (If you're puzzled by this, read Dietz et al. (2012), "Understanding Integer Overflow in C/C++". Or even you're not puzzled: it's an excellent paper!)

If it were me, I'd do something like this:

#include <limits.h>

int safe_add(int a, int b) {
    if (a > 0 && b > INT_MAX - a) {
        /* handle overflow */
    } else if (a < 0 && b < INT_MIN - a) {
        /* handle underflow */
    return a + b;

but I'm not entirely sure what the point of having separate cases for overflow and underflow is.

I also use Clang's -fsanitize=undefined when building for test.

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Quite a lengthy paper, yet you make it seem so simple here. :-) What are INT_MAX and INT_MIN equal to? –  Jamal Dec 11 '13 at 23:17
§ in the C99 standard defines INT_MIN to be the "minimum value for an object of type int" and INT_MAX to be the "maximum value for an object of type int". –  Gareth Rees Dec 11 '13 at 23:20
Ah, I see. And those macros come from <limits.h>. –  Jamal Dec 11 '13 at 23:22
Yes, that's right. Typical values are INT_MIN = −2^32 = −2147483648 and INT_MAX = 2^32 − 1 = 2147483647, but the values can vary depending on your processor and compiler, so the only way to make code portable is to use the macros. –  Gareth Rees Dec 11 '13 at 23:27
@Jamal Actually, with <limits>. std::numeric_limits<int>::min() and std::numeric_limits<int>::max() to be precise. –  Yuushi Dec 12 '13 at 0:16
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