C++ – Handling Undefined Behavior When Intermediate Result of Expression Overflows

c++integer-overflow

This question is a result of another SO question.

Example Code

#include <iostream>

int main()
{
    unsigned long b = 35000000;
    int i = 100;
    int j = 30000000;
    unsigned long n = ( i * j ) / b; // #1
    unsigned long m = ( 100 * 30000000 ) / b; // #2
    std::cout << n << std::endl;
    std::cout << m << std::endl;
}

Output

85
85

Compiling this code with g++ -std=c++11 -Wall -pedantic -O0 -Wextra gives the following warning:

9:28: warning: integer overflow in expression [-Woverflow]

Questions

Am I correct in thinking that #1 and #2 invoke undefined behavior because the intermediate result 100 * 30000000 does not fit into an int? Or is the output I am seeing well-defined?
Why do I only get a warning with #2?

Best Answer

Yes, it is undefined behaviour, and the result you get is usually¹ different if unsigned long is a 64-bit type.

¹ It's UB, so there are no guarantees.

Related Solutions

C++ – Risks of Integer Overflow

As pointed out by @Xeo in the comments (I actually brought it up in the C++ chat first):
Undefined behavior really means it and it can hit you when you least expect it.

The best example of this is here: Why does integer overflow on x86 with GCC cause an infinite loop?

On x86, signed integer overflow is just a simple wrap-around. So normally, you'd expect the same thing to happen in C or C++. However, the compiler can intervene - and use undefined behavior as an opportunity to optimize.

In the example taken from that question:

#include <iostream>
using namespace std;

int main(){
    int i = 0x10000000;

    int c = 0;
    do{
        c++;
        i += i;
        cout << i << endl;
    }while (i > 0);

    cout << c << endl;
    return 0;
}

When compiled with GCC, GCC optimizes out the loop test and makes this an infinite loop.

C++ – Is Signed Integer Overflow Still Undefined Behavior?

is still overflow of these types an undefined behavior?

Yes. Per Paragraph 5/4 of the C++11 Standard (regarding any expression in general):

If during the evaluation of an expression, the result is not mathematically defined or not in the range of representable values for its type, the behavior is undefined. [...]

The fact that a two's complement representation is used for those signed types does not mean that arithmetic modulo 2^n is used when evaluating expressions of those types.

Concerning unsigned arithmetic, on the other hand, the Standard explicitly specifies that (Paragraph 3.9.1/4):

Unsigned integers, declared unsigned, shall obey the laws of arithmetic modulo 2^n where n is the number of bits in the value representation of that particular size of integer

This means that the result of an unsigned arithmetic operation is always "mathematically defined", and the result is always within the representable range; therefore, 5/4 does not apply. Footnote 46 explains this:

46) This implies that unsigned arithmetic does not overflow because a result that cannot be represented by the resulting unsigned integer type is reduced modulo the number that is one greater than the largest value that can be represented by the resulting unsigned integer type.

Best Answer

Related Solutions

C++ – Risks of Integer Overflow

C++ – Is Signed Integer Overflow Still Undefined Behavior?

Related Question