C++ Integer Overflow – When Does It Become Undefined Behavior?

c++integer-overflowundefined-behavior

This is an example to illustrate my question which involves some much more complicated code that I can't post here.

#include <stdio.h>
int main()
{
    int a = 0;
    for (int i = 0; i < 3; i++)
    {
        printf("Hello\n");
        a = a + 1000000000;
    }
}

This program contains undefined behavior on my platform because a will overflow on the 3rd loop.

Does that make the whole program have undefined behavior, or only after the overflow actually happens? Could the compiler potentially work out that a will overflow so it can declare the whole loop undefined and not bother to run the printfs even though they all happen before the overflow?

_{(Tagged C and C++ even though are different because I'd be interested in answers for both languages if they are different.)}

Best Answer

If you're interested in a purely theoretical answer, the C++ standard allows undefined behaviour to "time travel":

[intro.execution]/5: A conforming implementation executing a well-formed program shall produce the same observable behavior as one of the possible executions of the corresponding instance of the abstract machine with the same program and the same input. However, if any such execution contains an undefined operation, this International Standard places no requirement on the implementation executing that program with that input (not even with regard to operations preceding the first undefined operation)

As such, if your program contains undefined behaviour, then the behaviour of your whole program is undefined.

Related Solutions

C Programming – Understanding Integer Overflow and Undefined Behavior

While the historical reason signed overflow was specified as undefined behavior was probably these bogus legacy representations (ones complement/sign-magnitude) and overflow interrupts, the modern reason for it to remain undefined behavior is optimization. As J-16 SDiZ hinted at, the fact that signed overflow is undefined behavior allows the compiler to optimize out some conditionals whose algebraic truth (but not necessarily representation-level truth) are already established by a previous branch. It may also allow the compiler to algebraically simplify some expressions (especially those involving multiplication or division) in ways that could give different results than the originally-written order of evaluation if a subexpression contains an overflow, since the compiler is allowed to assume that overflow does not happen with the operands you've given it.

The other huge example of undefined behavior for the purpose of permitting optimization is the aliasing rules.

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.

Best Answer

Related Solutions

C Programming – Understanding Integer Overflow and Undefined Behavior

C++ – Risks of Integer Overflow

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