CODEWITHSUNDEEP
cbuffer-overflowinteger-overflow
Mark Ezberg
Mark Ezberg

Reputation: 665

Sum of INT_MAX and INT_MAX

If I declare two max integers in C:

int a = INT_MAX;
int b = INT_MAX;

and sum them into the another int:

int c = a+b;

I know there is a buffer overflow but I am not sure how to handle it.

Upvotes: 2

Views: 1406

Answers (2)

Lundin
Lundin

Reputation: 213799

For the calculation to be meaningful, you would have to use a type large enough to hold the result. Apart from that, overflow is only a problem for signed int. If you use unsigned types, then you don't get undefined overflow, but well-defined wrap-around.

In this specific case the solution is trivial:

unsigned int c = (unsigned int)a + (unsigned int)b; // 4.29 bil

Otherwise, if you truly wish to know the signed equivalent of the raw binary value, you can do:

int c = (unsigned int)a + (unsigned int)b;

As long as the calculation is carried out on unsigned types there's no danger (and the value will fit in this case - it won't wrap-around). The result of the addition is implicitly converted through assignment to the signed type of the left operand of =. This conversion is implementation-defined, as in the result depends on signed format used. On 2's complement mainstream computers you will very likely get the value -2.

Upvotes: 1

Emoun
Emoun

Reputation: 2507

This causes undefined behavior since you are using signed integers (which cause undefined behavior if they overflow).

You will need to find a way to avoid the overflow, or if possible, switch to unsigned integers (which use wrapping overflow).

One possible solution is to switch to long integers such that no overflow occurs. Another possibility is checking for the overflow first:

if( (INT_MAX - a) > b) {
    // Will overflow, do something else
}

Note: I'm assume here you don't actually know the exact value of a and b.

Upvotes: 3

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