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From C traps and pitfalls
If a and b are two integer variables, known to be non-negative then to test whether
a+bmight overflow use:if ((int) ((unsigned) a + (unsigned) b) < 0 ) complain();
I didn't get that how comparing the sum of both integers with zero will let you know that there is an overflow?
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Write a “C” function, int addOvf(int* result, int a, int b) If there is no overflow, the function places the resultant = sum a+b in “result” and returns 0. Otherwise it returns -1. The solution of casting to long and adding to find detecting the overflow is not allowed.
So overflow can be detected by checking Most Significant Bit(MSB) of two operands and answer. But Instead of using 3-bit Comparator Overflow can also be detected using 2 Bit Comparator just by checking Carry-in(C-in) and Carry-Out(C-out) from MSB's. Consider N-Bit Addition of 2's Complement number.
An integer overflow or wraparound happens when an attempt is made to store a value that is too large for an integer type. The range of values that can be stored in an integer type is better represented as a circular number line that wraps around.
The code you saw for testing for overflow is just bogus.
For signed integers, you must test like this:
if (a^b < 0) overflow=0; /* opposite signs can't overflow */
else if (a>0) overflow=(b>INT_MAX-a);
else overflow=(b<INT_MIN-a);
Note that the cases can be simplified a lot if one of the two numbers is a constant.
For unsigned integers, you can test like this:
overflow = (a+b<a);
This is possible because unsigned arithmetic is defined to wrap, unlike signed arithmetic which invokes undefined behavior on overflow.
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