Integer Overflow

March 18, 2022 · 1 min · Gray King

tags: C/C++

In some problems, we need to detect is our result overflow in a 32-bit integer. The key ideas is check our value before it becomes bigger.

For example:


// INT_MAX 2147483647
// INT_MIN -2147483648

// INT_MAX's suffix is 7
if (res > INT_MAX / 10 || (res == INT_MAX / 10 && pop > 7)) {
	return 0;
}
// INT_MIN's suffix is -8
if (res < INT_MIN / 10 || (res == INT_MIN / 10 && pop < -8)) {
	return 0;
}
res = res * 10 + pop;

Our final result need a 10 times current value and plus a value, then we check:

If our result greater than INT_MAX / 10, which means overflow.
Then if our result equals to INT_MAX / 10, we continue compare the plus value to the suffix of INT_MAX.

Links to this note

LeetCode101: 8. String to Integer (atoi)

tags: Tricky,LeetCode101,Integer Overflow The key idea is how to detect integer overflow, it’s same to: LeetCode101: 7. Reverse Integer. class Solution { public: int myAtoi(string s) { auto iter = s.begin(); int base = 1, r = 0, p = 0; // Skip whitespace for (; iter != s.end() && *iter == ' '; ++iter) { } // negative or positive if (*iter == '-' || *iter == '+') { if (*iter == '-') { base = -1; } ++iter; } for (; iter !...

LeetCode101: 7. Reverse Integer

tags: Tricky,Stack,LeetCode101,Integer Overflow The key is how to detect integer overflow without store to a larger size integer. For this purpose, we could detect integer overflow before carry: The maximal of INT_MAX before carry is \(\frac{INT\_MAX}{10}\). We continue compare pop with the suffix of INT_MAX, 7, if maximal before carry is equal to \(\frac{INT\_MAX}{10}\). The minimal of INT_MIN before carry is \(\frac{INT\_MIN}{10}\) too. We continue compare pop with the suffix of INT_MIN, -8, if minimal before carry is equal to \(\frac{INT\_MIN}{10}\)....