1.1.4 Numerical Overflow

• Integer type has overflow problem, and float has precision problem. So prefer to use long long and double as your numerical type.
• In C and C++, you can use limits.h or <limit> to get the all the type limit information.
  INT_MAX //use in C 
  INT_MIN 
  //use in C++ 
  cout<< std::numeric_limits<int>::lowest() << ’\t’ 
  cout<< std::numeric_limits<int>::max() << ’\n’;
• For integer addition or subtraction, It just a round-trip. When you reach a position in the circle, how to understanding depends on its context and type.
  unsigned int ui = 1; 
  int i = -2; 
  // i+ui will stop in one position in the round clock. 
  //how to interpret it depends on the programe context; 
  int j = i+ui; // as int interpret this position CORRECT 
  (ui+i)<6 //  as unsigned interpret this position. ERROR!
• Judge it before calculation.
  if ((x > 0) && (a > INT_MAX - x)) /* ‘a + x‘ would overflow */; 
  // a is point, x>0 clockwise turn, then it will overflow 
   
  if ((x < 0) && (a < INT_MIN - x)) /* ‘a + x‘ would underflow */; 
  // a is point, x<0 anti-clockwise turn,  so it will underflow 
  //It’s easy to understand if you draw a clock figure. 
   
  if ((x < 0) && (a > INT_MAX + x)) /* ‘a - x‘ would overflow */; 
  if ((x > 0) && (a < INT_MIN + x)) /* ‘a - x‘ would underflow */; 
   
  if (a > INT_MAX / x) /* ‘a * x‘ would overflow */; 
  if ((a < INT_MIN / x)) /* ‘a * x‘ would underflow */; 
  // need to check for -1 for two’s complement machines 
  if ((a == -1) && (x == INT_MIN)) /* ‘a * x‘ overflow */ 
  if ((x == -1) && (a == INT_MIN)) /* ‘a * x‘ (or ‘a / x‘) overflow */
• Judge it after calculation
  uint32 a,b; 
  //assign values 
  uint32 result = a + b; 
  if (result < a) { 
      //Overflow 
  }
• Clang 3.4+ and GCC 5+ offer checked arithmetic builtins. They offer a very fast solution to this problem, especially when compared to bit-testing safety checks.
  unsigned long b, c, c_test; 
  if (__builtin_umull_overflow(b, c, &c_test)){ 
      // returned non-zero: there has been an overflow 
  }
• When you do some calculation, you can have some tricks to avoid overflow. n! the last three digit. You need use mod to keep last two digits in each calculation.
  (a+b)/2  //a+b maybe overflow 
  a/2+b/2 +(a&b&1);