You do not need to compare operators and loop control statements to determine the size of A and B in int type.

When I saw this question, I first thought of A-B and then determined whether the highest sign bit is 1 or 0 .. I think everyone will think of this:

Int max (int A, int B) {int MAX [2] = {a, B}; return MAX [(a-B) & 0x80000000)> 31]; // operation. set all the 31 digits after the highest bit to 0, and remove the 31 digits from the right to remove the symbol}

In general, the above Code can produce correct results, but when a is the minimum value of the int type and B is the maximum value of the int type, the subtraction will be out of bounds, and the symbol bit is lost. In fact, there is no such extreme example, because the number of two different symbols is easy to cross-border. So on the basis of this Code, I first judge the symbol bits of A and B, and increase the limit (the symbol is the same, it is reduced, so it is safe, no problem; if the symbol is the opposite, if A is a positive number, B is a negative number, and the maximum or minimum value is obtained directly), so the following code is available:

# Include "stdio. H "# include" stdlib. H "int min (int A, int B) // first judge the symbol bit. When the symbol is the same, the two numbers subtract {return (A> 31 = 0) & (B> 31 = 1) | (A> 31) ^ (B> 31) = 0) & (a-B)> 31) = 0 ))? B: A; // printf ("% s \ n", (A> 31 = 0) & (B> 31 = 1 )) | (A> 31) ^ (B> 31) = 0) & (a-B)> 31) = 0 ))? "A> = B": "A <B");} int main (void) {int A = 0x80000000; // int type minimum int B = 0x7fffffff; // maximum value of int type printf ("% d \ n", min (A, B); System ("pause"); Return 0 ;}