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CS版 - Any efficient way to compare two numbers like this...
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话题: number1话题: bits话题: two话题: number2话题: right
进入CS版参与讨论
1 (共1页)
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发帖数: 1
1
I want to compare two numbers
two bits by two bits from the right to left under the Intel platform
In detail,
if the first right two bits of number 1 > first right two bits of number 2,
then number1 > number2
else,
if the first right two bits of number 1 > first right two bits of number 2,
then number1 > number2
else, and so on
E.g., number1 = 0x001000
number2 = 0x110000
In general order, number1 But according to the above comparison rule, number1 should be bigger
1 (共1页)
进入CS版参与讨论
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话题: number1话题: bits话题: two话题: number2话题: right