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Is there a short way to check/get for least significant bit in a 32-bit integer, in MIPS? It is obviously set for the odd numbers and an algorithm which checks for the whole number is odd or even can decide for this. But I just wonder is there a better way to do this...
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The value at the least significant bit position = x & 1. The value of the isolated least significant 1 = x & -x. The zero-based index of the isolated least significant 1 = log2(x & -x)
In a binary number, the bit furthest to the left is called the most significant bit (msb) and the bit furthest to the right is called the least significant bit (lsb). The MSB gives the sign of the number (sign bit) , 0 for positive and 1 for negative.
In computing, the least significant bit is the bit which is farthest to the right and holds the least value in a multi-bit binary number. As binary numbers are largely used in computing and other related areas, the least significant bit holds importance, especially when it comes to transmission of binary numbers.
2 Bit Positions, Types, and Defaults. Put the LSB (least significant bit) on the right side of the (horizontal) register map and the MSB (most significant bit) on the left side. This allows for easier writing of the hex values. Number the first bit in the register to be 0, not 1.
andi $t0, $s0, 1 will get the least significant bit of $s0 and put it in $t0.
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