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编写下面代码的另一种方法是什么？
li $t1, 1 # load 1 in register $t1 sll $t1, $t1, 23 # shift value of 1 to the 24th bit to get 1000 0000 0000 0000 0000 0000
我担心这段代码实际上并没有将 1 移到第 24 位。那么另一种写法是什么？
提前致谢。