我目前正在做一个作业分配,它将一个位值(0x0或0x1)插入到第n个位置。MIPS程序集没有正确移动
.macro insert_to_nth_bit($regD, $regS, $regT, $maskReg)
# regD: bit pattern in which to be inserted at nth position
# regS: position n (0-31)
# regT: the bit to insert (0x0 or 0x1)
# maskReg: temporary mask
# Let's say regD = 00000101
sllv $maskReg, $maskReg, $regS # Let's say regS = 2, the second position at regD(0)
# maskReg = 00000010 after shifting
not $maskReg, $maskReg # maskReg = 11111101
and $regD, $regD, $maskReg # regD = 00000101
sllv $regT, $regT, $regS # regT = 00000001, we want to insert 1 into the 2nd position at regD
# regT = 00000010 after shifting
or $regD, $regD, $regT # 00000101 OR 00000010 = 00000111. The bit is what i wanted
.end_macro
这是我写来测试它的宏观
.text
.globl main
la $t0, 0x00000101 #t0 = 00000101
la $t1, 2 # nth position = 2
la $t2, 0x1 # insert 0x1
la $t3, 1 # maskReg = 00000001
insert_to_nth_bit($t0, $t1, $t2, $t3)
print_int($t0)
exit
print_int
和exit
另外两个小宏
结果,我得到的是261,这是00000105之后,我转换为十六进制。当我调试它时,我注意到当谈到第一班时,00000001和左移2变成了00000005,这使整个事情变得糟糕。我不知道我的宏的逻辑是错误的还是我测试宏的方式是错误的,所以它搞砸了我的输出?
1左移两次是4,或者说到0x101你得到0x105。问题是什么? –
@SamiKuhmonen我想我只是想通了,而不是将2传入$ t1,我应该将4传入$ t1。我只是想知道这个宏的逻辑是正确的吗? – user21478621