汇编MIPS如何逐位检查该值？

发布于 2025-01-13 20:47:50 字数 833 浏览 1 评论 0原文

我正在创建一个程序，供用户输入一个整数，然后检查每个位并计算其二进制值中有多少个 1。因此，如果我输入 4673，我应该得到“4”作为输出，因为有 4 个。这是我下面的代码，由于某种原因我只得到“0”作为输出。我的猜测是我没有正确加载“andi”和“srl”的每一位。我一步步检查，当涉及到 andi 和 srl $t0 时，它的值永远不会为 1，所以我不能一点一点地移动？

 .data
Msg: .asciiz "Enter an integer: "



.text
 # Print the first message
 li $v0, 4
 la $a0, Msg
 syscall

 # Prompt the user to enter the first integer
 li $v0, 5
 syscall

 # Store the first integer in $t0
 move $t0, $v0

 addi $t3, $zero, 1

main:
 bgt $t3, 31, exit
 addi $t3, $t3, 1
 andi $t0, $v0, 1
 srl $t0, $t0, 1
 bne $t0, $zero, count 
 j main


count:
 addi, $t1, $t1, 1
 # Shift to the next bit and then go back to main
 j main
 
exit:

# Tell the interpreter to get read to print an integer
 li $v0, 1
 add $a0, $zero, $t1
 
 #Print the integer
 syscall
 
 # End the program
 li $v0, 10
 syscall

原文

I am creating a program for a user to enter an integer and then check each bit and count how many 1's is in it's binary value. So if I input 4673 I should get "4" as an output because there is 4 ones. This is the code I have below, for some reason I am only getting "0" as an output. My guess is I am not properly loading each bit with the "andi" and "srl". I check it step by step and when it comes to andi and srl $t0 never holds a value of 1, so I must not be shifting bit by bit?

 .data
Msg: .asciiz "Enter an integer: "



.text
 # Print the first message
 li $v0, 4
 la $a0, Msg
 syscall

 # Prompt the user to enter the first integer
 li $v0, 5
 syscall

 # Store the first integer in $t0
 move $t0, $v0

 addi $t3, $zero, 1

main:
 bgt $t3, 31, exit
 addi $t3, $t3, 1
 andi $t0, $v0, 1
 srl $t0, $t0, 1
 bne $t0, $zero, count 
 j main


count:
 addi, $t1, $t1, 1
 # Shift to the next bit and then go back to main
 j main
 
exit:

# Tell the interpreter to get read to print an integer
 li $v0, 1
 add $a0, $zero, $t1
 
 #Print the integer
 syscall
 
 # End the program
 li $v0, 10
 syscall

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黎歌 2025-01-20 20:47:50

您的循环中有这条指令 andi $t0, $v0, 1 。但是 $v0 在循环内永远不会改变，所以你总是得到相同的值。无论该值是 0 还是 1，下一行的 srl 之后都将是 0。

整个位计数循环可以用这样的东西代替：

li $t1, 0  # number of 1-bits
count_ones:
    andi $t2, $t0, 1            # t2 = input & 1
    addu $t1, $t1, $t2          # count += (input & 1)
    srl $t0, $t0, 1             # input >>= 1
    bne $t0, $zero, count_ones  # loop until no 1-bits left

请注意，有更有效的方法可以做到这一点，而根本不需要任何循环。请参阅如何计算数量32 位整数中的设置位？

You've got this instruction andi $t0, $v0, 1 in your loop. But $v0 never changes within the loop, so you're always getting the same value. And regardless of whether that values was 0 or 1, it's going to be 0 after the srl on the following line.

The whole bit-counting loop could be replaced by something like this:

li $t1, 0  # number of 1-bits
count_ones:
    andi $t2, $t0, 1            # t2 = input & 1
    addu $t1, $t1, $t2          # count += (input & 1)
    srl $t0, $t0, 1             # input >>= 1
    bne $t0, $zero, count_ones  # loop until no 1-bits left

Note that there are more efficient ways of doing this, without any loops at all. See How to count the number of set bits in a 32-bit integer?

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