汇编 IF 语句，多个条件触发 [重复]

Question

我正在为一项任务开发“Craps”（一种骰子游戏），但我对代码中的一件事感到困惑。我试图弄清楚为什么当 beq if 语句触发时，win 提示和 lost 提示都会触发。在掷骰子的情况下，如果滚动玩家掷出 7 或 11，他们就赢了，并且不知何故，获胜条件也会触发代码底部的失败条件提示。

感谢您的帮助

.text
main:

li $v0, 4                               # Load system call code 4 into register $v0; System call code 4 represents print_string.
la $a0, heading                         # Load address of string to print.
syscall                                 # Perform system call.
        
# Prompt for the integer to enter
li $v0, 4
la $a0, prompt
syscall

# Read the integer and save it in $s0
li $v0, 5
syscall
move $s0, $v0

# Output the text
li $v0, 4
la $a0, output
syscall

# Output the value
li $v0, 1
move $a0, $s0
syscall





li $t0, 3 # t0 is a constant 10
li $t5, 0 # t1 is our counter (i)
loop:
beq $t5, $s0, end # if t1 == 10 we are done





##################body of loop
    # Output the text
li $v0, 4
la $a0, rng
syscall


.text
    li $a1, 6  #Here you set $a1 to the max bound.
    li $v0, 42  #generates the random number.
    syscall

    add $a0, $a0, 1  #Here you add the lowest bound
    li $v0, 1   #1 print integer

    syscall
   
    move $t1, $a0
   
       # Output the text
li $v0, 4
la $a0, registerMoved
syscall

# Output the value
li $v0, 1
move $a0, $t1
syscall

    # Output the text
li $v0, 4
la $a0, rng
syscall

#generation of second random number
    li $a1, 6  #Here you set $a1 to the max bound.
    li $v0, 42  #generates the random number.
    syscall

    add $a0, $a0, 1  #Here you add the lowest bound
    move $t2, $a0

    syscall
   
# Output the value
li $v0, 1
move $a0, $t2
syscall
   
       # Output the text
li $v0, 4
la $a0, registerMoved
syscall

# Output the value
li $v0, 1
move $a0, $t2
syscall

add $t3, $t1, $t2

       # Output the text
li $v0, 4
la $a0, combined
syscall

li $v0, 1
move $a0, $t3
syscall

li $t1, 7

 beq $t1, $a0, WIN  # if(input==7 output win)
li $t2, 11






###### end of body of loop



addi $t5, $t5, 1 # add 1 to t1
j loop # jump back to the top
end:


 # Exit the program
li $v0, 10
syscall

.text

WIN: 

   # Output the text
li $v0, 4
la $a0, winString
syscall


END_WIN:

.text
LOSE: 
   # Output the text
li $v0, 4
la $a0, loseString
syscall


END_LOSE:
    
.data
heading: .asciiz "Welcome to the game of craps\n"
registerMoved: .asciiz "\nThe register was copied to another address: "
prompt: .asciiz "Enter a number of rolls: "
die_roll: .asciiz "The dice roll is: "
win: .asciiz "You have won"
loss: .asciiz "You have lost"
output: .asciiz "\n You entered for the number of rolls :"
rng: .asciiz "\nThe number generated was: "
combined: .asciiz "\nThe combined value is: "
winString: .asciiz "The rolling player wins "
loseString: .asciiz "The rolling player loses "```

Answer 1

标签不会改变执行流程。插入一个名为END_WIN: 的标签绝不会导致函数或块“结束”；它只是将当前地址与一个从未使用过的名为END_WIN 的符号相关联。在WIN 之后的系统调用之后，执行落入LOSE，我怀疑这不是你想要的。

如果你想执行到别的地方，你需要显式地写一个跳转指令。