0

好吧只是为了测试我有这个代码

    .data
    # This shows you can use a .word and directly encode the value in hex
    # if you so choose
num1:   .word 0x3F800000
num2:   .float 1234.567


num3: .float 45.67834
num4: .float 0.0004
result: .word 0
string: .asciiz "\n"

    .text
main:
    la $t0, num1
    lwc1 $f2, 4($t0)
    lwc1 $f4, 8($t0)
    lwc1 $f6, 12($t0)


    # Print out the values of the summands

    li $v0, 2
    mov.s $f12, $f2
    syscall

    li $v0, 4
    la $a0, string
    syscall

    li $v0, 2
    mov.s $f12, $f4
    syscall

    li $v0, 4
    la $a0, string
    syscall
    li $v0, 4
    la $a0, string
    syscall

    # Do the actual addition

    add.s $f12, $f2, $f6
    add.s $f12, $f12, $f4


    # Transfer the value from the floating point reg to the integer reg

    swc1 $f12, 8($t0)
    lw $s0, 8($t0)

    # At this point, $f12 holds the sum, and $s0 holds the sum which can
    # be read in hexadecimal

    li $v0, 2
    syscall
    li $v0, 4
    la $a0, string
    syscall

    # This jr crashes MARS
    # jr $ra

我有这个

    add.s $f12, $f2, $f6
    add.s $f12, $f12, $f4

我试着交换命令说

    add.s $f12, $f4, $f6
    add.s $f12, $f12, $f2

但结果是一样的

我检查了维基百科以获取不同的浮点加法示例，但这总是 1280.2457

http://en.wikipedia.org/wiki/Floating_point

他们发生了这种情况：

 a = 1234.567, b = 45.67834, c = 0.0004
 (a + b) + c:
     1234.567   (a)
   +   45.67834 (b)
   ____________
     1280.24534   rounds to   1280.245
    1280.245  (a + b)
   +   0.0004 (c)
   ____________
    1280.2454   rounds to   1280.245  <--- (a + b) + c
 a + (b + c):
   45.67834 (b)
 +  0.0004  (c)
 ____________
   45.67874
   1234.567   (a)
 +   45.67874 (b + c)
 ____________
   1280.24574   rounds to   1280.246 <--- a + (b + c)

对我来说并没有发生，它只适用于我尝试的任何值

4

1 回答 1

0

纳米，我明白了

a=0.00004
b=45.67840
c=1234.567
于 2013-04-08T21:12:50.090 回答