代码之家 › 专栏 › 技术社区 › Vipoon

计算sin(x)的mips程序

mips assembly

Vipoon · 技术社区 · 9 年前

我试图通过在MIPS上创建一个程序来计算这一点。但是,它输出错误的大数字!我不知道我犯的愚蠢错误在哪里。我创建了三个函数,一个计算阶乘,另一个计算幂,第三个大函数是sin函数。

#read integer
li $v0, 5
syscall

addi $a0, $v0, 0 #argument of sin function, a0 = x
li $t4, 1 #n starting from 1
addi $s7, $v0, 0 #sum = x
jal sin
j end

sin:
    Loop:
    slti $t5, $t4, 5 #t0 < 5 (n < 5)
    beq $t5, 0, exitLoop
    addi $sp, $sp, -8 # adjust stack for 2 items
    sw $ra, 4($sp) # save return address
    sw $a0, 0($sp) # save argument
    li $a0, -1 #argument of pow function, number = -1
    addi $a1, $t4, 0 #argument of pow function, power = t4
    jal power
    addi $s0, $v0, 0 #s0 = v0 (return value of pow)
    lw $a0, 0($sp) # restore original x

    sll $a1, $t4, 1 #a1 = 2n 
    addi $a1, $a1, 1 #a1 = 2n+1
    jal power
    addi $s1, $v0, 0 #s0 = v0 (return value of pow)

    sll $a0, $t4, 1 #a0 = 2n
    addi $a0, $a0, 1 #a0 = 2n + 1
    jal factorial
    addi $s2, $v0, 0 #s0 = v0 (return value of pow)
    lw $a0, 0($sp) # restore original n
    lw $ra, 4($sp) # and return address

    mult $s0, $s1 # LO = (-1)^n * x^(2n+1)
    mflo $s3 # S3 = LO
    div $s3, $s2 # s3 / (2n+1!)
    mflo $s3
    add $s7, $s7, $s3 #sum = sum + s3
    addi $t4, $t4, 1 #n++

    j Loop  
    exitLoop:
addi $v1, $s7, 0
addi $sp, $sp, 8
jr $ra

power:
    addi $t0 $a0, 0 #t0 = a0
    li $t1, 0 #i = 0
    loop:
    slt $t3, $t1, $a1 #if i < n
    beq $t3, 0, exit 
    mult $t0, $a0 #t0 * a0
    mflo $t0 #LO = t0
    addi $t1, $t1, 1
    j loop
    exit:
    addi $v0, $t0, 0
    jr $ra

factorial:
    li $t0, 1
    bgt $a0, $t0, L1
    li $v0, 1
    jr $ra
    L1:
    addi $sp, $sp, -8 # adjust stack for 2 items
    sw $ra, 4($sp) # save return address
    sw $a0, 0($sp) # save argument
    addi $a0, $a0, -1 # decrement n
    jal factorial # recursive call
    lw $a0, 0($sp) # restore original n
    mul $v0, $a0, $v0 # multiply to get result
    lw $ra, 4($sp) # and return address
    addi $sp, $sp, 8 # pop 2 items from stack
    jr $ra # and return
end:
li $v0, 1
addi $a0, $v1, 0
syscall

1 回复 | 直到 9 年前

Craig Estey 9 年前

正如Jester提到的,您应该使用浮点运算。

您还可以利用序列中的每个幂项和每个阶乘项都是前一项的简单增量这一事实,从而大大简化事情。无需从头开始重新计算,也无需递归计算阶乘很更容易使用循环]。

根本不需要单独的功能。这是C中的函数。这是一个相对简单的asm翻译:

double
qsin(double x)
{
    double x2;
    double cur;
    int neg;
    double xpow;
    double n2m1;
    double nfac;
    int iters;
    double sum;

    // square of x
    x2 = x * x;

    // values for initial terms where n==0:
    xpow = x;
    n2m1 = 1.0;
    nfac = 1.0;
    neg = 1;

    sum = 0.0;

    // NOTES:
    // (1) with the setup above, we can just use the loop without any special
    //     casing
    // (2) this _will_ do an unnecessary calculation [that gets thrown away] on
    //     the last iteration, but it's a tradeoff for simplicity
    // (3) when translated to asm, it should be easy to restructure to avoid
    //     this (i.e. just harder to express in a loop here)
    for (iters = 5;  iters > 0;  --iters) {
        // calculate current value
        cur = xpow / nfac;

        // apply it to sum
        if (neg < 0)
            sum -= cur;
        else
            sum += cur;

        // now calculate intermediate values for _next_ sum term

        // get _next_ power term
        xpow *= x2;

        // go from factorial(2n+1) to factorial(2n+1+1)
        n2m1 += 1.0;
        nfac *= n2m1;

        // now get factorial(2n+1+1+1)
        n2m1 += 1.0;
        nfac *= n2m1;

        // flip sign
        neg = -neg;
    }

    return sum;
}