是什么导致此代码输出多行（mips程序集）？

时间：2016-03-27 23:17:03

标签： algorithm assembly mips bresenham

注意：请不要低估这个新手的问​​题以帮助学习。对于那些要求帮助被贬低的人来说，这是相当令人沮丧的。

您好我一直在尝试在mips程序集中实现bresenham的算法，但是我使用的是不完整的算法，即不包括错误等。我偶然发现了使用该算法的某人mips汇编代码但是我不明白代码中的什么使程序输出多行。 main_for是导致代码输出多行的部分原因吗？

.data  
bitmap :    .word 0x10040000
bitmap_x_max :  .word 256 # width in pixels
bitmap_y_max :  .word 256 # heigh in pixels
s_zero :    .float 0.0
s_meio :    .float 0.5
s_neg_one :     .float -1.0
w_one:      .word 1
w_neg_one:  .word -1
w_two:      .word 2
w_68:       .word 256
.text

#valores fixos em cada chamda, values fixed 
li   $a0, 32    
li   $a1, 32
addi $sp, $sp, -4
li $t0, 0x000FFFFF 
sw $t0, 0($sp)# coloca cor na pilha, saves colour to memory


move $t0, $zero #i = 0
lw   $s0, w_68 #final do loop
main_for:
    slt $t1, $t0, $s0 #atual < máximo?  
    beqz $t1, main_for_end

    li   $a0, 128
    li   $a1, 128
    move $a2, $t0
    li   $a3, 0

    jal desenhaLinha_bom




    addi $t0, $t0, 32
    j main_for

main_for_end:   

li $t0, 0x00000FFF 
sw $t0, 0($sp)# coloca cor na pilha, saves colour to stack?


move $t0, $zero #i = 0
lw   $s0, w_68 #final do loop





main_for2_end:  


li $v0, 10
syscall

SetPixel:
#recebe posicaoo x em $a0, y em $a1 e cor em $a2
# x position is in $a0, y position is in $a1, colour is in $a2
# pinta o bit_map com a cor indicada no pixel indicado, paint bit_map with colour indicated

addi $sp, $sp, -8
sw   $t0, 4($sp)
sw   $t1, 0($sp)

lw $t0, bitmap # endereco base, base address
lw $t1, bitmap_x_max# x_max: n de colunas maxima, bitmap width

# t1 = y*n_linhas
mult $a1, $t1
mflo $t1 # t1 = y*colunas

# t1 = y*n_colunas + x
add $t1, $t1, $a0 # t1 = y*n_colunas + x
sll $t1, $t1, 2 # t1 = 4*(y*n_colunas + x)
add $t1, $t1, $t0# t1 = base + 4*(y*n_colunas + x)

sw $a2, 0($t1)


lw   $t0, 4($sp)
lw   $t1, 0($sp)
addi $sp, $sp, 8

jr $ra



desenhaLinha_bom:
#desenha uma linha de (x0,y0)=(a0,a1) ate (x1,y1)=(a2,a3)
#a cor esta em 0($sp) , colour is at 0($sp)
#altera fp! CUIDADO!!!, be careful using $fp

move $fp, $sp # guarda ponto inicial da pilha, save starting point
addi $sp, $sp, -52 # abre espaco para 5 variaveis a serem restauradas
sw $t4, 48($sp)
sw $t3, 44($sp)
sw $t2, 40($sp)
sw $t1, 36($sp)
sw $t0, 32($sp)
sw $s6, 28($sp)
sw $s5, 24($sp)
sw $s4, 20($sp)
sw $s3, 16($sp)
sw $s2, 12($sp)
sw $s1, 8($sp)
sw $s0, 4($sp)
sw $ra, 0($sp)

    # register           | variable
# | registrador |   variável    |
# | t0      |   x_atual     |   
# | t1      |   y_atual     |
# | t2      |   erro_atual  |   
# | t3      |   erro_2|aux1 |
# | t4      |   aux2        |
# | s0      |   cor     |
# | s1      |   x_final     |
# | s2      |   y_final     |
# | s3      |   dx      |
# | s4              dy|-dy      |
# | s5      |   sx      |
# | s6      |   sy      |


lw $s0, 0($fp) #carrega a cor em s0, colour is in $s0

#calcula DX e DY, calculate dx and dy
sub $s3, $a2, $a0 #s3 = x1 - x0
abs $s3, $s3 #s3 = dx = |x1 - x0|

sub $s4, $a3, $a1 #s2 = y1 - y0
abs $s4, $s4 #s4 = dy = |y1 - y0|

# if x0 < x1 then sx := 1 else sx := -1
slt $t0, $a0, $a2 # t0 = (x0 < x1)?
beq $t0, $zero, desenhaLinha_if1_else  
desenhaLinha_if1_then:
    lw $s5, w_one
    j desenhaLinha_if1_exit
desenhaLinha_if1_else:
    lw $s5, w_neg_one
desenhaLinha_if1_exit:

# if y0 < y1 then sy := 1 else sy := -1
slt $t0, $a1, $a3 # t0 = (y0 < y1)?
beq $t0, $zero, desenhaLinha_if2_else  
desenhaLinha_if2_then:
    lw $s6, w_one
    j desenhaLinha_if2_exit
desenhaLinha_if2_else:
    lw $s6, w_neg_one
desenhaLinha_if2_exit:

sub $t2, $s3, $s4 #  err := dx-dy

lw $t0, w_neg_one
mult $t0, $s4
mflo $s4 # s4 = -dy

move $t0, $a0 # t0 = x = x0 (estado inicial)
move $t1, $a1 # t1 = y = y0 (estado inicial)
move $s1, $a2 # s1 = x1 (x_fim)
move $s2, $a3 # s2 = y1 (y_fim)



  desenhaLinha_for1:
    #plot(x0,y0)
    move $a0, $t0
    move $a1, $t1
    move $a2, $s0
    jal SetPixel # plot(x_atual, y_atual)

    #testa saida do loop
    sub $t3, $s1, $t0 #t3 = x_final - x_atual
    sub $t4, $s2, $t1 #t4 = y_final - y_atual
    add $t3, $t3, $t4 #se (x_atual == x_final) && (y_atual == y_final), t3 =                 0
    beqz $t3, desenhaLinha_for1_end # if x0 = x1 and y0 = y1 exit loop

    lw $t3, w_two
    mult $t3, $t2
    mflo $t3 #t3 = err2 = 2*Erro_atual 

    slt $t4, $s4, $t3 # -dy < err2 ?
    beqz $t4, desenhaLinha_for1_if1_end
    desenhaLinha_for1_if1_then:
        add $t2, $t2, $s4 # t2 = erro_atual = #erro_atual + (-dy)
        add $t0, $t0, $s5 # x_atual += sx
    desenhaLinha_for1_if1_end:

    slt $t4, $t3, $s3 # err2 < dx ?
    beqz $t4, desenhaLinha_for1_if2_end
    desenhaLinha_for1_if2_then:
        add $t2, $t2, $s3 # t2 = erro_atual = erro_atual + dx
        add $t1, $t1, $s6 # y_atual += sy
    desenhaLinha_for1_if2_end:

    j desenhaLinha_for1 #loop




desenhaLinha_for1_end:



lw $t4, 48($sp)
lw $t3, 44($sp)
lw $t2, 40($sp)
lw $t1, 36($sp)
lw $t0, 32($sp)
lw $s6, 28($sp)
lw $s5, 24($sp)
lw $s4, 20($sp)
lw $s3, 16($sp)
lw $s2, 12($sp)
lw $s1, 8($sp)
lw $s0, 4($sp)
lw $ra, 0($sp)
addi $sp, $sp, 52

jr $ra

2 个答案:

答案 0 :(得分：1)

offtopic：你不应该使用Bresenham的算法来学习MIPS，从较小的东西开始，或者当你需要解释个别指令正在做什么时，很难帮助你。< / p>

ontopic： 是的，你是对的，main_for是代码绘制多行的地方，w_68是控制多少行（循环执行的次数）。 w_68存储在$ s0中，并在每个循环中与$ t0进行比较（在0中初始化并在小于$ s0（w_68）时递增。

desenhaLinha_bom是绘制线的地方（实际的Bresenham）。 Haven没有检查那里实现的完整算法，但是如果可以的话，它是Bresenham算法的改进版本（不是天真的算法，因为它是在一个合并的octants中更难掌握单循环）

希望它有所帮助！

答案 1 :(得分：1)

（必须有更有效的学习方法（MIPS）汇编，而不是试图找出经过评论的源列表。）

假设标签是入口点的候选者，并且例程按照他们的名字建议。
有SetPixel: x/y position in $a0/$a1, colour in $a2，从desenhaLinha_bom: line from (x0,y0)=(a0,a1) to (x1,y1)=(a2,a3), colour is at 0($sp)调用（通过设置单个PIXEL绘制一行）（根据cjj20＆＃39的注释，cjj20计算出来的数量）。
desenhaLinha_bom仅在代码中从第.text（执行可能会开始执行）到li $v0, 10 syscall被调用，这可能会终止它;在一个名为main_for:的标签和向后跳转到它之间。在这个（＆＃34; main＆＃34;）循环的开头，t0（初始化为0 - current_x？）与s0/w_68进行比较（对于相等，看起来除了防御之外）。绘制线后，t0 / current_x增加32，这可能会或可能不会等于s0/w_68。
（所以我认为cjj20在假设main_for负责多行时是正确的。）
（您可能会注意到至少两个完全相似的答案 - 几小时后，几小时后。）

相关问题
最新问题