assembly 构建有向图[已关闭]

wwodge7n  于 2023-01-09  发布在  其他
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建立一个有向图,给定n =节点数(〈= 100),N条线,其中将是每个mod的键数,M条线与每个节点的键节点。例如:
4//节点数
2//节点0有2个键(1和2)
2//节点1有2个键(2和3)
1//节点2有2个键(有3个键)
0//节点3有2个键(没有键)
1//债券
节点0的2//
2//债券
节点1的3//
3//节点的连接2
给定此输入,输出应为:
0 1 1
0 0 1 1
0 0 0 1
0 0 0 0
我在asm att中编写了以下代码:

.data
    n: .space 4
    v: .space 400
    i: .space 4
    m: .space 400
    Leg: .space 4
    columnIndex: .space 4
    lineIndex: .space 4
    formatScanf: .asciz "%ld"
    formatPrintf: .asciz "%ld "
    newLine: .asciz "\n"
.text
.global main
main:
    
    pusha
    pushl $n
    pushl $formatScanf
    call scanf
    popl %ebx
    popl %ebx
    popa

    xor %ecx, %ecx
    lea v, %esi

forcitire: (=this function creates a vector for the the number of bonds)

    cmp %ecx, n
    je afisare

    pusha
    pushl $i
    pushl $formatScanf
    call scanf
    popl %edx
    popl %edx
    popa
    movl i, %eax
    mov %eax, (%esi, %ecx, 4)

    inc %ecx
    jmp forcitire

afisare: (=resets ecx)

    xor %ecx, %ecx

et_constructie: (=construct the matrix)
    
    cmp %ecx, n
    je et_rest
    mov (%esi, %ecx, 4), %ebx
    
    et_leg:
    cmp $0, %ebx
    je et_inc
    pusha
    pushl $Leg
    pushl $formatScanf
    call scanf
    popl %edx
    popl %edx
    popa
    
    movl $0, %edx
    movl %ebx, %eax
    mull n
    
    addl Leg, %eax
    
    lea m, %edi
    movl $1 , (%edi, %eax, 4)
    
    dec %ebx
    

    jmp  et_leg
    
    et_inc:
    incl %ecx
    jmp et_constructie
    

et_rest: (resets ecx)
 xor %ecx, %ecx

et_afis_matr: (=prints the graph matrix)
    movl $0, lineIndex
    for_lines:
    movl lineIndex, %ecx
    cmp %ecx, n
    je et_exit
    movl $0, columnIndex
        for_columns:
            movl columnIndex, %ecx
            cmp %ecx, n
            je cont

    movl lineIndex, %eax
    movl $0, %edx
    mull n
    addl columnIndex, %eax

    lea m, %edi
    movl (%edi, %eax, 4), %ebx
    pusha
    pushl %ebx
    pushl $formatPrintf
    call printf
    popl %ebx
    popl %ebx
    popa
    
    pusha
    pushl $0
    call fflush
    popl %ebx
    popa
    
    incl columnIndex
        jmp for_columns
        
    cont: (=goes to the next line)
        movl $4, %eax
        movl $1, %ebx
    movl $newLine, %ecx
    movl $2, %edx
    int $0x80
    incl lineIndex
    jmp for_lines
        
        
        et_exit:
           movl $1, %eax
               movl $0, %ebx
               int $0x80

代码打印
0 0 0 0
0 0 1 1
0 1 1
0 0 0 0

6qftjkof

6qftjkof1#

所以我把ebx移到了eax中,而不是ecx移到了eax中,这就解决了我的问题,在ebx中是特定节点的键数,ecx是与节点数比较的指数,节点数是我在矩阵中得到正确指数所需要的。

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