-3

嘿伙计们，我一直在尝试运行此代码，但我不断收到此错误消息“预期缩进块”我是 python 新手，我不明白问题出在哪里。根据我的知识，我已经正确缩进了代码。

import sys, string
wordsize = 24 # everything is a word
numregbits = 3 # actually +1, msb is indirect bit
opcodesize = 5
memloadsize = 1024 # change this for larger programs
numregs = 2**numregbits
opcposition = wordsize - (opcodesize + 1) # shift value to position opcode
reg1position = opcposition - (numregbits + 1) # first register position
reg2position = reg1position - (numregbits + 1)
memaddrimmedposition = reg2position # mem address or immediate same place as reg2
startexecptr ==0
def regval ( rstr ): # help with reg or indirect addressing
    if rstr.isdigit():
        return ( int( rstr ) )
    elif rstr[0] == '*':
        return int ( rstr[1:]  + (1<<numregbits)) 
    else:
        return 0 # should not happen
        mem = [0] * memloadsize # this is the memory load executable
# instruction mnemonic, type: (1 reg, 2 reg, reg+addr, immed, pseudoop), opcode
        opcodes = {'add': (2, 1),'sub': (2, 2),
           'dec': ( 1, 3), 'inc': ( 1, 4 ),
           'ld': (3, 7), 'st': (3, 8), 'ldi': (3, 9),
           'bnz': (3, 12), 'brl': (3, 13),
           'ret': ( 1, 14 ),
           'int': (3, 16), 'sys': (3, 16),
           'dw': (4, 0), 'go':(3, 0), 'end': (0, 0) }       
        curaddr = 0 # start assembling to location 0
#for line in open(sys.argv[1], 'r').readlines(): # command line
        infile = open("in.asm", 'r')
# Build Symbol Table
        symboltable = {}
        for line in infile.readlines(): # read our asm code
                tokens = string.split( string.lower( line )) # tokens on each line
                firsttoken = tokens[0]
    if firsttoken.isdigit(): # if line starts with an address
        curaddr = int( tokens[0] ) # assemble to here
        tokens = tokens[1:]
    if firsttoken == ';': # skip comments
    if firsttoken == 'go': # start execution here
        startexecptr = ( int( tokens[ 1 ] ) & ((2**wordsize)-1)) # data
    if firsttoken[0] == '.':
        symboltable[firsttoken] = curaddr
        curaddr = curaddr + 1
        print symboltable
        infile.close()
infile = open("in.asm", 'r')
    for line in infile.readlines():
        tokens = string.split( string.lower( line )) # tokens on each line
        firsttoken = tokens[0]
    if firsttoken.isdigit(): # if line starts with an address
        curaddr = int( tokens[0] ) # assemble to here
        tokens = tokens[1:]
    if firsttoken == 'go': # start execution here
        startexecptr = ( int( tokens[ 1 ] ) & ((2**wordsize)-1))
    if firsttoken[0] == '.':
        symaddr = symboltable[firsttoken]
        tokens = tokens[1:]
        memdata = 0
        instype = opcodes[ tokens[0] ] [0]
        memdata = ( opcodes[ tokens[0] ] [1] ) << opcposition
    if instype == 4: # dw type
        memdata = ( int( tokens[ 1 ] ) & ((2**wordsize)-1))
    elif instype == 0: # end type
        memdata = memdata
    elif instype == 1: # dec,inc type, one reg
        memdata = memdata + (regval( tokens[1] ) << reg1position)
    elif instype == 2: # add, sub type, two regs
        memdata = memdata + ( regval( tokens[1] ) << reg1position ) + ( regval( tokens[2] ) << reg2position)
    elif instype == 3: # ld,st type
        token2 = tokens[2]
    if token2.isdigit():
        memaddr = int( tokens[2] )
    else:
        memaddr = symboltable[ token2 ]
        memdata = memdata + ( regval( tokens[1] ) << reg1position ) + memaddr
        mem[ curaddr ] = memdata # memory image at the current location
        curaddr = curaddr + 1
        outfile = open("a.out", 'w') # done, write it out
        outfile.write( 'go ' + '%d' % startexecptr ) # start execution here
        outfile.write( "\n" )
    for i in range(memloadsize): # write memory image
        outfile.write( hex( mem[ i ] ) + " " + '%d'%i)
        outfile.write( "\n" )
        outfile.close()
4

1 回答 1

1

您有两if行缩进到同一级别：

if firsttoken == ';': # skip comments
if firsttoken == 'go': # start execution here

您的意思是在第一个之后缩进所有内容吗？

如果您自己的编辑器将两行显示为不同的缩进级别，您需要停止混合制表符和空格；Python 对选项卡的解释与您的编辑器不同。

运行你的脚本，python -tt <scriptname>让 Python 告诉你在哪里误用了选项卡。

于 2013-09-30T00:00:56.230 回答