判斷二進制文件指令集方法

目前這裏只支持下面這幾種指令集的識別法。

廢話不多說，直接上代碼：

def check_arch(ff):
        '''ff爲文件名'''
        #可執行文件、鏈接庫、動態流、對象
        mime_kw = 'x-executable|x-sharedlib|octet-stream|x-object'  
	ISADict = {b'\x00':'No Specific Instruction Set',
			   b'\x02':'SPARC' ,
			   b'\x03':'x86',
			   b'\x08':'MIPS',
			   b'\x14':'PowerPC',
			   b'\x16':'S390',
			   b'\x28':'ARM',
			   b'\x2a':'SuperH',
			   b'\x32':'IA-64',
			   b'\x3e':'x86-64',
			   b'\xb7':'Arch64',
			   b'\xf3':'RISC-V'
			   }

	magic_mime = magic.from_file(ff, mime=True) #create the file's type  讀取文件類型
	magic_hit = re.search(mime_kw, magic_mime, re.I)
	if magic_hit:
		with open(ff, "rb") as f:
			byte = f.read(20)  ###
			for key, value in ISADict.items():
				if byte[5] == b'\x01' and byte[18] == key: ##魔術的格式
					return value
				else:
					if byte[5] == b'\x02' and byte[19] == key: 
						return value