Question

def abrir_archivo(nombre_archivo):
    try:
        archivo = open(nombre_archivo,  r )
    except FileNotFoundError:
        print("El archivo no fue encontrado")
    except:
        print("Error desconocido")
    else:
        print("Abriendo exitosamente")
    finally:
        print("Finalizando ejecucion")

abrir_archivo(1)

When I run the program, the else line is executed and prints "Opening Successfully", this happens with the numbers 0, 1 and 2. My question is why? Isn t it supposed to give an error? When I pass 3 or -1 or any other number, it does print "Unknown error" (I speak Spanish, if something is not understood... google translate is guilty uu)

Answer 1

这是因为: open <>a>功能接受档案描述者为file>> case,该词在进行分类时和在 >> POSIX标准、0、1和2对应标准产出错误。

with open(1,  w ) as stdout:
    stdout.write( foo ) # outputs foo to standard output

http://docs.python.org/3/library/io.html#io.IOBase.fileno”rel=“nofollow noreferer”>><>。档案物体的方法:


import sys

with open(sys.stdout.fileno(),  w ) as stdout:
    stdout.write( foo ) # outputs foo to standard output

如果是公开档案记录员,也可接受其他分类编号。

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