我想做一个简单的密码检查例程在Genie,但我被困在一个for循环。下面是我想模仿的python代码:
#-----------------------------------------------
# password_test.py
# example of if/else, lists, assignments,raw_input,
# comments and evaluations
#-----------------------------------------------
# Assign the users and passwords
users = ['Fred','John','Steve','Ann','Mary']
passwords = ['access','dog','12345','kids','qwerty']
#-----------------------------------------------
# Get username and password
usrname = raw_input('Enter your username => ')
pwd = raw_input('Enter your password => ')
#-----------------------------------------------
# Check to see if user is in the list
if usrname in users:
position = users.index(usrname) #Get the position in the list of the users
if pwd == passwords[position]: #Find the password at position
print 'Hi there, %s. Access granted.' % usrname
else:
print 'Password incorrect. Access denied.'
else:
print "Sorry...I don't recognize you. Access denied."这是我所能得到的:
[indent=4]
init
users: array of string = {"Fred","John","Steve","Ann","Mary"}
passwords: array of string = {"access","dog","12345","kids","qwerty"}
print "Enter user name"
var usrname = stdin.read_line()
print "Enter password"
var pwd = stdin.read_line()
var position = 1
var i = 1
for i=0 to i < users.length
if (users[i]==usrname)
position += 1
if pwd == passwords[position]
print "Hi there, %d. Access granted."
else
print "Password incorrect. Access denied."
else
print "Sorry...I don't recognize you. Access denied."但是,我在编译器上得到了错误:
$ valac evenmores.gs
evenmores.gs:15.18-15.18: error: syntax error, expected `do' but got `<' with previous identifier
for i=0 to i < users.length
^
Compilation failed: 1 error(s), 0 warning(s)我还尝试了这里中建议的for循环
for (i = 0; i < users.length; i++)但没有结果。我会接受一些帮助。谢谢。
发布于 2015-10-02 16:02:59
您应该删除var i = 1并使用for i:int = 0 to (users.length - 1)
这里有几点:
for循环像这样使用时,它只是生成一个数字序列。注意,要生成一个递减的数字序列,您需要使用downto而不是to。下面给出了对数组进行迭代的更好方法for循环时,您可能得到了错误"The name 'i' does not exist in the context of `main'",这就是添加var i = 1的原因。但是,您可以将变量声明为for循环的一部分,如上面所示。通常,对于基本类型(如string和int ),我更喜欢将类型显式化,但您也可以使用类型推断。for var i = 0 to (users.length -1)也能工作要迭代数组,最好使用for item in array语法。对于您的示例,这个示例如下所示:
[indent=4]
init
users: array of string = {"Fred","John","Steve","Ann","Mary"}
passwords: array of string = {"access","dog","12345","kids","qwerty"}
print "Enter user name"
usrname:string = stdin.read_line()
print "Enter password"
pwd:string = stdin.read_line()
position:int = 0
for var user in users
if (user==usrname)
if pwd == passwords[position]
print "Hi there, %s. Access granted.", usrname
else
print "Password incorrect. Access denied."
else
print "Sorry...I don't recognize you. Access denied."
position++在运行代码时,您将看到代码存在一个基本问题。我认为更好的解决办法是使用字典:
[indent=4]
init
var access = new dict of string,string
access[ "Fred" ] = "access"
access[ "John" ] = "dog"
access[ "Steve" ] = "12345"
access[ "Ann" ] = "kids"
access[ "Mary" ] = "qwerty"
print "Enter user name"
username:string = stdin.read_line()
print "Enter password"
pwd:string = stdin.read_line()
if !(username in access.keys)
print "Sorry...I don't recognize you. Access denied."
else
if pwd == access[ username ]
print "Hi there, %s. Access granted.", username
else
print "Password incorrect. Access denied."要点:
libgee才能工作,所以您需要安装Gee及其开发文件。要构建程序,请使用valac --pkg gee-0.8 my_example.gs!操作符和in关键字。还请注意.keysaccess[ username ]发布于 2016-12-15 21:50:56
init
users: array of string = {"Fred","John","Steve","Ann","Mary"}
passwords: array of string = {"access","dog","12345","kids","qwerty"}
print "Enter user name"
usrname:string = stdin.read_line()
print "Enter password"
pwd:string = stdin.read_line()
error:int = 0
cont:int = 0
for var user in users
if (user!=usrname)
error++
if error == (users.length)
print "No reconocido. Acceso denegado."
if (user==usrname)
position:int = cont
if pwd == passwords[position]
print "OK: Acceso Concedido."
else
print "Password incorrecta."
cont++https://stackoverflow.com/questions/32907351
复制相似问题