【发布时间】:2016-02-04 11:21:30
【问题描述】:
我正在使用此代码:
int handleArgs(int argc, char *argv[]) {
if(argc <= 1)
{return 0;}
else
{ // If no arguments, terminate, otherwise: handle args
for(int i=1; i<argc; i++) {
if (argv[i] == "-a" || argv[i] == "--admin")
{ // If admin argument
char *pwd = argv[i+1]; // i + 1 b/c we want the next argument; the password
if(pwd == "1729" || pwd == "GabeN")
{ // Verify Password
cout << "Sorry, console feature unavailable.\n" << endl;// Will replace with console function when I get to it
}
else
{
cout << "Wrong/No passkey, bruh.\n" << endl;
} // If the password is wrong
}
else if (argv[i] == "-v" || argv[i] == "--version")
{ // If user asks for version info
cout << "You are running\nDev0.0.0" << endl; // Tell the user the version
}
else if (argv[i]==" -h" || argv[i]=="--help")
{
cout << "Usage: " << argv[0] << " -[switch] argument(s)\n";
cout << " -a, --admin Open console view. Requires password\n";
cout << " -v, --version Print version and exit\n";
cout << " -h, --help Print this message and exit\n" << endl;
}
else {
cout << "Is you dumb?\n '" << argv[0] << " --help' for help" << endl; // Insult the user
}
}
}
return 1;
}
但是,每次我给它一个参数时,我都会收到无效的参数消息(最后一个 else 语句):
Is you dumb?
'main --help' for help
我是 C++ 新手,我不知道自己在做什么(错误)。谁能给我一些有用的见解?谢谢
--破碎代码
【问题讨论】:
-
你能告诉我你是如何传递参数的......仅供参考......
-
@MGP int main(int argc, char *argv[])
-
@MGP handleArgs(argc, argv);
-
我的意思是你是怎么打电话的......只是为了例如。文件名 arg1 arg2....有些像这样...对吗?
-
argv[i] == "-a"等不会像您认为的那样做。您需要改用strcmp。
标签: c++ command-line arguments