我必须使用以下规则构造一个有序容器（必须是可迭代的）：
如果条件为真，则容器为
{1,0}，否则为{0,1}
我有以下代码，但我发现它并不“优雅”：
vector<int> orderedSides;
if (condition)
{
orderedSides.push_back(1);
orderedSides.push_back(0);
}
else
{
orderedSides.push_back(0);
orderedSides.push_back(1);
}
有没有更好的方法（从简洁和性能的角度来看）？
答案 0 :(得分：16)
您可以实现以下内容：
public class LocationUtil extends Service implements LocationListener {
private static final String TAG = LocationUtil.class.getSimpleName();
public Context mContext;
// flag for GPS status
private boolean isGPSEnabled = false;
// flag for network status
private boolean isNetworkEnabled = false;
// flag for passive status
private boolean isPassiveEnabled = false;
private boolean canGetLocation = false;
private Location location; // location
private double latitude; // latitude
private double longitude; // longitude
// The minimum distance to change Updates in meters
private static final long MIN_DISTANCE_CHANGE_FOR_UPDATES = 0;
// The minimum time between updates in milliseconds
private static final long MIN_TIME_BW_UPDATES = 0;
// Declaring a Location Manager
protected LocationManager locationManager;
public LocationUtil(Context context) {
this.mContext = context;
getLocation();
}
public LocationUtil() {
}
public Location getLocation() {
try {
location = null;
locationManager = (LocationManager) mContext
.getSystemService(LOCATION_SERVICE);
// getting GPS status
isGPSEnabled = locationManager
.isProviderEnabled(LocationManager.GPS_PROVIDER);
// getting network status
isNetworkEnabled = locationManager
.isProviderEnabled(LocationManager.NETWORK_PROVIDER);
// getting passive status
isPassiveEnabled = locationManager
.isProviderEnabled(LocationManager.PASSIVE_PROVIDER);
if (!isGPSEnabled && !isNetworkEnabled && !isPassiveEnabled) {
showSettingsAlert();
} else {
this.canGetLocation = true;
if (checkPermission()) {
if (isNetworkEnabled) {
locationManager.requestLocationUpdates(
LocationManager.NETWORK_PROVIDER,
MIN_TIME_BW_UPDATES,
MIN_DISTANCE_CHANGE_FOR_UPDATES, this);
Log.d(TAG, "Network");
if (locationManager != null) {
location = locationManager
.getLastKnownLocation(LocationManager.NETWORK_PROVIDER);
if (location != null) {
latitude = location.getLatitude();
longitude = location.getLongitude();
return location;
}
}
}
// if GPS Enabled get lat/long using GPS Services
if (isGPSEnabled) {
if (location == null) {
locationManager.requestLocationUpdates(
LocationManager.GPS_PROVIDER,
MIN_TIME_BW_UPDATES,
MIN_DISTANCE_CHANGE_FOR_UPDATES, this);
Log.d(TAG, "GPS Enabled");
if (locationManager != null) {
location = locationManager
.getLastKnownLocation(LocationManager.GPS_PROVIDER);
if (location != null) {
latitude = location.getLatitude();
longitude = location.getLongitude();
return location;
}
}
}
}
if (isPassiveEnabled) {
if (location == null) {
locationManager.requestLocationUpdates(
LocationManager.PASSIVE_PROVIDER,
MIN_TIME_BW_UPDATES,
MIN_DISTANCE_CHANGE_FOR_UPDATES, this);
Log.d(TAG, "Passive Enabled");
if (locationManager != null) {
location = locationManager
.getLastKnownLocation(LocationManager.PASSIVE_PROVIDER);
if (location != null) {
latitude = location.getLatitude();
longitude = location.getLongitude();
return location;
}
}
}
}
}
}
} catch (Exception e) {
e.printStackTrace();
}
return location;
}
/**
* Stop using GPS listener
* Calling this function will stop using GPS in your app
*/
public void stopUsingGPS() {
if (locationManager != null) {
locationManager.removeUpdates(LocationUtil.this);
}
}
private boolean checkPermission() {
int permission = ContextCompat.checkSelfPermission(mContext, ACCESS_FINE_LOCATION);
if (permission == PackageManager.PERMISSION_GRANTED) {
return true;
} else {
return false;
}
}
/**
* Function to get latitude
*/
public double getLatitude() {
if (location != null) {
latitude = location.getLatitude();
}
// return latitude
return latitude;
}
/**
* Function to get longitude
*/
public double getLongitude() {
if (location != null) {
longitude = location.getLongitude();
}
// return longitude
return longitude;
}
/**
* Function to check GPS/wifi enabled
*
* @return boolean
*/
public boolean canGetLocation() {
return this.canGetLocation;
}
/**
* Function to show settings alert dialog
* On pressing Settings button will lauch Settings Options
*/
public void showSettingsAlert() {
AlertDialog.Builder alertDialog = new AlertDialog.Builder(mContext);
// Setting Dialog Title
alertDialog.setTitle("GPS is settings");
// Setting Dialog Message
alertDialog.setMessage("GPS is not enabled. Do you want to go to settings menu?");
// On pressing Settings button
alertDialog.setPositiveButton("Settings", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int which) {
Intent intent = new Intent(Settings.ACTION_LOCATION_SOURCE_SETTINGS);
mContext.startActivity(intent);
}
});
// on pressing cancel button
alertDialog.setNegativeButton("Cancel", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int which) {
dialog.cancel();
}
});
// Showing Alert Message
alertDialog.show();
}
@Override
public void onLocationChanged(Location location) {
getLocation();
}
@Override
public void onProviderDisabled(String provider) {
}
@Override
public void onProviderEnabled(String provider) {
}
@Override
public void onStatusChanged(String provider, int status, Bundle extras) {
getLocation();
}
@Override
public IBinder onBind(Intent arg0) {
return null;
}
}
比明确的 if 子句稍短。
正如下面提到的@Deduplicator，我们可能会以更简洁的方式重写第二行：
vector<int> orderedSides(2, 0);
(condition ? orderedSides.front() : orderedSides.back()) = 1;
答案 1 :(得分：5)
QGraphicsView我认为它不是更高效，但我发现它更优雅。
答案 2 :(得分：1)
你可以在效率和避免重复之间做出妥协，先用条件初始化第一个，然后用第一个初始化第二个。
vector<int> orderedSides(1, bool(condition)) ;
orderedSides.push_back(!orderedSides.back());
答案 3 :(得分：0)
ln -s ~/Desktop
我知道这需要花费一些成本。作为候选人之一。
答案 4 :(得分：0)
您可以从数组中填充std::vector，即使在C ++ 98中也是如此。
以下是一个例子：
#include <iostream>
#include <vector>
int main() {
bool condition = false;
std::cout << "condition is: " << std::boolalpha << condition << '\n';
int arr[][2] = {{0,1}, {1,0}};
int index = condition;
std::vector<int> v(arr[index], arr[index]+2);
for (int i = 0; i < v.size(); i++)
std::cout << v[i] << ' ';
std::cout << '\n';
}
输出结果为：
$ g++ tt.cc && ./a.out
condition is: false
0 1
供参考：
答案 5 :(得分：0)
如果构建元素（问题中的int，无论现实生活中的是什么）都是免费且无副作用的：
static const int data[] = { 0, 1, 0 };
std::vector<int> orderedSides (data+condition, data+condition+2);
完整的程序示例：
#include <iostream>
#include <vector>
std::vector<int> make(bool cond)
{
static const int data[] = { 0, 1, 0 };
return std::vector<int> (data+cond, data+cond+2);
}
std::ostream& operator<<(std::ostream& os, const std::vector<int>& v)
{
return os << "{ " << v[0] << ", " << v[1] << " }";
}
int main()
{
std::cout << "true: " << make(true) << "\n"
<< "false: " << make(false) << "\n";
}
打印：
true: { 1, 0 }
false: { 0, 1 }