我想在eclipse插件(WTP的官方插件)中修复一个错误.我在本地更改了源代码,调试了它 - 一切都很好.
现在我想将此更改传播到我的eclipse安装,但我遇到了问题.似乎有不止一种方法可以达到这个目的,例如:
这个站点推荐了片段,但是Eclipse FAQ忽略了这一点.
但我被困住了,似乎没有办法对我有用.
更新:
我按照建议尝试创建并安装功能补丁.安装完成后,将安装功能补丁,但不会安装包含补丁的插件.插件的先前/现有版本仍然存在且处于活动状态.
我想知道为什么会这样吗?这件事与签名官方插件有关吗?是否有日志可以查看未安装修补插件的原因?
最终更新:
问题是我修补了一个父功能(已经在eclipse安装细节中显示),而不是直接包含该插件的直接功能.在为修补指定正确/"最低级别"功能后,一切都按预期工作.
当Wifi连接重新启动时,仅具有带动作的广播意图NETWORK_STATE_CHANGED_ACTION(其常量值是 android.net.wifi.STATE_CHANGE)是正常的吗?即,当Wifi断开连接时,我没有得到这个意图.
更新:我最感兴趣的是> = 2.2 Froyo
我在泛型方法中使用有界参数测试了一些东西,我发现了一些奇怪的行为.
如果有人能够在下面的代码片段中解释这两个错误,那将会很棒.
想象一下,有两个类Class1,Class2都来自a BaseClass. Class2实现一个接口.
在Class1,我有一个方法,Class2以下列方式返回一个实例:
public class Class2 extends BaseClass implements Interface {
@Override
public void method() {
System.out.println("test"); //$NON-NLS-1$
}
}
public class Class1 extends BaseClass {
public <T extends BaseClass & Interface> T getTwo() {
return new Class2();
// Error: Type mismatch: cannot convert from Class2 to T
}
public static void main(String[] args) {
Interface two = new Class1().getTwo();
// Error: Bound mismatch: The generic method …Run Code Online (Sandbox Code Playgroud) #include <unistd.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main() {
int i =10;
/* initialize random seed: */
srand(time(NULL));
while(i--){
if(fork()==0){
/* initialize random seed here does not make a difference:
srand(time(NULL));
*/
printf("%d : %d\n",i,rand());
return;
}
}
return (EXIT_SUCCESS);
}
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打印相同(每次运行时不同)数量10次 - 预计?我有一个更复杂的代码片段,每个分叉的进程依次运行 - 没有区别
for k,v in targets.iteritems():
price= str(v['stockprice'])
Bids = str(u''.join(v['OtherBids']))
Bids = Bids.split(',')
# create a list of unique bids by ranking
for a, b in zip(float(price), Bids):
try:
del b[next(i for i, e in enumerate(b) if format(e, '.4f') == a)]
except StopIteration:
pass
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我从我的字典中提取数据,但似乎所有这些都是unicode.我怎么能摆脱unicode废话?
在VS C++代码中,如果我没有选择任何内容或选择完整行并按下注释选择(Ctrl + K + Ctrl + C),那么它将使用//注释整行
int x = 5;
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按Ctrl + K + Ctrl + C后不选择任何内容或选择全行.
// int x = 5;
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现在,如果我选择该行的某些部分并再次按下注释按钮,则仅评论所选文本(粗体表示已选中)
int x = 5 ;
按下Ctrl + K + Ctrl + C并选择x = 5.
int /*x = 5*/;
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包含多条线
int x = 5;
int y = 2;
int z = x*5;
评论后快捷方式
int/* x = 5;
int y = 2;
int z =*/ x * 5;
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我想要的是
//int x = 5;
//int y = 2; …Run Code Online (Sandbox Code Playgroud) 我安装了多个版本的Boost(Windows 7/MinGW).我需要使用一个特定的(1.53.0).
我在CMakeFiles.txt文件中定义了BOOST_ROOT: SET(BOOST_ROOT C:/boost_1_53_0/)但是我一直收到这个错误:
> cmake .
BOOST_ROOT=C:/boost_1_53_0/
CMake Error at C:/Program Files (x86)/CMake 2.8/share/cmake-2.8/Modules/FindBoost.cmake:1191 (message):
Unable to find the requested Boost libraries.
Boost version: 1.48.0
Boost include path: C:/Boost/include/boost-1_48
Detected version of Boost is too old. Requested version was 1.53 (or
newer).
The following Boost libraries could not be found:
boost_filesystem
No Boost libraries were found. You may need to set BOOST_LIBRARYDIR to the
directory containing Boost libraries or BOOST_ROOT to the location of
Boost.
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我还将BOOST_ROOT定义为环境变量,但结果相同.
为什么cmake还在寻找旧版本?
我在偏好活动中注册广播接收器,并从(唤醒)IntentService发送(同步)广播.显然onReceive在服务的线程上运行.这是我的错吗?是否记录了行为?
偏好活动:
public final class SettingsActivity extends BaseSettings {
private static CharSequence sMasterKey;
private CheckBoxPreference mMasterPref;
// Receiver
/** If the master preference is changed externally this reacts */
private BroadcastReceiver mExternalChangeReceiver =
new ExternalChangeReceiver();
public static void notifyMonitoringStateChange(Context ctx,
CharSequence action, boolean isToggling) {
final LocalBroadcastManager lbm = LocalBroadcastManager
.getInstance(ctx);
Intent intent = new Intent(ctx, ExternalChangeReceiver.class);
intent.setAction(action.toString());
intent.putExtra(TOGGLING_MONITORING_IN_PROGRESS, isToggling);
lbm.sendBroadcastSync(intent);
}
@Override
protected void onStart() {
super.onStart();
final LocalBroadcastManager lbm = LocalBroadcastManager
.getInstance(this);
lbm.registerReceiver(mExternalChangeReceiver, new IntentFilter(
ac_toggling.toString()));
}
@Override …Run Code Online (Sandbox Code Playgroud) multithreading android broadcastreceiver localbroadcastmanager
当我没有明确指定应该使用密钥时,对元组列表(字典键,密钥是随机字符串的值对)进行排序更快(编辑:在@Chepner的评论中添加了operator.itemgetter(0)和密钥版现在更快!):
import timeit
setup ="""
import random
import string
random.seed('slartibartfast')
d={}
for i in range(1000):
d[''.join(random.choice(string.ascii_uppercase) for _ in range(16))] = 0
"""
print min(timeit.Timer('for k,v in sorted(d.iteritems()): pass',
setup=setup).repeat(7, 1000))
print min(timeit.Timer('for k,v in sorted(d.iteritems(),key=lambda x: x[0]): pass',
setup=setup).repeat(7, 1000))
print min(timeit.Timer('for k,v in sorted(d.iteritems(),key=operator.itemgetter(0)): pass',
setup=setup).repeat(7, 1000))
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得到:
0.575334150664
0.579534521128
0.523808984422 (the itemgetter version!)
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但是,如果我创建一个自定义对象key=lambda x: x[0]显式传递sorted使其更快:
setup ="""
import random
import string
random.seed('slartibartfast')
d={}
class A(object):
def __init__(self):
self.s …Run Code Online (Sandbox Code Playgroud) 这个:
import numpy as np
a = np.array([1, 2, 1])
w = np.array([[.5, .6], [.7, .8], [.7, .8]])
print(np.dot(a, w))
# [ 2.6 3. ] # plain nice old matrix multiplication n x (n, m) -> m
import tensorflow as tf
a = tf.constant(a, dtype=tf.float64)
w = tf.constant(w)
with tf.Session() as sess:
print(tf.matmul(a, w).eval())
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结果是:
C:\_\Python35\python.exe C:/Users/MrD/.PyCharm2017.1/config/scratches/scratch_31.py
[ 2.6 3. ]
# bunch of errors in windows...
Traceback (most recent call last):
File "C:\_\Python35\lib\site-packages\tensorflow\python\framework\common_shapes.py", line 671, in _call_cpp_shape_fn_impl
input_tensors_as_shapes, status) …Run Code Online (Sandbox Code Playgroud)