QPushButton
可以有图标,但我需要设置动画图标.这该怎么做?我创建了新的类,QPushButton
但是如何将icon替换QIcon
为QMovie
?
我可以为widget添加阴影:
QGraphicsDropShadowEffect *bodyShadow = new QGraphicsDropShadowEffect;
bodyShadow->setBlurRadius(9.0);
bodyShadow->setColor(QColor(0, 0, 0, 160));
bodyShadow->setOffset(4.0);
ui->widget->setGraphicsEffect(bodyShadow);
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但是这个阴影只会在左下方.我需要围绕小部件的影子.如何添加?
我正在尝试分配一些内存,但有时会出现"内存不足"的错误.cudaMemGetInfo说我需要更多的内存.所以,内存碎片问题.有可能解决这个问题吗?是不是可以将元素一个接一个地存放在存储器中,并且可以分解为我可以放在内存中的少量和平?
我创建了对话框并尝试添加一些别名来更改它的标题和文本:
import QtQuick 2.0
import Ubuntu.Components 0.1
import Ubuntu.Components.Popups 0.1
Item {
signal confirmed;
signal canceled;
property alias title: dialog.title
property alias text: dialog.text
signal show(variant caller);
Component {
id: dialogComponent
Dialog {
id: dialog
title: "Exit Game"
text: "Are you sure that you want to exit?"
Button {
text: "cancel"
onClicked:
{
canceled();
PopupUtils.close(dialog);
}
}
Button {
text: "confirm"
color: "orange"
onClicked:
{
confirmed();
PopupUtils.close(dialog);
}
}
}
}
onShow:
{
PopupUtils.open(dialogComponent, caller)
}
}
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但我有问题:
qrc:/ConfirmationDialog.qml:8:23: Invalid …
Run Code Online (Sandbox Code Playgroud) 在我的代码中,我在循环中创建16x16按钮,这需要几秒钟.
onCreateField:
{
for(var i=0;i<fieldWidth;i++)
{
for(var j=0;j<fieldHeight;j++)
{
createButton(i, j);
}
}
}
function createButton(x, y)
{
__buttonX = x;
__buttonY = y;
__component = Qt.createComponent("GameButton.qml");
if(__component != null)
continueButtonCreation();
else
__component.ready.connect(continueButtonCreation);
}
function continueButtonCreation()
{
var button = __component.createObject(field, {"row": __buttonY, "column": __buttonX});
if (button == null) {
// Error Handling
console.log("Error creating object");
return;
}
updateValveState.connect(button.stateUpdated);
button.buttonClicked.connect(buttonClicked);
field.clearField.connect(button.release);
}
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创建按钮的功能运行时,应用程序冻结.我希望在此函数运行时显示加载动画.那么,如何在并行线程中运行此函数以避免冻结?
最好的问候,尼克
我可以删除此元素的背景和交替颜色,但我找不到如何删除它周围的边框。是否可以?
如何在QML中使用视网膜支持?如何选择正确的尺寸和正确的图像分辨率?应用程序需要在视网膜和非视网膜设备上工作.
看看这段代码:
#include <stdio.h>
#include <omp.h>
int main()
{
long i, j;
#pragma omp for
for(i=0;i<=100000;i++)
{
for(j=0;j<=100000;j++)
{
if((i ^ j) == 5687)
{
//printf("%ld ^ %ld\n", i, j);
break;
}
}
}
}
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所以,结果:
robotex@robotex-work:~/Projects$ gcc test.c -fopenmp -o test_openmp
robotex@robotex-work:~/Projects$ gcc test.c -o test_noopenmp
robotex@robotex-work:~/Projects$ time ./test_openmp
real 0m11.785s
user 0m11.613s
sys 0m0.008s
robotex@robotex-work:~/Projects$ time ./test_noopenmp
real 0m13.364s
user 0m13.253s
sys 0m0.008s
robotex@robotex-work:~/Projects$ time ./test_noopenmp
real 0m11.955s
user 0m11.853s
sys 0m0.004s
robotex@robotex-work:~/Projects$ time ./test_openmp
real 0m15.048s
user 0m14.949s …
Run Code Online (Sandbox Code Playgroud) 我使用yield和task来异步获取四个jsons:
@gen.engine
def get_user_data(self, sn, snid, fast_withdrawals):
end_timestamp = time.time()
start_timestamp = end_timestamp - CONFIG.LOYALITY_LEVELS.PERIOD
active_apps_response, total_payments_response, payments_for_period_response, withdrawals_response = yield [
gen.Task(self.http_client.fetch, self.__get_active_apps_url(sn, snid)), gen.Task(self.http_client.fetch, self.__get_total_payments_url(sn, snid)),
gen.Task(self.http_client.fetch, self.__get_payments_sum_for_period_url(sn, snid, start_timestamp, end_timestamp)),
gen.Task(self.http_client.fetch, self.__get_total_withdrawals_url(sn, snid, fast_withdrawals))
]
active_apps = self.__active_apps_handler(active_apps_response)
total_payments = self.__get_total_payments_handler(total_payments_response)
payments_for_period = self.__payments_sum_for_period_handler(payments_for_period_response)
withdrawals = self.__get_total_withdrawals_handler(withdrawals_response)
yield gen.Return(active_apps, total_payments, payments_for_period, withdrawals)
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但是如果我使用yield而不是返回上层函数也变成了生成器而且我也不能使用return.那么,如何在没有调用函数生成器的情况下从龙卷风函数返回结果呢?我正在使用Python 2.7
我将我的游戏移植到QML并且有一个困难.看这张图片:
三角形是一个精灵,线条是力矢量.我可以在QML中显示精灵,但我无法理解如何绘制矢量.它的长度可以超过精灵大小.
我想我可以用Rectangle
用height
的1px的,但我不知道如何将它旋转.
力量可以在每帧变化,因此绘图性能必须良好.