我正在使用C++开发BlackJack游戏,我在其中有以下代码,我收到错误
typedef struct
{
int value;
char suit[8];
char name[8];
}Deck;
Deck Cards[52] = {{ 1,"Ace","Hearts"},{ 2, "Two","Hearts"}, { 3, "Three", "Hearts"}, { 4, "Four","Hearts"}, { 5,"Five","Hearts"},{ 6,"Six", "Hearts06"},
{ 7,"Seven","Hearts"},{ 8,"Eight","Hearts"},{ 9,"Nine","Hearts"},{ 10,"Ten","Hearts"},{10,"Jack","Hearts"},{10,"Queen","Hearts"},{10,"King","Hearts"},
{ 1,"Ace","Clubs"},{2, "Two", "Clubs"},{3,"Three","Clubs"},{4,"Four","Clubs"},{5,"Five","Clubs"},{6,"Six","Clubs"},{7,"Seven","Clubs"},{8,"Eight","Clubs"},
{ 9,"Nine","Clubs"},{10,"Ten","Clubs"},{10,"Jack","Clubs"},{10,"Queen","Clubs"},{10,"King","Clubs"},
{ 1,"Ace","Diamonds"},{2,"Two","Diamonds"},{3,"Three","Diamonds"},{4,"Four","Diamonds"},{5,"Five","Diamonds"},{6,"Six","Diamonds"},{7,"Seven","Diamonds"},
{ 8,"Eight","Diamonds"},{9,"Nine","Diamonds"},{10,"Ten","Diamonds"},{10,"Jack","Diamonds"},{10,"Queen","Diamonds"},{10,"King","Diamonds"},
{ 1,"Ace","Spades"},{ 2,"Two","Spades"},{3,"Three","Spades"},{4,"Four","Spades"},{5,"Five","Spades"},{6,"Six","Spades"},{7,"Seven","Spades"},
{ 8,"Eight","Spades"},{ 9,"Nine","Spades"},{10,"Ten","Spades"},{10,"Jack","Spades"},{10,"Queen","Spades"},{10,"King","Spades"}};
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该错误是
Main.c:39:127: error: initializer-string for array of chars is too long [-fpermissive]
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第39行是上面发布的代码中的最后一行
请帮我弄清楚编译器抛出错误的原因
我有自己的json文件"list.json",用于下面的示例信息列表.我的json文件位于Xcode内部,我想向表中显示我的信息,请给我一些提示并帮助实现,如何解析本地json并在表中加载信息.
[
{
"number": "1",
"name": "jon"
},
{
"number": "2",
"name": "Anton"
},
{
"number": "9",
"name": "Lili"
},
{
"number": "7",
"name": "Kyle"
},
{
"display_number": "8",
"name": "Linda"
}
]
Run Code Online (Sandbox Code Playgroud) 我试图用来UISearchController在iOS 11中与范围栏一起显示搜索栏。
这是我用来设置搜索控制器的代码
let searchController = UISearchController(searchResultsController: nil)
searchController.delegate = self
navigationItem.searchController = searchController
navigationItem.hidesSearchBarWhenScrolling = false
searchController.searchResultsUpdater = self
searchController.dimsBackgroundDuringPresentation = false
searchController.searchBar.showsScopeBar = true
searchController.searchBar.scopeButtonTitles = ["1", "2", "3", "4"]
searchController.searchBar.delegate = self
definesPresentationContext = true
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我想要一个带有始终可见的范围栏的搜索栏。加载视图控制器时,上面的代码可以正常工作,并显示搜索栏和范围栏。
但是,一旦搜索控制器变为活动状态然后被关闭,iOS将在关闭搜索控制器时隐藏范围栏,并且仅显示搜索栏。
我试图通过在中添加以下代码来解决此问题didDismissSearchController,但合并范围栏和搜索栏位于彼此顶部,而不是合并范围栏位于搜索栏下方(如下图所示)。将此代码添加到searchBarTextDidBeginEditing(_ searchBar: UISearchBar)或searchBarTextDidEndEditing(_ searchBar: UISearchBar)无效。
searchController.searchBar.showsScopeBar = true
searchController.searchBar.sizeToFit()
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我一直试图让它成为当你将鼠标悬停在盒子中的任何部分上时,盒子悬停(div)将淡化到不透明度0.7,这样它看起来大部分是黑色但很透明.但我所做的一切都在努力.而且我不想为它分配ID,因为会有更多的盒子.
这是我想要的代码:http://pastebin.com/3ZRcrx57
谢谢
我有以下代码,但它不起作用,任何人都可以帮助我
<form id="myform" action="#">
<h3>Registration Form</h3>
<div id="inputs">
<!-- username -->
<label for="username">Username</label>
<input id="username" title="Must be at least 8 characters."/>
<br />
<!-- password -->
<label for="password">Password</label>
<input id="password" type="password" title="Make it hard to guess." />
<br />
<!-- email -->
<label for="email">Email</label>
<input id="email" title="We won't send you any marketing material." />
<br />
<!-- message -->
<label for="body">Message</label>
<textarea id="body" title="What's on your mind?"></textarea>
<br />
<!-- message -->
<label for="where">Select one</label>
<select id="where" title="Select one of …Run Code Online (Sandbox Code Playgroud) 有谁知道如何获取页面中存在的所有HTML标记?我只需要获取没有ID或其他属性的标签,并创建一种树形结构.喜欢用Javascript或JQuery做到这一点.
例如,此HTML代码:
<html>
<head>
<title>
Example Page
</title>
</head>
< body>
<h1 style="somestyle">
Blabla
</h1>
<div id="id">
<table id="formid">
<tr>
<td>
</td>
</tr>
</table>
</div>
</body>
</html>
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应该返回:
html
头部
标题
正文
h1
div
表
tr
td
我试图从.cshtml文件中调用按钮单击:
<input type="button" id="ImageHosting" value="To Image Hosting" onclick="ImageHosting_Click()"/>
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这是.js文件:
function ImageHosting_Click() {
$("#ImageHosting").click(function () {
alert("test");
});
}
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我无法收到警报消息.知道为什么吗?
这可能已被问过,但我似乎无法找到答案.我希望改变jQuery数据表的页脚的措辞,其中显示"显示0到0的0条目"只是"0条目".谢谢您的帮助!
我正在尝试在hibernate中执行简单的sql事务时收到错误.它说,我正在使用hibernate 4.3derby dialect has been deprecated
我试图谷歌,我发现我需要版本特定的方言.你能告诉我在哪里以及如何获得它?
您可能希望从以下位置查看错误日志:
.
谁能告诉我该怎么办?
我在Wordpress网站上使用Zurb Foundation框架.我遇到的问题是响应式顶栏.当它进入移动模式时,菜单不起作用.它不会在点击时下拉.我已经下载了他们的html模板,完全复制了代码但仍然无效.
有谁知道如何修理它.
我正在使用JNA为C++编写的DLL 编写Java接口,以控制小型设备.
在翻译数据类型时,我遇到了
const char*** someVariable
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有人可以向我解释这意味着什么以及如何在Java中重新创建?
我已阅读指针,我正在使用JNA文档将C++类型映射到Java,但我找不到对末尾带有三个星号的类型的引用.
这应该被解释为String数组的指针吗?
我想在我的网站上使用Google搜索API创建Google搜索表单,但我无法使用API推荐搜索字词,任何人都可以了解如何使用Google搜索API来推荐Google中的字词.
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