如果我的代码通常像这样:
char* log = new char[logLength];
glGetProgramInfoLog(..., ..., log)
//Print Log
delete [] log;
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如何使用C++ 11智能指针实现相同的结果?谁知道在我有机会删除那段记忆之前会发生什么.
所以我想我需要向下转换为C风格的指针?
我的网站上有一些外部脚本存在一些奇怪的问题.我终于把它归结为这个片段了.
<script type="text/javascript" src="//www.googleadservices.com/pagead/conversion.js" onload="console.log('conversion')"></script>
<script type="text/javascript" src="//ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js" onload="console.log('jquery')"></script>
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在firefox中加载这个只有第二个脚本加载,而使用chrome,都加载.这是在OSX上使用最新的浏览器.
现在,如果我将crossorigin属性添加到两个脚本,那么它将停止在chrome中使用此错误Script from origin 'http://www.googleadservices.com' has been blocked from loading by Cross-Origin Resource Sharing policy: No 'Access-Control-Allow-Origin' header is present on the requested resource. Origin 'http://www.example.com:3000' is therefore not allowed access.
现在的脚本是
<script type="text/javascript" src="//www.googleadservices.com/pagead/conversion.js" onload="console.log('12123')" crossorigin async></script>
<script type="text/javascript" src="//ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js" onload="console.log('123')" async crossorigin></script>
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我对这种行为上的差异感到震惊.这是因为firefox比CHrome更严格吗?这是我设置的地方吗?或者这是Firefox/Chrome方面的错误?
另外,我应该与供应商讨论如何为CORS设置js?令我印象深刻的是谷歌失败了,但我还有另一个资源破裂.
通过权利,MDN声称script标签不受同一原产地政策的限制.
我的要求如下
我目前的SSH配置是
Host shortname1?
Hostname %h.prod.xyz.com
Host test-myname
Hostname combo.test-myname.xyz.com
Host *.xyz.com
Hostname %h
Hostname *.xyz.com
User myuser
Port 12345
IdentityFile ~/.ssh/id_rsa
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确切的用例是
ssh shortname1a; ssh shortname1b; ssh test-myname; ssh combo.test-myname.xyz.com
Host *.xyz.com 在最后一个用例调试之后,这是最近添加的,这似乎打破了前2个.
这是正确的方式还是有更简洁的方式?
我是Perl的新手但是我在使用perl模块方面有点专长,我曾经在HTML :: table extract上工作但是当我处理Mechanize模块时,它一直在说我无法通过包"在运行时找到对象方法"new"以下计划
#!user/Perl/bin;
use strict;
use warnings;
use WWW::Mechanize;
my $URL ="https://accounts.google.com"
my $username = <>;
my $password = <>;
my $mech =new www::Mechanize();
$mech -> get($url);
$mech -> form_name('formname');
$mech -> field ('username' =>$username);
$mech -> field ('password' =>$password);
$mech -> click('signin');
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我也参加过堆栈溢出成员给出的一些指导方针,并按照以下步骤安装了Mechanize :: Firefox
cpan WWW::Mechanize::Firefox
cpan MozRepl
cpan MozRepl::RemoteObject
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但是我在运行代码时遇到了同样的错误
请帮帮我..
谢谢N @ veen
任务是
- name: ensure rvm install path is writable by the set owner:group
file: path='{{ rvm1_install_path }}' state=directory recurse=yes owner='{{ rvm1_user }}' group='{{ rvm1_group }}'
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我明白了
TASK: [rvm_io.rvm1-ruby | ensure rvm install path is writable by the set owner:group] ***
failed: [54.68.205.15] => {"failed": true, "gid": 111, "group": "jenkins", "item": "", "mode": "0755", "owner": "jenkins", "path": "/var/lib/jenkins/.rvm", "size": 4096, "state": "directory", "uid": 106}
msg: chgrp failed
FATAL: all hosts have already failed -- aborting
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文件夹已经存在.
我在用
sudo: true
sudo_user: jenkins …Run Code Online (Sandbox Code Playgroud) 我试图用ansible设置我的dotfiles / machine bootstrap东西,并且我想安装一堆GUI工具,但前提是我在支持它的机器上。我ansible_os_family == 'Debian'用来检测apt-get可以工作。不知道如何检查是否安装了例如gtk或类似的东西。
我正在研究以下问题:
编写一个方法runningTotal,它返回一个新的ArrayIntList,它包含原始列表的运行总计.换句话说,新列表中的第i个值应该存储原始列表的元素0到i的总和.
我陷入了第二种方法的最后一部分(返回).没有参数的方法.
public class ArrayIntList {
private int[] elementData;
private int size;
}
// when client calls : test = ArrayIntList.runningTotal(test);
// the folowing method works fine
public static ArrayIntList runningTotal(ArrayIntList other) {
other.elementData[0] = other.elementData[0];
for(int i = 1; i < other.size; i++){
other.elementData[i] = other.elementData[i]+ other.elementData[i-1];
}
return other;
}
// when client calls: test = test.runningTotal();
public ArrayIntList runningTotal() {
elementData[0] = elementData[0];
for(int i = 1; i < size; i++){
elementData[i] = elementData[i]+ elementData[i-1];
} …Run Code Online (Sandbox Code Playgroud) 我刚试过以下代码
class MailBox{
};
template<typename T>
void foo(T){
cout << "In foo" << endl;
}
template<typename T>
void foo1(T){
foo(T);
}
main()
{
MailBox m;
std::vector<MailBox> m1;
foo1(m1);
}
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我们在编译时遇到错误
test1.cpp: In function âvoid foo1(T)â:
test1.cpp:15: error: expected primary-expression before â)â token
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知道如何解决这个问题吗?
我试图为回文实现一个解决方案,我认为我的逻辑是正确的,但我的程序陷入了无限循环,我收到"Prep.exe已停止工作"的错误消息
int main()
{
string word, reverse;
cout << "Please enter a word to test for palindrome : ";
cin >> word;
cout << "Your word is: "<< word << endl;
int i = 0;
int size = word.length() - 1;
while (size >= 0)
{
reverse[i++] = word[size--];
//cout << reverse[i++];
}
cout << "The reversed word is: " << reverse << endl;
if (word == reverse)
cout << "It is palindrome" << endl;
else
cout << "It is …Run Code Online (Sandbox Code Playgroud) 我正在编写一个程序,我在其中继承了C++中的另一个类并覆盖了几个方法.因为我只想在这些方法中添加一行,所以我尝试在基类中调用该方法并在其下面添加一行.我收到以下错误.
"Unhandled exception at 0x00FA4456 in Ch.12.exe: 0xC0000005: Access violation reading location 0x67525A64."
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这是最接近(据我所知),我可以在java中获得"超级".
该类的简化版本
using namespace std;
#include <iostream>
#include <string>
#include "dateType2.h"
class extDateType : public dateType
{
private:
string monthString;
void updateMonthString()
{
string months[12] = {"January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"};
monthString = months[getMonth()];
}
public:
void printDateString()
{
string months[12] = {"January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"};
cout << months[getMonth()];
cout << " …Run Code Online (Sandbox Code Playgroud) 设置简单:功能和功能实现有n个原型。函数指针有一大堆。每个函数都列在此数组中。使用gcc进行编译时,某些仍然会导致-Wunused函数。
码:
void foo1(void);
void foo2(void);
void bar1(void);
void bar2(void);
/* and their implementations */
void (*functions[])(void) = { foo1, foo2, bar1, bar2 };
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这就是设置的样子(只是一个例子)!现在,使用gcc编译时,其中一个foo / bar函数会导致-Wunused函数警告。其他人没有。为什么?
所以这是我到目前为止,我不知道为什么程序没有响应我想要的方式.继续显示"avg2可能尚未初始化".有任何想法吗??
if (a < b && a < c) {
System.out.println("The lowest score was: " + a);
System.out.println("The average without the lowest score is: " + ((b + c) / 2));
avg2 = ((b + c) / 2);
}
if (b < a && b < c) {
System.out.println("The lowest score was: " + b);
System.out.println("The average without the lowest score is: " + ((a + c) / 2));
avg2 = ((a + c) / 2);
}
if (c < …Run Code Online (Sandbox Code Playgroud) 写一个函数isReverse的定义,它的前两个参数是相等大小的整数数组,第三个参数是一个整数,表示每个数组的大小.当且仅当一个数组与另一个数组相反时,该函数返回true.("Reverse"在这里表示相同的元素,但顺序相反.)
int isReverse(int array1[], int array2[], int size)
{
int i;
for (i=0;i<size;i++)
{
if(array1[i] == array2[size-1])
return 0;
else
return 1;
}
}
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我一直在收到错误.它出什么问题了.
c++ ×4
ansible ×2
c ×2
java ×2
arrays ×1
c++11 ×1
config ×1
cors ×1
dry ×1
exception ×1
firefox ×1
gcc-warning ×1
html ×1
if-statement ×1
inheritance ×1
perl ×1
prototype ×1
ruby ×1
rvm ×1
script-tag ×1
ssh ×1
templates ×1
while-loop ×1