小编Kli*_*tel的帖子

我正在尝试将两个故事板用于我的iOS项目,但我无法让代码切换到相应的故事板.相反,代码绝对没有.是否我没有正确设置控制开关的设置？不属于iPhone 5的设备的主要故事板称为MainStoryboard.iphone 5适当的布局称为iphone5.我想这可能是项目配置设置问题.(这是在我的appdelegate.m文件中).

-(void)initializeStoryBoardBasedOnScreenSize {

    if ([UIDevice currentDevice].userInterfaceIdiom == UIUserInterfaceIdiomPhone)
    {    // The iOS device = iPhone or iPod Touch


        CGSize iOSDeviceScreenSize = [[UIScreen mainScreen] bounds].size;

        if (iOSDeviceScreenSize.height == 480)
        {   // iPhone 3GS, 4, and 4S and iPod Touch 3rd and 4th generation: 3.5 inch screen (diagonally measured)

            // Instantiate a new storyboard object using the storyboard file named Storyboard_iPhone35
            UIStoryboard *iPhone35Storyboard = [UIStoryboard storyboardWithName:@"MainStoryboard" bundle:nil];

            // Instantiate the initial view controller object from the storyboard
            UIViewController *initialViewController = …

这是一个我试图编写的简单游戏,然而,它并不像预期的那样高效......这是我的目标.阵列无法正常工作,整体逻辑不完整.我们的目标是:如果骰子上的第一辊的总和为2,3,或12,你告诉他/她失去了球员; 如果总和是7或11,你告诉玩家他/她赢了; 如果第一次掷骰是任何其他数字(4,5,6,8,9,10),则告诉玩家他/她必须再次滚动.现在扩展该程序如下 - 如果玩家被告知在第一次滚动后再次滚动(如果它是4,5,6,8,9,10),保存该数字,称之为"点".现在继续滚动,直到发生两件事之一 - 如果玩家在再次滚动"点数"之前滚动7,则玩家输掉并且转弯结束; 如果玩家在7滚动之前匹配"该点",则玩家获胜并且转弯结束.如果除了7或"点"之外的任何数字都没有发生任何事情并且玩家只是继续滚动.告诉玩家他们是赢还是输,每次掷骰的结果,以及获得结果需要多少卷.

#include <iostream>
#include <ctime>

using namespace std;

int main()
{
    int die1, 
        die2, 
        sum, 
        point,
        rollChoice;
    static int rollCount;

    int *rolls = new int[];

    rollCount=1;
    point=0;

    srand(time(0));

    cout<<"Enter 1 to roll: ";
    cin>>rollChoice;

    if(rollChoice==1)
    {

        for (int i=0; i<INT_MAX; i++)
        {
            die1=rand()%10;
            die2=rand()%10;

            sum=die1+die2;

            rolls[sum];

            if(rolls[i] == 2 || rolls[i] == 3 || rolls[i] == 12)
            {
                cout<<"\nYou have lost!"<<endl;

                rollCount++;

                cout<<"\n";
                cin>>rollChoice;
            }

            else if(rolls[i] == 7 || rolls[i] == 11) …

即使其他变量的输出与程序结束时的概率变量位于相同的放置区域,变量"概率"也总是为零.我觉得它与放置有关,但它可能是另一个初始化问题.概率永远不应该为零.

#include <iostream>
#include <cstdlib>  
#include <ctime>
#include <iomanip>

using namespace std;

int main() {

    int die1, 
        die2, 
        sum,
        turns,
        win=0,
        loss=0;
    double probability=0;
    int thepoint, rolls;
        srand(time(0));


    cout<<"How many turns would you like? ";
    cin>>turns;

    for(int i=0; i<turns; i++)
    {
        sum=0;

        die1=rand()%6;
        die2=rand()%6;
        sum=die1+die2;

        //cout<<"\nFirst die is a  "<<die1<<endl;
        //cout<<"Second die is a "<<die2<<endl;
        //cout<<"\n\n>>>Turn "<<i<<": You rolled "<<sum;


        switch (sum){
            case 2: case 3: case 12:
                //cout<<"You have lost this turn with 2 3 or 12!"<<endl;
                loss++;
                break; …

我有一个mysql函数来转义字符串.我继续被一个永无止境的错误所困扰.饲料吐出:

警告:mysqli_real_escape_string()期望参数1为mysqli,在第39行的/home/shipstud/public_html/post_auth.php中给出null.

任何关于如何解决这个问题的想法将不胜感激.我已附上以下相关代码:

//connect to server and database
$db=mysqli_connect('***','***','***','***');

// check connection
if (mysqli_connect_errno()) {
    echo "Connect failed";
    exit();
}



//parameter checking
$username = safe(stripslashes(trim($_POST['username'])));


//sanitize input parameters
function safe($value)
{
    $secureString = mysqli_real_escape_string($db, $value);

   return $secureString;
}