小编Jus*_*ien的帖子

我相信他们是不同事物的密码,但我不确定是什么.当在终端连接到MySQL时,我输入/usr/LOCAL/mysql/BIN/mysql -h host -u username -p然后提示我输入密码,密码是''.但是当使用PHP连接到MySQL时,我使用以下代码并且它可以工作

DEFINE('DB_HOST', 'localhost'); 
DEFINE('DB_USER', 'root'); 
DEFINE('DB_PASS', 'root'); 

$dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASS,) or 
die('could not connect: '. mysqli_connect_error() );

如果我要使用DEFINE('DB_PASS', '');它返回"拒绝访问用户'root'@'localhost'(使用密码:NO)",为什么似乎有两个单独的密码？

int main(int argc, const char * argv[])
{

    int age = 40;
    float gpa = 3.25f;
    char grade ='A';
    double fun = 2.000043f;
    char companyName[20] = "O'Brien Enterprises";

    int *pAge = &age;
    int *pGpa = &gpa;
    int *pGrade = &grade;
    int *pFun = &fun;
    int *pCompanyName = &companyName;

    printf("Value of variables through pointers:\n");
    printf("age = %i\n", *pAge);
    printf("gpa = %f\n",*pGpa);
    printf("grade = %c\n", *pGrade);
    printf("fun = %d\n", *pFun);
    printf("companyName = %s\n", *pCompanyName);


    return 0;
}

当我运行此代码时,Xcode会回复大量的警告和错误.当声明和初始化除了age它之外的所有变量的指针时incompatible pointer …