Question

由于某些原因我不知道，程序在尝试执行时会崩溃

mysql_query(mysql,"CREATE TABLE writers(name VARCHAR(25))");

虽然查询成功执行并且表格确实已经创建，但是窗口抱怨程序停止响应但是没有。

这是我的主要功能。

int main()
{
    MYSQL* mysql;
    mysql_init(mysql);
    mysql_real_connect(mysql,"localhost","root","xxxx","test",0,NULL,0);
    mysql_query(mysql,"CREATE TABLE writers(name VARCHAR(25))"); // 'Program stops responding' without actually crashing.
    mysql_close(mysql);
    getchar();
    return 0;
}

Answer 1

您对API的使用是错误的，您正在将悬空指针传递给mysql_init而不会检查单个返回值。不要那样做。

您可以像这样转换代码：

MYSQL* mysql = mysql_init(NULL);
if (!mysql) {
  // HANDLE THE ERROR
}
if (!mysql_real_connect(mysql,"localhost","root","xxxx","test",0,NULL,0)) {
  // HANDLE THE ERROR
}
if (mysql_query(mysql,"CREATE TABLE writers(name VARCHAR(25))")) {
  // HANDLE THE ERROR
}
mysql_close(mysql); // No error checking here, that's ok,
                    // it doesn't report errors
                    // But just for this call. All others => check.
getchar();
return 0;

每次执行查询时程序都会挂起

1 个答案: