iphone - 存储到 'var name' 的值永远不会被读取

Question

我在编译器警告方面遇到了一些困难。代码如下:

    const char *sql;
switch (fromVersion) {
    case 0:
    {
        sql = "ALTER TABLE abiliator_options ADD COLUMN new_column00 TEXT NOT NULL DEFAULT 'migrated from version 0'";
    }
    case 1:
    {
        sql = "ALTER TABLE abiliator_options ADD COLUMN new_column01 TEXT NOT NULL DEFAULT 'migrated from version 1'";
    }
    case 2:
    {
        sql = "ALTER TABLE abiliator_options ADD COLUMN new_column02 TEXT NOT NULL DEFAULT 'migrated from version 2'";
    }
    case 3:
    {                       
        sql = "ALTER TABLE abiliator_options ADD COLUMN new_column03 TEXT NOT NULL DEFAULT 'migrated from version 3'";
    }
    case 4:
    {                       
        sql = "ALTER TABLE abiliator_options ADD COLUMN new_column04 TEXT NOT NULL DEFAULT 'migrated from version 4'";
    }
    case 5:
    {                       
        sql = "ALTER TABLE abiliator_options ADD COLUMN new_column05 TEXT NOT NULL DEFAULT 'migrated from version 5'";
    }
}


sqlite3_stmt *selectstmt;
if(sqlite3_prepare_v2(database, sql, -1, &selectstmt, NULL) == SQLITE_OK) 
{
    if (sqlite3_step(selectstmt) == SQLITE_DONE) 
    {           
        NSLog(@"Alter statement successful");
        [self setDatabaseSchemaVersion];
    }
    else {

        NSLog(@"Failed to alter the table with message '%s'.", sqlite3_errmsg(database));
    }
}
else {
    NSLog(@"Failed to prepare the statement with message '%s'.", sqlite3_errmsg(database));

}
sqlite3_finalize(selectstmt);

第一个警告是对于所有 sql 变量分配，存储到“sql”的值永远不会被读取，除了开关中的最后一个(情况 5)。情况 5 不会导致警告。

第二个警告是函数调用参数是未初始化的值。这是为了if(sqlite3_prepare_v2(数据库, sql, -1, &selectstmt, NULL) == SQLITE_OK)声明。

预先感谢您的提示。

Answer 1

你错过了每个案例之后的休息时间。仅使用最后一个分配。