This interview was conducted for 1 year or more exp. guys in C++.1) Firstly he asked me about the project and i was thoroughlygrilled.2)Then he asked to list the various types of storage classes and asked
1.Study the Following Points:a.One Cannot Take the address of a Bit Fieldb.bit fields cannot be arrayedc.Bit-Fields are machine Dependantd.Bit-fields cannot be declared as staticWhich of the Following
1. Pointer to structure.2. Static variable and difference b/w (const char *p,char const*p,const char* const p). 3. Pass by value & reference.4. String library functions(syntax).5. Write a program to compare
Then he told me to write a query. He wanted me to retrieve data from two tables. The main thing was that we should use " group by "any questions?
Latest Answer: normalization is a process to reduce the redundancy in databases.meaning finding a way to represent a particular data in such a way that with least representaion also it can be accesed in all the ways possible. There are a few ways of doing it like ...
Latest Answer: Non primary column direclty depended on the primarykey column is called a 3rd narmalization form ...
Latest Answer: A column in a table that does not uniquely identify rows in that table, but is used as a link to matching columns in other tables. A foreign key is a field in a relational table that matches the primary key column of another table. The foreign key can ...
Latest Answer: A Primary Key is a constraint which enables a unique identification of each tuple or record in a table.Based on primary key,the database system ensures that no duplication appear in a table ...
Latest Answer: There is no relationship concept in DBMS and RDBMS is Exclusively designed for relationship concept.DBMS supports single user whereas RDBMS supports multi user ...
If in a multi user system a static variable is initialized as 10 and first user increments its value by 1 and then the second user increments its value by one. What is the value of the variable now?
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