When we find a job embedded aspects, let us most headache is probably the kind of technical interview questions, because we can not touch the business proposition of law, do not know how to prepare today's advanced embedded Huaqing the instructor will face questions of major companies are aggregated, shared with everyone, I hope you can help small partners. With answers below Oh!
First, the choice (8 to 10 multiple choice)
1, the output of the following code:
#include
void Change (int * A, B int &, int C)
{
C * = A;
B = 30;
* A = 20 is;
}
int main ()
{
int A = 10, B = 20 is, C = 30;
Change (& A, B, C);
the printf ( "% D,% D,% D,", A, B, C );
return 0;
}
a 20,30,30
B 10,20,30
C 20,30,10
D 10,30,30
2, port default FTP and SMTP services are services ()
a 20 is the 25
B 21 is and 25
C and 20, 21 are 25
D 20 is 21
3, the linear table (a1, a2, ..., an ) when stored in a link, the time to access the i-th element of complexity to the position ()
AO (i)
BO (. 1 )
CO.'s (n-)
the DO (. 1-I)
4, there is a structure that looks like this:
struct A {
Long A1;
Short A2;
int A3;
int * A4;
};
Will calculated using sizeof (struct A) In 64-bit compiler size number?
A 24
B 28
C 16
D 18 is
. 5, comprising an ordered array of 20 elements do binary search, start of the array index is 1, then the lookup a [2] is the subscript sequence comparison of ()
a 9,5,4 , 2
B 10,5,3,2
C 9,6,2
D 20,10,5,3,2
. 6, a key 10 in order B- trees, each tree root node contained Maximum number of of (), allowing a minimum of () th.
10,5 A
B 9,4
C 8,3
D 7,6
. 7, the operating system uses buffering technology, by reducing () times of the CPU, and improve resource utilization.
A Interrupt
B access
C Control
D-dependent
[multiple choice]
8, on AVL trees and red-black tree, which of the following statements is correct?
Both belong to A balanced binary tree from
both B search, insert, delete the same time complexity
C comprises n inner height of the red-black tree nodes is O (log (n))
D of the JDK TreeMap is implemented in a AVL
9, Servlet life cycle can be divided into initialization phase, the operational phase and the destruction of three phases, the following procedure is part of the initialization phase ().
A loading Servlet class data and the corresponding .class
B creates serletRequest servletResponse objects and
C to create ServletConfig object
D Servlet creates an object
10, Linux execution LS, which will cause the system call ()
A nmap
B Read
C the execve
D the fork
Second, the short answer
1 , defined briefly embedded systems, applications and features?
a: embedded system definition: application-centric, computer technology, software and hardware can be cut configuration, function, reliability, cost, size, power consumption a special computer system, there are strict constraints.
Embedded system applications: used in military equipment, information terminals, automotive electronics, industrial manufacturing, aerospace and other fields.
Embedded System features: specificity, can be cut, real-time, high reliability, low power consumption.
2. What is the role of the keyword static?
A: In C, the static keyword has three distinct functions:
1) In the body of the function is declared as a static variable to maintain its function is called in this process unchanged.
2) in the module (but function in vitro), a variable is declared as static can be accessed within the module by a function, but other functions can not be accessed outside of the module. It is a local global variables.
3) Within the module, it is declared as a static function may only be called by other functions within the module. That is, this function is limited to use in its declaration module local scope.
Most candidates correctly answer the first part, the second part of the answer correct part, with very few people understand the third. This is a serious weakness in a candidate, because he obviously does not understand the benefits and importance of localized data and code range.
Answers: 1 ~ 5: ACCAB 6 ~ 10: BA ABC ACD BC
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