嵌入式面試問題及解答(英文)

1、.Basic c question.1.Question:where in memory the variables are stored?Local variables, global variables, static.AnswerLocal variables sit in Stack.Global and static goto Data segment.Dynamic memory comes from Heap.2.Questioncan you explian the meaning for the follwoing programchar *c1, *c2, *c3, *c4

2、, *c5 ;char analysis8 = 'a', 'n', 'a', 'l', 'y', 's', 'i' ,'s'int main()c5 = c4 = analysis;+c4;c3 = &analysis6;c2 = c5 + 2 ;c1 = &analysis7 - 3 ;printf ("%ct%ct%ct%ct%c", *c1,*c2,*c3,*c4,*c5);return 0;Answerc1, c2, c3,

3、 c4 and c5 are pointers (which can hold addresses). analysis is an array variable which is holding the string "analysis". >c5 = c4 = analysis;The starting address of the array is given to c5 and c4. Hence they point to first character 'a' in the array.>+c4;c4 is incremented b

4、y 1. Hence points to 'n' in the array.>c3 = &analysis6;c3 is given the address of 7th character (count from 0) of the array. Hence c3 points to character 'i'. >c2 = c5 + 2 ;.c5 points to first character. plus 2 means, c2 points to 3rd character 'a' (second 'a

5、9; in the string).>c1 = &analysis7 - 3 ;c1 points to 3rd character from the end (not counting the last character).c1 holds the address. *c1 means the data strored at c1. Since c1 points to 3rd character from the end (that is 5th character - count starts from 0), *c holds character 'y'

6、. Hence *c1 will print 'y' on the screen.Similarly for other pointer variables *c2, *c3, *c4 and *c53.Question:a=5 b=10 c=7(a>c)?a:(b>c)?b:c)Answer: 104Question : How do you declare an array of N pointers to functions returning pointers to functions returning pointers to characters?A.

7、char *(*(*aN)()();B. Build the declaration up incrementally, using typedefs:C. Use the cdecl program, which turns English into C and vice versa:D. All of the above.Answer : D5Question :void main()int count=10,*temp,sum=0;temp=&count;*temp=20;temp=∑*temp=count;printf("%d %d %d "

8、,count,*temp,sum);Answer : 20; 20; 20;.6Question :void main()int i=7;printf("%d",i+*i+);Answer : 49.Note: Don t change a variable twice in one expression.7Question : The number of syntax errors in the program?int f()void main()f(1);f(1,2);f(1,2,3);f(int i,int j,int k)printf("%d %d %d&

9、quot;,i,j,k);Answer : None.8Question : void main()float j;j=1000*1000;printf("%f",j);A. 1000000B. OverflowC. ErrorD. None of the aboveAnswer :D.9Question : Give the output of the programvoid main()unsigned i=1; /* unsigned char k= -1 => k=255; */signed j=-1; /* char k= -1 => k=65535

10、*/* unsigned or signed int k= -1 =>k=65535 */if(i<j)printf("less");elseif(i>j)printf("greater");elseif(i=j)printf("equal");Answer : less10Give the output of the programvoid main()char *s="12345sn"printf("%d",sizeof(s);Answer : 411Question :

11、Give the output of the programvoid main()int i;for(i=1;i<4,i+)switch(i)case 1: printf("%d",i);break;case 2:printf("%d",i);break;case 3:printf("%d",i);break;.switch(i) case 4:printf("%d",i);Answer : 1,2,3,412.Question : How to pass two arguments to a functio

12、n prompted to by function pointerA. g -> (1,2)B. *g(1,2)C. (*g)(1,2)D. g(1,2)Answer : C13Question: Can you have constant volatile variable?Answer:YES. We can have a const volatile variable.a volatile variable is a variable which can be changed by the extrenal events (like an interrput timers will

13、 increment the voltile varible. If you dont want you volatile varibale to be changed thendeclare them as“ const volatile”.14.Question:study the code:#include<stdio.h>void main()const int a=100;int *p;p=&a;(*p)+;printf("a=%dn(*p)=%dn",a,*p);What is printed?A)100,101B)100,100C)101,

14、101 D)None of the above.Answer CEmbedded C1. Using the #define statement, how would you declare a manifest constant that returns the number of seconds in a year? Disregard leap years in your answer.Answer#define SECONDS_PER_YEAR (60 * 60 * 24 * 365)ULI'm looking for several things here:Basic kno

15、wledge of the #define syntax (for example, no semi-colon at the end, the need to parenthesize, and so on)An understanding that the pre-processor will evaluate constant expressions for you. Thus, it is clearer, and penalty-free, to spell out how you are calculating the number of seconds in a year, ra

16、ther than actually doing the calculation yourselfA realization that the expression will overflow an integer argument on a 16-bit machine-hence the need for the L, telling the compiler to treat the variable as a LongAs a bonus, if you modified the expression with a UL (indicating unsigned long), then

17、 you are off to a great start. And remember, first impressions count!2. Write the "standard" MIN macro-that is, a macro that takes two arguments and returns the smaller of the two arguments.Answer:#define MIN(A,B)(A)<= (B) ? (A) : (B)The purpose of this question is to test the following

18、:Basic knowledge of the #define directive as used in macros. This is important because until the inline operator becomes part of standard C, macros are the only portable way of generating inline code. Inline code is often necessary in embedded systems in order to achieve the required performance lev

19、elKnowledge of the ternary conditional operator. This operator exists in C because it allows the compiler to produce more optimal code than an if-then-else sequence. Given that performance is normally an issue in embedded systems, knowledge and use of this construct is important Understanding of the

20、 need to very carefully parenthesize arguments to macrosI also use this question to start a discussion on the side effects of macros, for example, what happens when you write code such as:.least = MIN(*p+, b);3. Infinite loops often arise in embedded systems. How does you code an infinite loop in C?

21、 There are several solutions to this question. My preferred solution is:Answer:while(1)?5. Using the variable a, give definitions for the following: a) An integerb) A pointer to an integerc) A pointer to a pointer to an integerd) An array of 10 integerse) An array of 10 pointers to integersf) A poin

22、ter to an array of 10 integersg) A pointer to a function that takes an integer as an argument and returns an integerh) An array of ten pointers to functions that take an integer argument and return an integerAnswers :a) int a; / An integerb) int *a; / A pointer to an integerc) int *a; / A pointer to

23、 a pointer to an integerd) int a10; / An array of 10 integerse) int *a10; / An array of 10 pointers to integersf) int (*a)10; / A pointer to an array of 10 integersg) int (*a)(int); / A pointer to a function a that takes an integer argument and returns an integerh) int (*a10)(int); / An array of 10

24、pointers to functions that take an integer argument and return an integerPeople often claim that a couple of these are the sorts of thing that one looks up in textbooks-and I agree. While writing this article, I consulted textbooks to ensure the syntax was correct. However,I expect to be asked this

25、question (or something close to it) when I'm being interviewed. Consequently, I make sure I know the answers, at least for the few hours of the interview. Candidates who don't know all the answers (or at least most of them) are simply unprepared for the interview. If they can't be prepar

26、ed for the interview, what will they be prepared for?.6. What are the uses of the keyword static?Answer:Static has three distinct uses in C:A variable declared static within the body of a function maintains its value between function invocationsA variable declared static within a module, (but outsid

27、e the body of a function) is accessible by all functions within that module. It is not accessible by functions within any other module. That is, it is a localized globalFunctions declared static within a module may only be called by other functions within that module. That is, the scope of the funct

28、ion is localized to the module within which it is declaredMost candidates get the first part correct. A reasonable number get the second part correct, while a pitiful number understand the third answer. This is a serious weakness in a candidate, since he obviously doesn't understand the importan

29、ce and benefits of localizing the scope of both data and code.7. What do the following declarations mean?const int a;int const a;const int *a;int * const a;int const * const a ;AnswerThe first two mean the same thing, namely a is a const (read-only) integer. The third means a is a pointer to a const

30、 integer (that is, the integer isn't modifiable, but the pointer is). The fourth declares a to be a const pointer to an integer (that is, the integer pointed to by a is modifiable, but the pointer is not). The final declaration declares a to be a const pointer to a const integer (that is, neithe

31、r the integer pointed to by a, nor the pointer itself may be modified).8. What does the keyword volatile mean? Give three different examples of its use.Answer:A volatile variable is one that can change unexpectedly. Consequently, the compiler can make no assumptions about the value of the variable.

32、In particular, the optimizer must be careful to reload the variable every time it is used instead of holding a copy in a register. Examples of volatile variables are:.Hardware registers in peripherals (for example, status registers) Non-automatic variables referenced within an interrupt service rout

33、ineVariables shared by multiple tasks in a multi-threaded application9.Can a pointer be volatile ? Explain.Yes, although this is not very common. An example is when an interrupt service routine modifies a pointer to a buffer10.What's wrong with the following function?:int square(volatile int *pt

34、r)return *ptr * *ptr;Answers:This one is wicked. The intent of the code is to return the square of the value pointed to by *ptr . However, since *ptr points to a volatile parameter, the compiler will generate code that looks something like this:int square(volatile int *ptr)int a,b;a = *ptr;b = *ptr;

35、return a * b;Because it's possible for the value of *ptr to change unexpectedly, it is possible for a and b to be different. Consequently, this code could return a number that is not a square! The correct way to code this is:long square(volatile int *ptr)int a;a = *ptr;return a * a;.11. Embedded

36、 systems always require the user to manipulate bits in registers or variables. Given an integer variable a, write two code fragments. The first should set bit 3 of a. The second should clear bit 3 of a. In both cases, the remaining bits should be unmodified.Answer:Use #defines and bit masks. This is

37、 a highly portable method and is the one that should be used.My optimal solution to this problem would be:#define BIT3(0x1<< 3)static int a;void set_bit3(void) a |= BIT3;void clear_bit3(void) a &= BIT3;12Embedded systems are often characterized by requiring the programmer to access a speci

38、fic memory location. On a certain project it is required to set an integer variable at the absolute address 0x67a9 to the value 0xaa55. The compiler is a pure ANSI compiler. Write code to accomplish this task.Answer:int *ptr;ptr = (int *)0x67a9;*ptr = 0xaa55;13. What does the following code output a

39、nd why? void foo(void)unsigned int a = 6;int b = -20;(a+b > 6) ? puts("> 6") :puts("<= 6");The answer is that this outputs "> 6." The reason for this is that expressions involving signed and unsigned types have all operands promoted to unsigned types. Thus ?

40、0 becomes a very large.positive integer and the expression evaluates to greater than 6. This is a very important point in embedded systems where unsigned data types should be used frequently.14.What does the following code fragment output and why?char *ptr;if (ptr = (char *)malloc(0) = NULL)puts(&qu

41、ot;Got a null pointer");elseputs("Got a valid pointer");Answer: Got a valid pointerThis is a fun question. I stumbled across this only recently when a colleague of mine inadvertently passed a value of 0 to malloc and got back a valid pointer! That is, the above code will output "

42、Got a valid pointer." I use this to start a discussion on whether the interviewee thinks this is the correct thing for the library routine to do. Getting the right answer here is not nearly as important as the way you approach the problem and the rationale for your decision.15.Typedef is freque

43、ntly used in C to declare synonyms for pre-existing data types. It is also possibleto use the preprocessor to do something similar. For instance, consider the following code fragment:#define dPSstruct s *typedefstruct s * tPS;The intent in both cases is to define dPS and tPS to be pointers to struct

44、ure s. Which method, if any, is preferred and why?The answer is the typedef is preferred. Consider the declarations:dPS p1,p2;tPS p3,p4;The first expands to:struct s * p1, p2;which defines p1 to be a pointer to the structure and p2 to be an actual structure, which is probably not what you wanted. Th

45、e second example correctly defines p3 and p4 to be pointers.Obscure syntax16. C allows some appalling constructs. Is this construct legal, and if so what does this code do?.int a = 5, b = 7, c;c = a+b;This code is treated as:c = a+ + b;Thus, after this code is executed, a = 6, b = 7, and c = 12.Real-time question1. What is real-time system?"A real-time system is one in which the correctnes