嵌入式面试问题及解答(英文)word版.doc

Basic c question.
1.
Question:
where in memory the variables are stored?
Local variables, global variables, static.
Answer
Local variables sit in Stack.
Global and static goto Data segment.
Dynamic memory comes from Heap.
2.
Question
can you explian the meaning for the follwoing program
char *c1, *c2, *c3, *c4, *c5 ;
char analysis[8] = {'a', 'n', 'a', 'l', 'y', 's', 'i' ,'s'};
int main()
{
c5 = c4 = analysis;
++c4;
c3 = &analysis[6];
c2 = c5 + 2 ;
c1 = &analysis[7] - 3 ;
printf ("%c\t%c\t%c\t%c\t%c", *c1,*c2,*c3,*c4,*c5);
return 0;
}
Answer
c1, c2, c3, 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 by 1. Hence points to 'n' in the array.
>c3 = &analysis[6];
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' in the string).
>c1 = &analysis[7] - 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'.
Hence *c1 will print 'y' on the screen.
Similarly for other pointer variables *c2, *c3, *c4 and *c5
3.
Question:
a=5 b=10 c=7
(a>c)?a:((b>c)?b:c)
Answer: 10
4
Question : How do you declare an array of N pointers to functions returning
pointers to functions returning pointers to characters?
A. char *(*(*a[N])())();
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 : D
5
Ques