|
Do not award
half marks. Do not deduct marks for trivial syntactic errors. Alternative
correct answers should be given credit, unless otherwise indicated
in the marking scheme.
(a)(i)Describe the purpose of a storage
class modifier in C,quoting an
example.[2 marks ]
•A
storage class modifier is used to alter the way the storage is created
(1 mark)
•Any one of const, extern, static, volatile, auto or register
(1 mark
)(ii)Describe the purpose
of a type modifier in C,quoting an
example.[2 marks ]
•A
type modifier is used to modify the particular type of the data involved.
(1 mark)
•Any one of signed unsigned long or short (1 mark)
(b)(i)Give a declaration for an array
of 5 integers,with all the elements initialised to 0.[2 marks ]
The
declaration should read as follows:
int MyArray[5]= {0, 0, 0, 0, 0};
And the following marking scheme should be used:
•A correct declaration;(1 mark)
•A correct initialisation.(1 mark)
(ii)Declare a two dimensional array of characters,of size 2 ×2,initialised
with the
values a, b, c, d [2marks ]
The declaration should read as follows:
char MyArray[2][2]= {’a’, ’b’, ’c’,
’d’};
And the following marking scheme should be used:
•A correct declaration;(1 mark)
•A correct initialisation.(1 mark)
(c)Assuming x y and z are all declared
as int state the value of z in the following
expressions,given the values of x and y
(i)z =x && y where x= 65535 and y= 0 [1mark ]
(ii)z =x && y where x= 65535 and y= 0 [1mark ]
(iii)z =x << y where x= 3 and y= 2 [1mark ]
(iv)z =x >> y where x= 3 and y= 2 [1mark ]
(i)0 [1 mark ]
(ii)1 [1 mark ]
(iii)0 [1 mark ]
(iv)12 [1 mark ]
(d)Give the result of applying the arithmetic
operators in the following expressions,
when x= 4 y= 8 and z= 12
(i)(z % x)+y [1 mark ]
(ii)(y + z)%x [1 mark ]
(iii)y* = x [1 mark ]
(iv)y* = ++x [1 mark ]
(i)8 [1 mark ]
(ii)0 [1 mark ]
(iii)32 [1 mark ]
(iv)40 [1 mark ]
(e)Give the output of the procedure FooBar
the definition of which is given below:
void foobar() {
int *x, *y;
int a[4]= {1, 2, 3, 4};
x= &a[1];
y= a;
*(x-1)= 5;
*y= 6;
x++;
y= (++y)+1;
(*y)++;
x= x+2;
printf("%d %d %d %d\n", a[0], a[1], a[2], a[3]);
}
[4 marks ]
The output is
as follows:
6 2 4 4
And one mark should be awarded for each correct item.[4 marks ]
(f)Write a procedure,called CharCount
the signature of which is given below,which takes in three parameters,a
string s, a character c and an index i
The procedure should calculate how many times c appears in s and return
this value as the index i
You may not use any C library functions.
void CharCount(const char s[], const char* c, int* i); [4 marks ]
A sample definition
of CharCount follows:
void CharCount(const char s[ ], const char* c, int* i)
{
*i= 0;
while (*s!= ’\0’) {
if(*s == *c) (*i)++;
s++;
}
}
And the following marking scheme should be used:
•Ensuring the value of the index is reset to zero;(1 mark)
•Declaring a suitable loop to iterate over the string;(1 mark)
•With a suitable test to ensure the endof the string has not been
reached; (1 mark)
•Incrementing the index if the characters are equal.(1 mark) [4 marks
]
(g)Given the following definition of
a simple linked list in C,define a recursive function, called search
which takes three parameters,a linked list l, a value v and an index
i. The function should return a reference to the the first element
in the list which matches the value passed in,and the index i should
indicate how many list elements have been traversed.
typedef struct node {
struct node* next;
int datum;
} list_element;
list_element *front, *back;
[6 marks ]
A sample definition
of search follows:
list_element* search(const list_element* l, int v, int* i)
{
list_element* result;
*i= (*i)++; if(l== NULL) {
return NULL;
} else {
if(l->datum== v) {
return l;
}else {
return search(l->next,
v, i);
}
}
}
And the following marking scheme shouldbe used:
•Declaring an appropriate function signature;(1 mark)
•Declaring a suitable conditional structure to ensure the base case
(where l->next == NULL has not been reached;(1 mark)
•Incrementing the index on each recursion;(1 mark)
•Testing the current list element to see if the value matches;(1 mark)
•Returning a reference to it if it does;(1 mark)
•Recursing with the next list element if it does not;(1 mark) [6 marks
]
|