(Q-1)
main()
{
printf("%d", out);
}
int out=100;
Answer:
Compiler error: undefined symbol out in function main.
Explanation:
The rule is that a variable is available for use from the point of declaration. Even though a is a global
variable, it is not available for main. Hence an error.
(Q-2)
main()
{
extern out;
printf("%d", out);
}
int out=100;
Answer:
100
Explanation:
This is the correct way of writing the previous program.
(Q-3)
main( )
{
int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
printf("%u %u %u %d \n",a,*a,**a,***a);
printf("%u %u %u %d \n",a+1,*a+1,**a+1,***a+1);
}
Answer:
100, 100, 100, 2
114, 104, 102, 3
Explanation:
The given array is a 3-D one. It can also be viewed as a 1-D array.
2 4 7 8 3 4 2 2 2 3 3 4
100 102 104 106 108 110 112 114 116 118 120 122
thus, for the first printf statement a, *a, **a give address of first element . since the indirection ***a gives
the value. Hence, the first line of the output.
for the second printf a+1 increases in the third dimension thus points to value at 114, *a+1 increments in second
dimension thus points to 104, **a +1 increments the first dimension thus points to 102 and ***a+1 first gets the
value at first location and then increments it by 1. Hence, the output.
(Q-4)
main( )
{
static int a[ ] = {0,1,2,3,4};
int *p[ ] = {a,a+1,a+2,a+3,a+4};
int **ptr = p;
ptr++;
printf("\n %d %d %d", ptr-p, *ptr-a, **ptr);
*ptr++;
printf("\n %d %d %d", ptr-p, *ptr-a, **ptr);
*++ptr;
printf("\n %d %d %d", ptr-p, *ptr-a, **ptr);
++*ptr;
printf("\n %d %d %d", ptr-p, *ptr-a, **ptr);
}
Answer:
111
222
333
344
Explanation:
Let us consider the array and the two pointers with some address
a
0 1 2 3 4
100 102 104 106 108
p
100 102 104 106 108
1000 1002 1004 1006 1008
ptr
1000
2000
After execution of the instruction ptr++ value in ptr becomes 1002, if scaling factor for integer is 2 bytes.
Now ptr - p is value in ptr - starting location of array p, (1002 - 1000) / (scaling factor) = 1,
*ptr - a = value at address pointed by ptr - starting value of array a, 1002 has a value 102 so the value
is (102 - 100)/(scaling factor) = 1, **ptr is the value stored in the location pointed by the pointer of
ptr = value pointed by value pointed by 1002 = value pointed by 102 = 1. Hence the output of the firs printf is 1, 1, 1.
After execution of *ptr++ increments value of the value in ptr by scaling factor, so it becomes1004. Hence,
the outputs for the second printf are ptr - p = 2, *ptr - a = 2, **ptr = 2.
After execution of *++ptr increments value of the value in ptr by scaling factor, so it becomes1004. Hence,
the outputs for the third printf are ptr - p = 3, *ptr - a = 3, **ptr = 3.
After execution of ++*ptr value in ptr remains the same, the value pointed by the value is incremented by the
scaling factor. So the value in array p at location 1006 changes from 106 10 108,. Hence, the outputs for the
fourth printf are ptr - p = 1006 - 1000 = 3, *ptr - a = 108 - 100 = 4, **ptr = 4.
(Q-5)
main( )
{
void *vp;
char ch = 'g', *cp = "goofy";
int j = 20;
vp = &ch;
printf("%c", *(char *)vp);
vp = &j;
printf("%d",*(int *)vp);
vp = cp;
printf("%s",(char *)vp + 3);
}
Answer:
g20fy
Explanation:
Since a void pointer is used it can be type casted to any other type pointer. vp = &ch stores address of char ch
and the next statement prints the value stored in vp after type casting it to the proper data type pointer.
the output is 'g'. Similarly the output from second printf is '20'. The third printf statement type casts it
to print the string from the 4th value hence the output is 'fy'.
(Q-6)
main ( )
{
static char *s[ ] = {"black", "white", "yellow", "violet"};
char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
p = ptr;
**++p;
printf("%s",*--*++p + 3);
}
Answer:
ck
Explanation:
In this problem we have an array of char pointers pointing to start of 4 strings. Then we have ptr which is a pointer to a pointer of type char and
a variable p which is a pointer to a pointer to a pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The next statement increment
value in p by 1 , thus now value of p = s+2. In the printf statement the expression is evaluated *++p causes gets value s+1 then the pre decrement is
executed and we get s+1-1 = s . the indirection operator now gets the value from the array of s and adds 3 to the starting address. The string is
printed starting from this position. Thus, the output is 'ck'.
(Q-7)
main()
{
int i, n;
char *x = "girl";
n = strlen(x);
*x = x[n];
for(i=0; i‹n; ++i)
{
printf("%s\n",x);
x++;
}
}
Answer:
(blank space)
irl
rl
l
Explanation:
Here a string (a pointer to char) is initialized with a value "girl". The strlen function returns the length
of the string, thus n has a value 4. The next statement assigns value at the nth location ('\0') to the first
location. Now the string becomes "\0irl" . Now the printf statement prints the string after each iteration it
increments it starting position. Loop starts from 0 to 4. The first time x[0] = '\0' hence it prints nothing
and pointer value is incremented. The second time it prints from x[1] i.e "irl" and the third time it
prints "rl" and the last time it prints "l" and the loop terminates.
(Q-8)
main()
{
int i=-1;
+i;
printf("i = %d, +i = %d \n",i,+i);
}
Answer:
i = -1, +i = -1
Explanation:
Unary + is the only dummy operator in C. Where-ever it comes you can just ignore it just because it has no
effect in the expressions (hence the name dummy operator).
(Q-9)
What are the files which are automatically opened when a C file is executed?
Answer: stdin, stdout, stderr (standard input,standard output,standard error).
(Q-10)
main(){
int * j;
void fun(int **);
fun(&j);
}
void fun(int **k) {
int a =0;
/* add a stmt here*/
}
Answer: *k = &a
Explanation:
The argument of the function is a pointer to a pointer.
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