41)
#include<stdio.h>
#include<stdio.h>
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
Initialization should not be done for structure members
inside the structure declaration
42)
#include<stdio.h>
#include<stdio.h>
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer:
Compiler Error
Explanation:
in the end of nested structure yy a member have to be
declared.
43) main()
{
extern int i; i=20; printf("%d",sizeof(i));
}
Answer:
Linker error: undefined symbol '_i'.
Explanation:
extern declaration specifies that the variable i is defined
somewhere else. The compiler passes the external variable to be resolved by the
linker. So compiler doesn't find an
error. During linking the linker
searches for the definition of i. Since it is not found the linker flags an
error.
44) 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.
45) main()
{
extern out;
printf("%d", out);
}
int out=100;
Answer:
100
Explanation:
This is the correct way of writing the previous program.
46) main()
{
show();
}
void show()
{
printf("I'm the greatest");
}
Answer:
Compier error: Type mismatch in redeclaration of show.
Explanation:
When the compiler sees the function show it doesn't know
anything about it. So the default return type (ie, int) is assumed. But when
compiler sees the actual definition of show mismatch occurs since it is
declared as void. Hence the error.
The solutions are as follows:
1. declare void show() in main() .
2. define show() before main().
3. declare extern void show() before the use of show().
47) 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.
48) main( )
{
int a[ ] = {10,20,30,40,50},j,*p;
for(j=0; j<5; j++)
{
printf(“%d” ,*a);
a++;
}
p = a;
for(j=0; j<5; j++)
{
printf(“%d ” ,*p);
p++;
}
}
Answer:
Compiler error: lvalue required.
Explanation:
Error is in line with statement a++. The operand must be an
lvalue and may be of any of scalar type for the any operator, array name only
when subscripted is an lvalue. Simply array name is a non-modifiable lvalue.
**49) 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.
50) main( )
{
char *q;
int j;
for (j=0; j<3; j++) scanf(“%s” ,(q+j)); for (j=0; j<3;
j++) printf(“%c” ,*(q+j)); for (j=0; j<3; j++) printf(“%s” ,(q+j));
}
Explanation:
Here we have only one pointer to type char and since we take
input in the same pointer thus we keep writing over in the same location, each
time shifting the pointer value by 1. Suppose the inputs are MOUSE, TRACK and VIRTUAL. Then for the first input
suppose the pointer starts at location 100
then the input one is stored as
M O U S E \0
When the second
input is given
the pointer is
incremented as j
value
becomes 1, so the input is filled in memory starting from
101.
M T R A C K \0
The third input
starts filling from the location 102
M T V I R T U A L \0
This is the final value stored .
The first printf prints the values at the position q, q+1
and q+2 = M T V The second printf prints three strings starting from locations
q, q+1, q+2 i.e MTVIRTUAL, TVIRTUAL and VIRTUAL.
51) 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’.
52) 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’.
53) 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
54) int i,j;
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.
for(i=0;i<=10;i++)
{ j+=5; assert(i<5);
}
Answer:
Runtime error: Abnormal program termination.
assert failed (i<5), <file name>,<line
number>
Explanation:
asserts are used during debugging to make sure that certain
conditions are satisfied. If assertion fails, the program will terminate
reporting the same. After debugging use,
#undef NDEBUG
and this will disable all the assertions from the source
code. Assertion is a good debugging tool to make use of.
55) 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).
56) What are
the files which are automatically opened when a C file is executed?
Answer:
stdin, stdout, stderr (standard input,standard
output,standard error).
57) what will be the position of the file marker?
a: fseek(ptr,0,SEEK_SET);
b: fseek(ptr,0,SEEK_CUR);
Answer :
a: The SEEK_SET sets the file position marker to the
starting of the file. b: The SEEK_CUR sets the file position marker to the
current position
of the file.
58) main()
{
char name[10],s[12];
scanf(" \"%[^\"]\"",s);
}
How scanf will execute?
Answer:
First it checks for the leading white space and discards
it.Then it matches with a quotation mark and then it reads all character upto another quotation
mark.
59) What is
the problem with the following code segment?
while ((fgets(receiving array,50,file_ptr)) != EOF)
;
Answer & Explanation:
fgets returns a pointer. So the correct end of file check is
checking for != NULL.
60) main()
{
main();
}
Answer:
Runtime error : Stack overflow.
Explanation:
main function calls itself again and again. Each time the
function is called its return address is stored in the call stack. Since there
is no condition to terminate the function
call, the call
stack overflows at
runtime. So it terminates the program and results in an
error.
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