Predict the output or
error(s) for the following:
44. main()
{
extern out;
printf("%d", out);
}
int out=100;
Answer:
100
Explanation:
This is the correct way of writing the previous program.
45. 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().
46. 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
47. 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.
48. 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
49. 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.
50. 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.
51. main()
{
int i, n;
char *x = girl;
n = strlen(x);
*x = x[n];
for(i=0; 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.
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