C Program Page 3

31:-
main()
{
int i=1;
while (i<=5) { printf("%d",i); if (i>2)
goto here;
i++;
}
}
fun()
{
here:
printf("PP");
}
Answer:
Compiler error: Undefined label 'here' in function mainExplanation:Labels have functions scope, in other words The scope of the labels is limited to functions . The label 'here' is available in function fun() Hence it is not visible in function main.

32:-
main()
{
static char names[5][20]={"pascal","ada","cobol","fortran","perl"};
int i;
char *t;
t=names[3];
names[3]=names[4];
names[4]=t;
for (i=0;i<=4;i++) printf("%s",names[i]); } Answer:
Compiler error: Lvalue required in function mainExplanation:Array names are pointer constants. So it cannot be modified.

33:-
void main()
{
int i=5;
printf("%d",i++ + ++i);
}
Answer:Output Cannot be predicted exactly.
Explanation:Side effects are involved in the evaluation of i

34:-
void main()
{
int i=5;
printf("%d",i+++++i);
}
Answer:Compiler Error
Explanation:The expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of operators.

35:-
#include
main()
{
int i=1,j=2;
switch(i)
{
case 1:
printf("GOOD");
break;
case j:
printf("BAD");
break;
}
}
Answer:
Compiler Error: Constant expression required in function main.Explanation:The case statement can have only constant expressions (this implies that we cannot use variable names directly so an error).
Note:Enumerated types can be used in case statements.

36:-
main()
{
int i;
printf("%d",scanf("%d",&i));// value 10 is given as input here
}
Answer: 1
Explanation:Scanf returns number of items successfully read and not 1/0. Here 10 is given as input which should have been scanned successfully. So number of items read is 1.

37:-
#define f(g,g2) g##g2
main()
{
int var12=100;
printf("%d",f(var,12));
}
Answer:100

38:-
main()
{
int i=0;
for(;i++;printf("%d",i)) ;
printf("%d",i);
}
Answer: 1
Explanation: before entering into the for loop the checking condition is "evaluated". Here it evaluates to 0 (false) and comes out of the loop, and i is incremented (note the semicolon after the for loop).

39:-
#include
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer: M
Explanation: p is pointing to character '\n'.str1 is pointing to character 'a' ++*p meAnswer:"p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10. then it is incremented to 11. the value of ++*p is 11. ++*str1 meAnswer:"str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98. both 11 and 98 is added and result is subtracted from 32. i.e. (11+98-32)=77("M");

40:-
#include
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s=malloc(sizeof(struct xx));
printf("%d",s->x);
printf("%s",s->name);
}
Answer:Compiler Error
Explanation: Initialization should not be done for structure members inside the structure declaration

41:-
#include
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

42:-
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.

43:-
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.

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, 2114, 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’.