C APTITUDE


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 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.
54)       int i,j;
            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.

61)       main()
            {
            char *cptr,c;
            void *vptr,v;
            c=10;  v=0;
            cptr=&c; vptr=&v;
            printf("%c%v",c,v);
            }
Answer:
Compiler error (at line number 4): size of v is Unknown.
Explanation:
You can create a variable of type void * but not of type void, since void is an empty type. In the second line you are creating variable vptr of type void * and v of type void hence an error.

62)       main()
            {
            char *str1="abcd";
            char str2[]="abcd";
            printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
            }
Answer:
2 5 5
Explanation:
In first sizeof, str1 is a character pointer so it gives you the size of the pointer variable. In second sizeof the name str2 indicates the name of the array whose size is 5 (including the '\0' termination character). The third sizeof is similar to the second one.

63)       main()
            {
            char not;
            not=!2;
            printf("%d",not);
            }
Answer:
0
Explanation:
! is a logical operator. In C the value 0 is considered to be the boolean value FALSE, and any non-zero value is considered to be the boolean value TRUE. Here 2 is a non-zero value so TRUE. !TRUE is FALSE (0) so it prints 0.

64)       #define FALSE -1
            #define TRUE   1
            #define NULL   0
            main() {
               if(NULL)
                        puts("NULL");
               else if(FALSE)
                        puts("TRUE");
               else
                        puts("FALSE");
               }
Answer:
TRUE
Explanation:
The input program to the compiler after processing by the preprocessor is,
            main(){
                        if(0)
                                    puts("NULL");
            else if(-1)
                                    puts("TRUE");
            else
                                    puts("FALSE");
                        }
Preprocessor doesn't replace the values given inside the double quotes. The check by if condition is boolean value false so it goes to else. In second if -1 is boolean value true hence "TRUE" is printed.

65)       main()
            {
            int k=1;
            printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");
            }
Answer:
1==1 is TRUE
Explanation:
When two strings are placed together (or separated by white-space) they are concatenated (this is called as "stringization" operation). So the string is as if it is given as "%d==1 is %s". The conditional operator( ?: ) evaluates to "TRUE".

66)       main()
            {
            int y;
            scanf("%d",&y); // input given is 2000
            if( (y%4==0 && y%100 != 0) || y%100 == 0 )
                 printf("%d is a leap year");
            else
                 printf("%d is not a leap year");
            }
Answer:
2000 is a leap year
Explanation:
An ordinary program to check if leap year or not.

67)       #define max 5
            #define int arr1[max]
            main()
            {
            typedef char arr2[max];
            arr1 list={0,1,2,3,4};
            arr2 name="name";
            printf("%d %s",list[0],name);
            }
Answer:
Compiler error (in the line arr1 list = {0,1,2,3,4})
Explanation:
arr2 is declared of type array of size 5 of characters. So it can be used to declare the variable name of the type arr2. But it is not the case of arr1. Hence an error.
Rule of Thumb:
#defines are used for textual replacement whereas typedefs are used for declaring new types.

68)       int i=10;
            main()
            {
             extern int i;
              {
                 int i=20;
                        {
                         const volatile unsigned i=30;
                         printf("%d",i);
                        }
                  printf("%d",i);
               }
            printf("%d",i);
            }
Answer:
30,20,10
Explanation:
'{' introduces new block and thus new scope. In the innermost block i is declared as,
            const volatile unsigned
which is a valid declaration. i is assumed of type int. So printf prints 30. In the next block, i has value 20 and so printf prints 20. In the outermost block, i is declared as extern, so no storage space is allocated for it. After compilation is over the linker resolves it to global variable i (since it is the only variable visible there). So it prints i's value as 10.

69)       main()
            {
                int *j;
                {
                 int i=10;
                 j=&i;
                 }
                 printf("%d",*j);
}
Answer:
10
Explanation:
The variable i is a block level variable and the visibility is inside that block only. But the lifetime of i is lifetime of the function so it lives upto the exit of main function. Since the i is still allocated space, *j prints the value stored in i since j points i.

70)       main()
            {
            int i=-1;
            -i;
            printf("i = %d, -i = %d \n",i,-i);
            }
Answer:
i = -1, -i = 1
Explanation:
-i is executed and this execution doesn't affect the value of i. In printf first you just print the value of i. After that the value of the expression -i = -(-1) is printed.

71)       #include<stdio.h>
main()
 {
   const int i=4;
   float j;
   j = ++i;
   printf("%d  %f", i,++j);
 }
Answer:
Compiler error
            Explanation:
i is a constant. you cannot change the value of constant

72)       #include<stdio.h>
main()
{
  int a[2][2][2] = { {10,2,3,4}, {5,6,7,8}  };
  int *p,*q;
  p=&a[2][2][2];
  *q=***a;
  printf("%d..%d",*p,*q);
}
Answer:
garbagevalue..1
Explanation:
p=&a[2][2][2]  you declare only two 2D arrays. but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. now q is pointing to starting address of a.if you print *q meAnswer:it will print first element of 3D array.

73)       #include<stdio.h>
main()
  {
    register i=5;
    char j[]= "hello";                    
     printf("%s  %d",j,i);
}
Answer:
hello 5
Explanation:
if you declare i as register  compiler will treat it as ordinary integer and it will take integer value. i value may be  stored  either in register  or in memory.

74)       main()
{
              int i=5,j=6,z;
              printf("%d",i+++j);
             }
Answer:
11
Explanation:
the expression i+++j is treated as (i++ + j)     

75)     char *someFun1()
            {
            char temp[ ] = “string";
            return temp;
            }
            char *someFun2()
            {
            char temp[ ] = {‘s’, ‘t’,’r’,’i’,’n’,’g’};
            return temp;
            }
            int main()
            {
            puts(someFun1());
            puts(someFun2());
            }
Answer:
            Garbage values.
Explanation:
            Both the functions suffer from the problem of dangling pointers. In someFun1() temp is a character array and so the space for it is allocated in heap and is initialized with character string “string”. This is created dynamically as the function is called, so is also deleted dynamically on exiting the function so the string data is not available in the calling function main() leading to print some garbage values. The function someFun2() also suffers from the same problem but the problem can be easily identified in this case.

76)   struct aaa
        {
         struct aaa *prev;
         int i;
         struct aaa *next;
        };
main()
{
 struct aaa abc,def,ghi,jkl;
 int x=100;
 abc.i=0;abc.prev=&jkl;
 abc.next=&def;
 def.i=1;def.prev=&abc;def.next=&ghi;
 ghi.i=2;ghi.prev=&def;
 ghi.next=&jkl;
 jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;
 x=abc.next->next->prev->next->i;
 printf("%d",x);
}
Answer:
2
Explanation:
                        above all statements form a double circular linked list;
abc.next->next->prev->next->i
this one points to "ghi" node the value of at particular node is 2.

77)       struct point
 {
 int x;
 int y;
 };
struct point origin,*pp;
main()
{
pp=&origin;
printf("origin is(%d%d)\n",(*pp).x,(*pp).y);
printf("origin is (%d%d)\n",pp->x,pp->y);
}
           
Answer:
origin is(0,0)
origin is(0,0)
Explanation:
pp is a pointer to structure. we can access the elements of the structure either with arrow mark or with indirection operator.
Note:
Since structure point  is globally declared x & y are initialized as zeroes
                       
78)       main()
{
 int i=_l_abc(10);
             printf("%d\n",--i);
}
int _l_abc(int i)
{
 return(i++);
}
Answer:
9
Explanation:
return(i++) it will first return i and then increments. i.e. 10 will be returned.

79)       main()
{
 char *p;
 int *q;
 long *r;
 p=q=r=0;
 p++;
 q++;
 r++;
 printf("%p...%p...%p",p,q,r);
}
Answer:
0001...0002...0004
Explanation:
++ operator  when applied to pointers increments address according to their corresponding data-types.

 80)      main()
{
 char c=' ',x,convert(z);
 getc(c);
 if((c>='a') && (c<='z'))
 x=convert(c);
 printf("%c",x);
}
convert(z)
{
  return z-32;
}
Answer:
Compiler error
Explanation:
declaration of convert and format of getc() are wrong.

81)       main(int argc, char **argv)
{
 printf("enter the character");
 getchar();
 sum(argv[1],argv[2]);
}
sum(num1,num2)
int num1,num2;
{
 return num1+num2;
}
Answer:
Compiler error.
Explanation:
argv[1] & argv[2] are strings. They are passed to the function sum without converting it to integer values. 

82)       # include <stdio.h>
int one_d[]={1,2,3};
main()
{
 int *ptr;
 ptr=one_d;
 ptr+=3;
 printf("%d",*ptr);
}
Answer:
garbage value
Explanation:
ptr pointer is pointing to out of the array range of one_d.

83)       # include<stdio.h>
aaa() {
  printf("hi");
 }
bbb(){
 printf("hello");
 }
ccc(){
 printf("bye");
 }
main()
{
  int (*ptr[3])();
  ptr[0]=aaa;
  ptr[1]=bbb;
  ptr[2]=ccc;
  ptr[2]();
}
Answer:
bye
Explanation:
ptr is array of pointers to functions of return type int.ptr[0] is assigned to address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to ccc.

84) char *someFun()
            {
            char *temp = “string constant";
            return temp;
            }
            int main()
            {
            puts(someFun());
            }
Answer:
            string constant
Explanation:
            The program suffers no problem and gives the output correctly because the character constants are stored in code/data area and not allocated in stack, so this doesn’t lead to dangling pointers.

85)     #include<stdio.h>
main()
{
FILE *ptr;
char i;
ptr=fopen("zzz.c","r");
while((i=fgetch(ptr))!=EOF)
printf("%c",i);
}
Answer:
contents of zzz.c followed by an infinite loop 
            Explanation:
The condition is checked against EOF, it should be checked against NULL.

86)       main()
{
 int i =0;j=0;
 if(i && j++)
            printf("%d..%d",i++,j);
printf("%d..%d,i,j);
}
Answer:
0..0
Explanation:
The value of i is 0. Since this information is enough to determine the truth value of the boolean expression. So the statement following the if statement is not executed.  The values of i and j remain unchanged and get printed.
     
87)       main()
{
 int i;
 i = abc();
 printf("%d",i);
}
abc()
{
 _AX = 1000;
}
Answer:
1000
Explanation:
Normally the return value from the function is through the information from the accumulator. Here _AH is the pseudo global variable denoting the accumulator. Hence, the value of the accumulator is set 1000 so the function returns value 1000.

88)       int i;
            main(){
int t;
for ( t=4;scanf("%d",&i)-t;printf("%d\n",i))
                        printf("%d--",t--);
                        }
            // If the inputs are 0,1,2,3 find the o/p
Answer:
            4--0
                        3--1
                        2--2     
Explanation:
Let us assume some x= scanf("%d",&i)-t the values during execution
                        will be,
          t        i       x
          4       0      -4
          3       1      -2
          2       2       0
         
89)       main(){
  int a= 0;int b = 20;char x =1;char y =10;
  if(a,b,x,y)
        printf("hello");
 }
Answer:
hello
Explanation:
The comma operator has associativity from left to right. Only the rightmost value is returned and the other values are evaluated and ignored. Thus the value of last variable y is returned to check in if. Since it is a non zero value if becomes true so, "hello" will be printed.

90)       main(){
 unsigned int i;
 for(i=1;i>-2;i--)
                        printf("c aptitude");
}
Explanation:
i is an unsigned integer. It is compared with a signed value. Since the both types doesn't match, signed is promoted to unsigned value. The unsigned equivalent of -2 is a huge value so condition becomes false and control comes out of the loop.

91)       In the following pgm add a  stmt in the function  fun such that the address of
'a' gets stored in 'j'.
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.
     
92)       What are the following notations of defining functions known as?
i.      int abc(int a,float b)
                        {
                        /* some code */
 }
ii.    int abc(a,b)
        int a; float b;
                        {
                        /* some code*/
                        }
Answer:
i.  ANSI C notation
ii. Kernighan & Ritche notation

93)       main()
{
char *p;
p="%d\n";
           p++;
           p++;
           printf(p-2,300);
}
Answer:
            300
Explanation:
The pointer points to % since it is incremented twice and again decremented by 2, it points to '%d\n' and 300 is printed.

94)       main()
          {
 char a[100];
 a[0]='a';a[1]]='b';a[2]='c';a[4]='d';
 abc(a);
}
abc(char a[]){
 a++;
             printf("%c",*a);
 a++;
 printf("%c",*a);
}
Explanation:
The base address is modified only in function and as a result a points to 'b' then after incrementing to 'c' so bc will be printed.
               
95)       func(a,b)
int a,b;
{
 return( a= (a==b) );
}
main()
{
int process(),func();
printf("The value of process is %d !\n ",process(func,3,6));
}
process(pf,val1,val2)
int (*pf) ();
int val1,val2;
{
return((*pf) (val1,val2));
 }
Answer:
The value if process is 0 !
Explanation:
The function 'process' has 3 parameters - 1, a pointer to another function  2 and 3, integers. When this function is invoked from main, the following substitutions for formal parameters take place: func for pf, 3 for val1 and 6 for val2. This function returns the result of the operation performed by the function 'func'. The function func has two integer parameters. The formal parameters are substituted as 3 for a and 6 for b. since 3 is not equal to 6, a==b returns 0. therefore the function returns 0 which in turn is returned by the function 'process'.

96)       void main()
{
            static int i=5;
            if(--i){
                        main();
                        printf("%d ",i);
            }
}
Answer:
 0 0 0 0
Explanation:
            The variable "I" is declared as static, hence memory for I will be allocated for only once, as it encounters the statement. The function main() will be called recursively unless I becomes equal to 0, and since main() is recursively called, so the value of static I ie., 0 will be printed every time the control is returned.

97)       void main()
{
            int k=ret(sizeof(float));
            printf("\n here value is %d",++k);
}
int ret(int ret)
{
            ret += 2.5;
            return(ret);
}
Answer:
 Here value is 7
Explanation:
            The int ret(int ret), ie., the function name and the argument name can be the same.
            Firstly, the function ret() is called in which the sizeof(float) ie., 4 is passed,  after the first expression the value in ret will be 6, as ret is integer hence the value stored in ret will have implicit type conversion from float to int. The ret is returned in main() it is printed after and preincrement.

98)       void main()
{
            char a[]="12345\0";
            int i=strlen(a);
            printf("here in 3 %d\n",++i);
}
Answer:
here in 3 6
Explanation:
            The char array 'a' will hold the initialized string, whose length will be counted from 0 till the null character. Hence the 'I' will hold the value equal to 5, after the pre-increment in the printf statement, the 6 will be printed.
           
99)       void main()
{
            unsigned giveit=-1;
            int gotit;
            printf("%u ",++giveit);
            printf("%u \n",gotit=--giveit);
}
Answer:
 0 65535
Explanation:
           
100)     void main()
{
            int i;
            char a[]="\0";
            if(printf("%s\n",a))
                        printf("Ok here \n");
            else
                        printf("Forget it\n");
}
Answer:
 Ok here
Explanation:
Printf will return how many characters does it print. Hence printing a null character returns 1 which makes the if statement true, thus "Ok here" is printed.
 

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