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Tuesday, June 10, 2014

C Aptitude-4

Predict the output or error(s) for the following:

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.

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

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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