Tuesday 13 August 2013

BasicLevelCPractice_Array_ SET1


Note :    All the programs are tested under Turbo C/C++ compilers. 

It is assumed that,

Ø  Programs run under DOS environment,

Ø  The underlying machine is an x86 system,

Ø  Program is compiled using Turbo C/C++ compiler.

The program output may depend on the information based on this assumptions (for example sizeof(int) == 2 may be assumed).

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



1.      main()

{

            char s[ ]="man";

            int i;

            for(i=0;s[ i ];i++)

            printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);

}

Answer:

                        mmmm

                       aaaa

                       nnnn

Explanation:

s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally  array name is the base address for that array. Here s is the base address. i is the index number/displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the  case of  C  it is same as s[i].


2.      main()

{

             int c[ ]={2.8,3.4,4,6.7,5};

             int j,*p=c,*q=c;

             for(j=0;j<5;j++) {

                        printf(" %d ",*c);

                        ++q;     }

             for(j=0;j<5;j++){

printf(" %d ",*p);

++p;     }

}

 

Answer:

                        2 2 2 2 2 2 3 4 6 5

            Explanation:

Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.

3.      VImp. main()

{

            char string[]="Hello World";

            display(string);

}

void display(char *string)

{

            printf("%s",string);

}

            Answer:

Compiler Error : Type mismatch in redeclaration of function display

            Explanation :

In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.

4.      VImp.#include<stdio.h>

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:

77       

Explanation:

p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.

 Now performing (11 + 98 – 32), we get 77("M");

 So we get the output 77 :: "M" (Ascii is 77).

5.      Imp. Need to check. #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:

SomeGarbageValue---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, it will print first element of 3D array.       

6.      Imp. #include<stdio.h>

main()

{

struct xx

{

      int x=3;

      char name[]="hello";

 };

struct xx *s;

printf("%d",s->x);

printf("%s",s->name);

}

            Answer:

Compiler Error

Explanation:

You should not initialize variables in declaration

7.      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 main

Explanation:

Array names are pointer constants. So it cannot be modified.

8.      #include<stdio.h>

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");

9.      #include<stdio.h>

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. 

10.  #include<stdio.h>

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.

 

 

 

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