Important Output Finding Questions Based On Randomize And Pointer Concept Class-12

Important Output Finding Questions Based On Randomize And Pointer Concept

Class-12

1. Find the output of the following program :

#include<iostream.h>
#include<ctype.h>
void Secret(char Msg[ ], int N);
void main( )
{ char SMS[ ]=”rEPorTmE”;
Secret(SMS,2);
cout<<SMS<<endl;
}
void Secret(char Msg[ ], int N)
{ for(int C=0;Msg[C]!=’\0’;C++)
if(C%2= =0)
Msg[C]=Msg[C]+N;
else if(isupper(Msg[C]))
Msg[C]=tolower(Msg[C]);
else
Msg[C]=Msg[C]-N;
}

Ans: teRmttoe

2. Study the following program and select the possible output from it:

#include<iostream.h>
#include<stdlib.h>
const int MAX=3;
void main( )
{ randomize( );
int Number;
Number=50+random(MAX);
for(int P=Number;P>=50;P--)
cout<<P<<”#”;
cout<<endl;
}

(i) 53#52#51#50# (ii) 50#51#52#
(iii) 50#51# (iv) 51#50#

Ans: (iv) 51#50#

Solution: MAX value is 3 That’s why random(MAX) can produce 0 or 1 or 2. (random(N) will produce no. between 1 to n-1) The Number value may be 50 or 51 or 52. The P value starts from Number, upto 50, each time decreases by 1. So Possible outputs are as follows:

52#51#50#
51#50#
50#

As the output 51#50# is available in given answers, so 51#50# is the answer.)

3.  Find the output of the following program;

#include<iostream.h>
#include<ctype.h>
void main( )
{ char Text[ ] = “Mind@work!”;
for(int I=0; Text[I]!=’\0’;I++)
{ if(!isalpha(Text[I]))
Text[I]=’*’;
else if(isupper(Text[I]))
Text[I]=Text[I]+1;
else
Text[I] = Text[I+1];
}
cout<<Text;
}

Ans): Output: Nnd@*Xrk!*

4. In the following program, find the correct possible output(s) from the options:

#include<stdlib.h>
#include<iostream.h>
void main( )
{ randomize( );
char City[][10]= {“DEL”,”CHN”,”KOL”,”BOM”,”BNG”};
int Fly;
for(int I=0; I<3;I++)
{ Fly=random(2) + 1;
cout<<City[Fly]<<”:”;}
}

Outputs:

(i) DEL : CHN : KOL:
(ii) CHN: KOL : CHN:
(iii) KOL : BOM : BNG:
(iv) KOL : CHN : KOL:

Ans): Since random(2) gives either 0 or 1, Fly value will be either 1 or 2.(random(n) gives you any number between 0 to n-1)City[1] is “CHN”. City[2] is “KOL”.Since I value from 0 to 2 (ie<3), 3 iterationswill takes place.So the possible output consists 3 strings separated by :, each of them may be either “CHN” or “KOL”.

So the possible output will be

(ii) CHN : KOL : CHN:
(iv) KOL :CHN : KOL:

5. Find the output of the following program:

#include<iostream.h>
#include<ctype.h>
void main( )
{ char Mystring[ ] = "what@OUTPUT!";
for(int I=0; Mystring[I]!='\0';I++)
{ if(!isalpha(Mystring[I]))
Mystring[I]='*';
else if(isupper(Mystring[I]))
Mystring[I]=Mystring[I]+1;
else
Mystring[I] =Mystring[I+1];
}
cout<<Mystring;}

Ans: Output:

hat@*PVUQVU*

6. In the following program, find the correct possible output(s) from the options:

#include<stdlib.h>
#include<iostream.h>
void main( )
{ randomize( );
char Area[ ][10]={“NORTH”,”SOUTH”,”EAST”,”WEST”};
int ToGo;
for(int I=0; I<3;I++)
{ ToGo=random(2) + 1;
cout<<Area[ToGo]<<”:”;
}}

2

Ans: Outputs:

(i) SOUTH : EAST : SOUTH :
(ii) NORTH : SOUTH : EAST :
(iii) SOUTH : EAST : WEST :
(iv) SOUTH : EAST : EAST :

Ans) Since random(2) gives either 0 or 1, ToGo value will be either 1 or 2.(random(n) gives you any number between 0 to n-1) Area[1] is “SOUTH”. Area[2] is “EAST”.Since I value from 0 to 2 (ie<3), 3 iterations will takes place. So the possible output consists 3 strings separated by :, each of them may be either “SOUTH” or “EAST”.

So the possible output will be

(i) SOUTH : EAST : SOUTH :
(iv) SOUTH : EAST : EAST :

7. Find the output of the following program.

#include<iostream.h>
#include<string.h>
#include<ctype.h>
void Convert(char Str[ ],int Len)
{ for(int Count=0;Count<Len;Count++)
{ if(isupper(Str[Count]))
Str[Count]=tolower(Str[Count]);
else if (islower(Str[Count]))
Str[Count]=toupper(Str[Count]);
else if(isdigit(Str[Count]))
Str[Count]=Str[Count]+1;
else Str[Count]=’*’;
}
}
void main( )
{ char Text[ ]=”CBSE Exam 2005”;
int Size = strlen(Text);
Convert(Text,Size);
cout<<Text<<endl;
for(int C=0,R=Size – 1;C<=Size/2;C++,R--)
{ char Temp=Text[C];
Text[C]=Text[R];
Text[R]=Temp;
}
cout<<Text<<endl;
}

Ans:Output: cbse*eXAM*3116

8. Find and write the output of the following C++ program code:

Note: Assume all required header files are already being included in

the program.

void main( )

{

int Ar[ ] = { 6 , 3 , 8 , 10 , 4 , 6 , 7} ;

int *Ptr = Ar , I ;

cout<<++*Ptr++ << '@' ;

I = Ar[3] - Ar[2] ;

cout<<++*(Ptr+I)<<'@'<<"\n" ;

cout<<++I + *Ptr++ << '@' ;

cout<<*Ptr++ <<'@'<< '\n' ;

for( ; I >=0 ; I -=2)

cout<<Ar[I] << '@' ;

}

Ans 7@11@

6@8@

11@3@

9. Find and write the output of the following C++ program code:

typedef char STRING[80];

void MIXNOW(STRING S)

{

 int Size=strlen(S);

 for(int I=0;I<Size;I+=2)

 {

 char WS=S[I];

S[I]=S[I+1];

 S[I+1]=WS;

}

for (I=1;I<Size;I+=2)

 if (S[I]>=’M’ && S[I]<=’U’)

 S[I]=’@’;

}

void main()

{

 STRING Word=”CBSEEXAM2019”;

MIXNOW(Word);

cout<<Word<<endl;

}

Ans. BCE@XEMA0291