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TCS NQT Input/Output Quiz- 4

Question 1

Time: 00:00:00

What is the output of the code?

int main()

{

    short int diff=1;

    switch(++diff-diff++){

        case 1:

        printf("one over subtraction");

        break;

        case 2:

        printf("two over subtraction");

        break;

        default:

        printf("None");

        break;

    }

    return 0;

}

two over subtraction

one over subtraction

None

No output

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SInce diff = 1
So the expression, ++diff-diff++ => 2-2 =>0 (As, ++diff will increment the diff by 1 so diff becomes 2 and then diff++ first do operation then increments the diff so we get output as 0)

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Question 2

Time: 00:00:00

Write the output of the following code in box?

#include <iostream>



int main(void)

{

int stones=5,pebbles=1;

stones-=pebbles+=stones;

printf("%d",stones);

return 0;

}

-1

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Initially, stones = 5, pebbles = 1
stones-=pebbles+=stones , as the assignment operator is used which has associativity from right to left so first, pebbles+=stones gets evaluated,
pebbles=pebbles+ stones => pebbles = 1+5 => pebbles = 6
Then, stones -= pebbles => stones = 5-6 => stones = -1.

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Question 3

Time: 00:00:00

What will be the output of the following code

#include <iostream>

int main()

{

int x=5;

int y=5;

if(!(x|y))

printf("x=y");

else

printf("x!=y");

return 0;

}

Run time error

Compile time error

x!=y

x=y

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| is the bitwise or operator and gives 1 if any of the bit is 1 otherwise 0
Since, inside if condition 5|5 leads to 5 and !5 = 0
So, the condition becomes false and the else block gets executed and prints x!=y.

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Question 4

Time: 00:00:00

What is the output of below code?

#include <stdio.h>

int main()

{

int cat;

for(cat=5;cat>=1;cat--);

printf("%d",cat);

    return 0;

}

54321

543210

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for(cat=5;cat>=1;cat--);

Since, ; is put after the for loop so after the complete execution of the for loop print statement will execute.

This for loop will execute 5 times and make cat =0

So, 0 will be printed.

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Question 5

Time: 00:00:00

What is the output of below code?

#include <iostream>

using namespace std;

int main()

{

int i;

for(i=5;i<=10;i++){

printf("%d",i);

}

return 0;

}

56789

567890

5678910

Error

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Since, the for loop will execute for the value 5, 6, 7, 8, 9, 10 when i becomes 11 the loop will terminate, so for every iteration it will print the corresponding i hence,
5678910 gets printed.

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Question 6

Time: 00:00:00

Wha is the output of below code?

#include <iostream>

using namespace std;

int main()

{

do{

  do{

    printf("A");

     break;

   }while(1); 

  printf("B");

}while(0);



return 0;

}

Print a infinitely

Print AB

Print B infinitely

Print A infinitely

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In the given code the nested do loop is used, inner do loop print A then terminates due to break statement and outer loop will print B then the condition of the outer loop is 0 so it will stop further execution so, we get output as AB.

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Question 7

Time: 00:00:00

What is the output of below code?

#include <iostream>

using namespace std;

int main()

{

int number=0141;

printf("%c",number);

return 0;

}

Compilation error

1100001

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Once you attempt the question then PrepInsta explanation will be displayed.

0141 means that in the octal base, %c will print the character value.

Octal value of a is 141 so, a get printed.

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Question 8

Time: 00:00:00

What is the output of below code?

#include <iostream>

using namespace std;

int main()

{

int number=0141;

printf("%c",number);

return 0;

}

Compilation error

1100001

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User Explanation

Once you attempt the question then PrepInsta explanation will be displayed.

0141 means that in the octal base, %c will print the character value.

Octal value of a is 141 so, a get printed.

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Question 9

Time: 00:00:00

What is the output of below code?

#include <iostream>

using namespace std;

int main()

{

int number=0141;

printf("%c",number);

return 0;

}

Compilation error

1100001

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PrepInsta Explanation
User Explanation

Once you attempt the question then PrepInsta explanation will be displayed.

0141 means that in the octal base, %c will print the character value.

Octal value of a is 141 so, a get printed.

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Question 10

Time: 00:00:00

What is the output of below code?

#include <iostream>

using namespace std;

int main()

{

int number=0141;

printf("%c",number);

return 0;

}

Compilation error

1100001

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PrepInsta Explanation
User Explanation

Once you attempt the question then PrepInsta explanation will be displayed.

0141 means that in the octal base, %c will print the character value.

Octal value of a is 141 so, a get printed.

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