Answer / sandeep gupta

A very interesting question where u need knowledge of
Computer architecture also.... :) :) :) :)
The floating point(FP) no. are stored differently in memory
as: mantissa × 2^exponent [M× 2^E]. M is 23 bit long and E
is 8 bit. 1 bit is for sign of number. The format is :
SEEE EEEE EMMM MMMM MMMM MMMM MMMM MMMM
Now binary of 5.375 is 101.011 which can be written in
normalized form(which has single 1 before decimal point) as
1.01011 × 2^2. This 1 and point(.) is always neglected while
storing the no. in registers. so mantissa is 01011 or
010110000000000000000000 × 2^E where E is 8 bit(0 to 255).
The actual value of the exponent is calculated by
subtracting 127 from the stored value (0 to 255) giving a
range of –127 to +128.
So here we need E=2 which we can get from 129(129-127=2)
whose binary is: 10000001.
Now above format becomes:
0100 0000 1010 1100 0000 0000 0000 0000 which is 40ac0000 in
hex. So stored format is: 00->00->ac->40

Answer / govind verma

i think mr sandeep this value stored in this manner
0100 0000 1010 1100 0000 0000 0000 0000 which is 40ac0000


float takn 4 byte in memory in program p=(char*)&a casting this to char type nw p(it char ponter its point to i byte value) point to lower byte of the a(5.375) which is 0000 0000
p[0]-> contain the fist lower byte data 00(bcoz print bye the hexa formate specifier withe 2 width)
p[1]->contain 00
p[2]-> contain ac
p[3]->contain 40

then output becom ...0000ac40

There are 21 people in a room. They have to form groups of 3 people each. How many combinations are possible? Write a C program to print the same.

1 Answers   TCS,

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main() { int *j; { int i=10; j=&i; } printf("%d",*j); }

9 Answers   HCL, Wipro,

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#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); }

1 Answers  

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2 Answers  

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# include <stdio.h> int one_d[]={1,2,3}; main() { int *ptr; ptr=one_d; ptr+=3; printf("%d",*ptr); }

1 Answers  

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A program that will create a movie seat reservation. The program will display the summary seats and its status. The user will be ask what seat no. to be reserved, then it will go back again to the summary to display the updated seat status. If the seat no. is already reserved then it will prompt an error message. And also if the input seat no is out of range then it will also prompt an error message. The program is continuously running. Termination of the program will depends on how the programmer will apply. Sample output: Movie Seats Reservation Summary of Seats: Seat 1: Available Seat 2: Available Seat 3: Available Seat 4: Available Seat 5: Available Enter seat no. (Press 0 to terminate Or the assigned seat capacity) : 1 Movie Seats Reservation Summary of Seats: Seat 1: Reserve Seat 2: Available Seat 3: Available Seat 4: Available Seat 5: Available Enter seat no. (Press 0 to terminate Or the assigned seat capacity) : 6 The Seat no. is out of range! Movie Seats Reservation Summary of Seats: Seat 1: Reserve Seat 2: Available Seat 3: Available Seat 4: Available Seat 5: Available Enter seat no. (Press 0 to terminate Or the assigned seat capacity) : 1 The Seat no. is already reserved! Movie Seats Reservation Summary of Seats: Seat 1: Reserve Seat 2: Available Seat 3: Available Seat 4: Available Seat 5: Available Enter seat no. (Press 0 to terminate Or the assigned seat capacity) : 0 GoodBye... Thank You!!!

0 Answers  

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Write a c program to search an element in an array using recursion

1 Answers   Wipro,

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main() { int i, j, *p; i = 25; j = 100; p = &i; // Address of i is assigned to pointer p printf("%f", i/(*p) ); // i is divided by pointer p } a. Runtime error. b. 1.00000 c. Compile error d. 0.00000

3 Answers   HCL,

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void main() { int x,y=2,z; z=(z*=2)+(x=y=z); printf("%d",z); }

4 Answers  

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main() { int i=5,j=6,z; printf("%d",i+++j); }

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Write a procedure to implement highlight as a blinking operation

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find A^B using Recursive function

2 Answers  

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