Question

1. An integer in C (int) is represented by 4 bytes (1 byte = 8 bits). Find the largest integer that can be handled by C. Verify that number by computer (try to printf that number, and printf that number +1). No programming is required for this. Hint: (a) Conversion from binary to decimal: 3 2 1 0 2 1101 =1 2 + 0 2 +1 2 +1 2 = 8 + 0 + 2 +1 10 =11 (b) If an integer is represented by 1 byte (8 bits), the maximum integer number is 1111111 (Seven 1's as one bit must be reserved for a sign). 6 5 0 2 1111111 =1 2 +1 2 ++1 2 1 2 1 7 =  − =128 −1 (c) The number looks like a telephone number in Dallas.

Answer 1

Answer:

Check the explanation

Explanation:

In C, int requires 4 butes to sotre a integer number. that means it requires 4*8 = 32 bits. The maximum number can be stored using 32 bits is

$(11111111111111111111111111111111)_2$

The first digit is used for sign. decimal equivalent to this number can be computed as

$(1111111111111111111111111111111)_2= 1\times2^{30}+1\times2^{29}+...+1\times2^0$

=$1\times2^{31}-1$

= $2147483647-1=2147483646$

That means int can store a up to 2147483646.

                      Testing with C code

#include<stdio.h>

int main()

{

  int a = 2147483647; // a with max number

  printf("%d\n",a);// printing max number

  a = a+1;// adding one to the number

  printf("%d\n",a);// printing the number after adding one

  return 0;

}

                THE OUTPUT

$ ./a.out

2147483647

-2147483648