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katrin [286]
3 years ago
10

If the processor has 32 address lines in its address bus; i.e. 32-bit address range, what is its address space or range; i.e. wh

at is the smallest and largest address that can be represented
Computers and Technology
1 answer:
Marysya12 [62]3 years ago
4 0

Answer:

The answer is "0 to 4294967295"

Explanation:

Given bit:

32 bits

Calculating smallest and largest bits:

In any processor, it works only on the 0 and 1 bit and it also stores 0 and 1 values. so, the smallest address in bit value is= 0 and the largest bit value can be defined as follows:

largest address value in bits:

\Longrightarrow 2^{32}-1\\\\\Longrightarrow 4294967296 -1\\\\\Longrightarrow 4294967295\\\\

smallest address= 0

largest address = 4294967295

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Answer:

Following are the statement in C++ language :

#include <iomanip> // header file  

using namespace std; // namespace

int main() // main method  

{

   int x1,x2,x3,x4,x5; // variable declaration

cout<<" Enter the 5 sucessive integer:";

cin>>x1>>x2>>x3>>x4>> x5; // taking 5 input from user

cout<<right;

// prints each  of  its own line and form a right-justified  

cout<<setw(5)<< x1 << "\n" << setw(5) << x2 << "\n" <<setw(5) << x3 << "\n" << setw(5) << x4 << "\n" << setw(5) << x5 << "\n";

return(0);    

}

Output:

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23

445

6

8

 

  45

  23

 445

   6

   8

Explanation:

Description of program is given below  

  • Declared a 5 integer value x1,x2,x3,x4,x5 respectively.
  • Read the 5 integer value from user by using cin funtion.
  • Finally Print these 5 value in its own line and form a right-justified by using setw() function on it
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Answer:

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Explanation:

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