Answer:
For 32 bits Instruction Format:
OPCODE DR SR1 SR2 Unused bits
a) Minimum number of bits required to represent the OPCODE = 3 bits
There are 8 opcodes. Patterns required for these opcodes must be unique. For this purpose, take log base 2 of 8 and then ceil the result.
Ceil (log2 (8)) = 3
b) Minimum number of bits For Destination Register(DR) = 4 bits
There are 10 registers. For unique register values take log base 2 of 10 and then ceil the value. 4 bits are required for each register. Hence, DR, SR1 and SR2 all require 12 bits in all.
Ceil (log2 (10)) = 4
c) Maximum number of UNUSED bits in Instruction encoding = 17 bits
Total number of bits used = bits used for registers + bits used for OPCODE
= 12 + 3 = 15
Total number of bits for instruction format = 32
Maximum No. of Unused bits = 32 – 15 = 17 bits
OPCODE DR SR1 SR2 Unused bits
3 bits 4 bits 4 bits 4 bits 17 bits
Answer:
#include <iostream>
using namespace std;
int main() {
int k;
double d;
string s;
cin >> k >> d >> s;
cout << s << " " << d << " " << k << "\n" << k << " " << d << " " << s; }
Explanation:
k is int type variable that stores integer values.
d is double type variable that stores real number.
s is string type variable that stores word.
cin statement is used to take input from user. cin takes an integer, a real number and a word from user. The user first enters an integer value, then a real number and then a small word as input.
cout statement is used to display the output on the screen. cout displays the value of k, d and s which entered by user.
First the values of k, d and s are displayed in reverse order. This means the word is displayed first, then the real number and then the integer separated again by EXACTLY one space from each other. " " used to represent a single space.
Then next line \n is used to produce a new line.
So in the next line values of k, d and s are displayed in original order (the integer , the real, and the word), separated again by EXACTLY one space from each other.
The program along with the output is attached.
Answer:
For this wild card mask 0.0.15.255 the ACE IP address will be 172.16.47.254
Explanation:
ACL is the access control list that is used to enlist the ip addresses that allowed or restricted to access the network. ACE is an IP address from the list ACL that has all rules and regulations related to access of network. The ACE could be in the range of IP address in ACL. ACL can be calculated with the help of initial IP address adding with wild card mask.
So
Initial IP address is = 172.16.32.0
Wild card mask =0.0.15.255
by adding above values we can find the last IP address of ACL.
after addition
Final IP address is = 172.16.47.255
The options that are available with question, Only option between the range is 172.16.47.254. So we can say that This is the only ACE IP address in options.
the reason HTML seems to work even if it has syntax errors is due to browser having built in ways to parse the code meaning it will still show but most likely look way different then you would want.
the code may turn pink due to syntax errors
Answer:
方法/步骤
右键单击以选择此计算机，然后有一个菜单来选择属性。
选择后，打开属性面板以找到我们的高级系统设置。
打开后，我们在系统设置下找到高级设置。
打开后，我们看到下面有一个环境变量选项。
打开后，我们在右下角看到一个删除选项。
单击删除到，在右下角后单击删除，然后单击确定的选项。 ...
本文未经授权摘自百度经验
