Some of the disadvantages of using layered protocols are the redundancy and overall lower performance.
Reimplementing everything from remote and wired to connectionless and association situated application correspondence, with each ringer and shriek of BGP, MPLS, multicast steering and so on would be a gigantic endeavor and potentially unmanageable in this solid across the board convention. I could envision specialty utilize situations where most extreme execution is imperative most importantly where an exceptionally basic framework might be wanted.

8 0

3 years ago

How will 5G technology affect the current classroom? What are some concerns about 5G technology? Is 5G technology worth the risk

notsponge [240]

Answer:Death is a risk of 5g towers

Explanation:Someone caught a video of birds falling dead out of the sky because of the radiation from the 5g towers. There is enough radiation to make your eyes melt before you get close to it.

4 0

3 years ago

Matching parentheses. An math expression may have a number of parentheses like (, ), [, ], { and }. Each openning parenthesis (,

Solnce55 [7]

C++ Code

#include
using namespace std;

bool match(const char exp[],const int s)
{
// declare a character array to perform stack operations
char stack[s];

// declare top and initialize to -1 and flag to 0
int top=-1,i,flag=0;

// visit all characters in the expression string
for(i=0;i {
// if the character is [ or ( or { then push it into stack
if(exp[i]=='[' || exp[i]=='(' || exp[i]=='{')
{
top++;
stack[top]=exp[i];
}
// if the character is ] or ) or } then check conditions
else if(exp[i]==']' || exp[i]==')' || exp[i]=='}')
{
// check stack is empty or not
if(top!=-1)
{
// check all possible failure conditions
if(exp[i]==')' && (stack[top] == '{' || stack[top]=='['))
{
flag = 1;
break;
}
else if(exp[i]==']' && (stack[top] == '{' || stack[top]=='('))
{
flag = 1;
break;
}
else if(exp[i]=='}' && (stack[top] == '(' || stack[top]=='['))
{
flag = 1;
break;
}
top--;
}
else
{
flag=1;
break;
}
}
}
// after visiting all characters of expression string check if stack is not empty and flag is 1. if any one of the condition is true return false. otherwise return true

if(top>=0 || flag==1)
return false;
else
return true;
}

int main()
{

// declare character array to store expression
char exp[10000];
cout<<"Enter an Expression"<
// read expression from user
cin.getline(exp, 10000);
int s=0;

// find the length of the expression string
for(int i=0;exp[i]!='\0';i++)
{
s++;
}

// call the match function
bool status = match(exp,s);

// print the result based on value returned by match() function
if(status == 1)
cout<<"true"< else
cout<<"false"<
}

Sample Input/Output is attached

3 0

2 years ago