Answer:
The answer is "Option B"
Explanation:
Conversion of type is a way of changing an entity from one data type to another. It is used to make the variable process properly by a function in computer coding.
In the C++ language, the result of the multitype value added or any other option is the result between those variables as a large type at that time, that's why choice b is correct.
Answer:
True
Explanation:
IPv6 Is a later version of IP addresses, used to solve the problem of the limited number of IPv4 addresses in the network.
Just like IPv4, IPv6 can also is configured to a device statically and dynamically. Dynamic IPv6 configuration could be a stateless autoconfiguration SLAAC, a stateless DHCPV6 or a stateful DHCPV6.
The IPv6 address is configured with a prefix and a prefix length and a EUI generated 64 bit interface or a random interface id by the device.
Answer:
E)nXML
Explanation:
XML is a short for Extensible Markup Language. It is a markup language like the Hypertext markup language (HTML) and are both used in the development of web applications. However while the HTML describes the content of the Web page that is the graphics, images and videos and how they are displayed, the XML handles the description of data and information formats, their storage and the transportation and sharing over the internet.
Answer: eCommerce is to reach the more and right customers at the right time so that more orders can be placed and in turns, high revenue can be generated.
Explanation:
Answer:
#include <iostream>
#include <string>
#include <stack>
#include <math.h>
using namespace std;
int main() {
string s;
double n=0;
int position=0;
stack<int> wholeNumbers;
cout<<"Enter a decimal number:";
cin>>s;
string::iterator counter = s.begin();
while(*counter!='.' && counter!=s.end()){
wholeNumbers.push(*counter-48);
counter++;
position=position+1;
}
for(int i=0;i<position;i++){
n=n+(wholeNumbers.top()*pow(10,i));
wholeNumbers.pop();
}
position=-1;
if(counter!=s.end()){
counter++;
}
while(counter!=s.end()){
n=n+((*counter-48)*pow(10,position));
position=position-1;
counter++;
}
cout<<n;
}
Explanation:
- Inside the while loop, push the push a number to the wholeNumbers stack by subtracting it with 48.
- Increment the counter and position variable by 1 inside the while loop.
- Count the number of digit, push each digit to top of stack and find the end of the number,
- Run a for loop up to the value of position variable and pop a value from the wholeNumbers stack.
