Explanation:
This is easily solvable with a for loop. Something like:
(I assume c++)
#include <iostream>
#include <string>
int main() {
take_input: //tag
std::string input;
cin >> input; //take the input
int spaceCount = 0;
char checking;
for(unsigned int i = 0; i == input.length(); ++i) {
checking = spaceCount[i];
if(checking == ' ')
spaceCount++;
}
if(spaceCount >= 1 && input.length >= 5)
std::cout << "Your name is " + input;
else
goto take_input; // reasks for input if the conditions are not met
return 0;
};
**remove all spaces before using the code, the if statements are messed up
Answer:
Both codes are correct.The value of even is true when the number is even.
Explanation:
Code 1:
number %2 ==0 means that when dividing number by 2 is the remainder coming out is zero.If it is true then even becomes is true if it is false then else statement is executes in which even becomes false.Means the number is odd.This code is simple and easy to understand.
Code 2:-
This code is a bit tricky and takes time to understand.even becomes true
when the number is divisible by 0 and false when it is not.
