The syntax is truly dependent on the programming language, as different languages have different API's.
But, the most standard form of accessing an array length is via a method or property by dot notation.
For example -
for(int i = 0; i <= m_Array.Length; i++) {...}
This code is from C#, using the Length property.
I haven’t read it, but just by the title it was probably talking about an incredible machine that was new to people.
Answer:
Option 4 is the correct answer to the above question.
Explanation:
In the Linux operating system, The etc folder is listed on the root directory. When any user wants to go in the root directory of the Linux operating system from anywhere, then he needs to follow the command of the name of the folder where he wants to go with the '/' symbols. The above question states that the user wants to go to the etc folder of the root directory, then he needs to press the command '/etc' with the help of cd command because cd command is used to open any folder. So the option 4 is the correct, while the other is not because--
- Option 1 states about cd which gives the error.
- Option 2 states about cd /home/mary/etc which take the user to the etc folder which is in the mary folder.
- Option 3 states about the cd etc which is used to take the user on the etc folder which is in the current folder.
- Option 5 states about the cd\etc which gives the error.
Answer:
public class MagicSquare {
public static void main(String[] args) {
int[][] square = {
{ 8, 11, 14, 1},
{13, 2, 7,12},
{ 3, 16, 9, 6},
{10, 5, 4, 15}
};
System.out.printf("The square %s a magic square. %n",
(isMagicSquare(square) ? "is" : "is not"));
}
public static boolean isMagicSquare(int[][] square) {
if(square.length != square[0].length) {
return false;
}
int sum = 0;
for(int i = 0; i < square[0].length; ++i) {
sum += square[0][i];
}
int d1 = 0, d2 = 0;
for(int i = 0; i < square.length; ++i) {
int row_sum = 0;
int col_sum = 0;
for(int j = 0; j < square[0].length; ++j) {
if(i == j) {
d1 += square[i][j];
}
if(j == square.length-i-1) {
d2 += square[i][j];
}
row_sum += square[i][j];
col_sum += square[j][i];
}
if(row_sum != sum || col_sum != sum) {
return false;
}
}
return d1 == sum && d2 == sum;
}
}
