```
#!/usr/local/bin/python3
foo = float( input( "Enter a number: " ) )
if( foo < 0.0 ):
print( "negative" )
elif( foo > 0.0 ):
print( "positive" )
else
print( "zero" )
exit( 0 )
```
Answer:
Destroying all identifiers connected to the data.
Explanation:
Identifiers are your virtual location when navigating, if you destroy your 'Virtual ID', nobody could identify you. Also, you could hide your identifiers, in this case, there's a small risk.
Answer:
The answer is "Option d"
Explanation:
The wardrobe engineering consists of choosing clothes that are so simple in design and better in the style that they are acceptable to the existing show-up and perfect for the future years. This design Originally coined Dressing for Confidence author John Molloy, which describe that clothes and accessories can also be used to create a certain picture, and wrong choices can be described as follows:
- In option a, It's right, that we know about current fashions, but selecting suitable career apparel is not important, that's why it is incorrect.
- In option b, It is wrong because whatever you wear is not to make choices of any sort.
- In option c, It is wrong because it can't provide promotion.
Google c++ tutorial. Cplusplus and Tutorials Point both have some good resources
Answer:
- public static String bothStart(String text1, String text2){
- String s = "";
-
- if(text1.length() > text2.length()) {
- for (int i = 0; i < text2.length(); i++) {
- if (text1.charAt(i) == text2.charAt(i)) {
- s += text1.charAt(i);
- }else{
- break;
- }
- }
- return s;
- }else{
- for (int i = 0; i < text1.length(); i++) {
- if (text1.charAt(i) == text2.charAt(i)) {
- s += text1.charAt(i);
- }else{
- break;
- }
- }
- return s;
- }
- }
Explanation:
Let's start with creating a static method <em>bothStart()</em> with two String type parameters, <em>text1 </em>&<em> text2</em> (Line 1).
<em />
Create a String type variable, <em>s,</em> which will hold the value of the longest substring that both inputs start with the same character (Line 2).
There are two possible situation here: either <em>text1 </em>longer than<em> text2 </em>or vice versa. Hence, we need to create if-else statements to handle these two position conditions (Line 4 & Line 13).
If the length of<em> text1</em> is longer than <em>text2</em>, the for-loop should only traverse both of strings up to the length of the <em>text2 </em>(Line 5). Within the for-loop, we can use<em> charAt()</em> method to extract individual character from the<em> text1</em> & <em>text2 </em>and compare with each other (Line 15). If they are matched, the character should be joined with the string s (Line 16). If not, break the loop.
The program logic from (Line 14 - 20) is similar to the code segment above (Line 4 -12) except for-loop traverse up to the length of <em>text1 .</em>
<em />
At the end, return the s as output (Line 21).
