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Sveta_85 [38]
2 years ago
13

Which is the preferred means of trapping user input for errors?

Computers and Technology
1 answer:
bogdanovich [222]2 years ago
8 0

Escaping is the preferred method used for trapping user inputs for errors.

<h3>HTML (HyperText Markup Language)</h3>

Most web pages users interact with today are designed based on the rules of HTML. In a web-based environment, a preferred means developers use to trap user input for errors is the escape string. This typically involves using mysql_real_escape_string() when coding.

You can learn more about escaping here brainly.com/question/14821549

#SPJ12

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This program will calculate the rise in ocean levels over 5, 10, and 50 years, Part of the program has been written for you. The
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Answer:

Here is the C++ program:

#include <iostream>  //to use input output functions

using namespace std;  //to identify objects cin cout

int main(){  //start of main method

     

   double risingLevel;  //declares a double type variable to hold rising level

   cin>>risingLevel;  //reads the value of risingLevel from user

   

   cout<<"Level: "<<risingLevel<<endl;  //displays the rising level

   cout << "Years: 5, Rise: " << risingLevel * 5<<endl;  //displays the rise in ocean level over 5 years

   cout << "Years: 10, Rise: " << risingLevel * 10<<endl;  //displays the rise in ocean level over 10 years

   cout << "Years: 50, Rise: " << risingLevel * 50<<endl;  //displays the rise in ocean level over 50 years

}

Explanation:

The program works as follows:

Lets say the user enters rising level 2.1 then,

risingLevel = 2.1

Now the first print (cout) statement :

cout<<"Level: "<<risingLevel<<endl;  prints the value of risingLevel on output screen. So this statement displays:

Level: 2.1          

Next, the program control moves to the statement:

   cout << "Years: 5, Rise: " << risingLevel * 5<<endl;

which computes the rise in ocean levels over 5 years by formula:

risingLevel * 5 = 2.1 * 5 = 10.5

It then displays the computed value on output screen. So this statement displays:

Years: 5, Rise: 10.5                                                                                                                          Next, the program control moves to the statement:

   cout << "Years: 10, Rise: " << risingLevel * 10<<endl;

which computes the rise in ocean levels over 10 years by formula:

risingLevel * 10 = 2.1 * 10 = 21

It then displays the computed value on output screen. So this statement displays:

Years: 10, Rise: 21                                                                                      

Next, the program control moves to the statement:

   cout << "Years: 50, Rise: " << risingLevel * 50<<endl;

which computes the rise in ocean levels over 50 years by formula:

risingLevel * 50 = 2.1 * 50 = 105

It then displays the computed value on output screen. So this statement displays:

Years: 50, Rise: 105                                                                                                                          Hence the output of the entire program is:

Level: 2.1                                                                                                                     Years: 5, Rise: 10.5                                                                                                           Years: 10, Rise: 21                                                                                                            Years: 50, Rise: 105

The screenshot of the program and its output is attached.

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