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Serhud [2]
3 years ago
5

Assume that x is a double variable that has been initialized. Write a statement that prints it out, guaranteed to have a decimal

point, but without forcing scientific (also known as exponential or e-notation).
Computers and Technology
1 answer:
Feliz [49]3 years ago
6 0

Answer:

cout <<showpoint << x; is  the statement which  prints the decimal point,        but without forcing scientific.

Explanation:

The showpoint function in c++ print the decimal point number without forcing scientific.The showpoint function set  the showpoint format flag for the str stream in c++.

Following are the program in c++

#include<iostream> //header file

using namespace std; // namespace

int main() // main function

{

   double x=90.67; // double variable

   cout <<showpoint << x; // display x  without forcing scientific

  return 0;

}

Output

90.6700

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5 0
3 years ago
Write a C++ program that overloads a function named LinearSearch that searches an array of data of either integer data type, flo
frutty [35]

Explanation:

#include <iostream>

#include <vector>

#include <iomanip>

using namespace std;

/*

Overloading means equal function name with different call prototype each.

It's a useful tool that lets you think once and write similar role code twice or more times keeping equal function name.

Please look at lines 12-14: they are overloaded call prototype versions of LinearSearch with identical mission (to know if a value is present in a set)

*/

//#define GENERICS

#ifndef GENERICS                                                                // Conditional compilation statement

bool LinearSearch(vector<int>   ,int);

bool LinearSearch(vector<float> ,float);

bool LinearSearch(vector<double>,double);

#else                                                                           // Conditional compilation statement (uncomment line 10 statement)

template <typename genType>

bool LinearSearch(vector<genType> v, genType value);                            // LinearSearch (generic type parameters version)

#endif

int main(int argc, char *argv[]) {                                              // Test code follows:    

vector<int> vInt    {1,                     -2,     3,          -4,     5};     // Test data (int)

vector<float> vFlo  {1.23456f,              -2.35f, 3.1416f,    -4.7f,  5.55f}; // Test data (float)

vector<double> vDou {1.234567890123456789,  -2.35,  3.1415927,  -4.7,   5.55};  // Test data (double)

cout <<3 << "        is " << (LinearSearch(vInt,3)      ?"":"not ") << "in: {";for(auto x:vInt) cout << setw(11) << x;cout << "}\n";

cout <<3.1416f<< "   is " << (LinearSearch(vFlo,3.1416f)?"":"not ") << "in: {";for(auto x:vFlo) cout << setw(11) << x;cout << "}\n";

cout <<3.14   << " is " << (LinearSearch(vDou,3.1416) ?"":"not ") << "in: {";for(auto x:vDou) cout << /*setprecision(9)<<*/setw(11) << x;cout << "}\n";

return 0;

}

#ifndef GENERICS

// Overloaded function code follows:

bool LinearSearch(vector<int> v, int value) {                                   // LinearSearch (integer parameters version)

bool found = false;                                                             // Until now we haven't found the value we are searching for,

for(auto x:v) found = found || x == value;                                      // but if it is here found becomes true.

return(found);

}

bool LinearSearch(vector<float> v, float value) {                               // LinearSearch (float parameters version)

bool found = false;

for(auto x:v) found = found || x == value;

return(found);

}

bool LinearSearch(vector<double> v, double value) {                             // LinearSearch (double float parameters version)

bool found = false;

for(auto x:v) found = found || x == value;

return(found);

}

/*

So far, so good. It works! But, despite there was little effort involved, redundant code remains.

Is there any solution? Yes! It's called generic programming.

*/

#else

/*

Consider the following code:

   - Template statement at first accepts a generic type as a parameter (genType)

   - genType is used for both parameters (array and value to search)

   - Template statements located after generic version of LinearSearch instructs compiler to generate each type version

Conclusions:    

   - Compared with overloaded versions generic code is more elegant and avoids completely redundancy

   - It enhances overloading while keeping its advantages

*/

template <typename genType>

bool LinearSearch(vector<genType> v, genType value) {   // LinearSearch (generic type parameters version)

bool found = false;                                     // Until now we haven't found the value we are searching for,

for(auto x:v) found = found || x == value;              // but if it is here found becomes true.

return(found);

}

template bool LinearSearch(vector<int>   ,int);

template bool LinearSearch(vector<float> ,float);

template bool LinearSearch(vector<double>,double);

#endif

Download cpp
3 0
3 years ago
Suppose you are given a sequence that is described by a formula, starting at index n=0. If you need to change the starting index
Arisa [49]

Answer:

n+1

Explanation:

Given

n = 0 --- starting index

Required

Change the starting index to n = 1

We have:

n = 0

To change the starting index to k, we simply rewrite as:

n+k

<em>In this case; k=1; so, the starting index will be: </em>n+1<em />

3 0
3 years ago
What are three limitations of computer?​
Lady bird [3.3K]

Answer:

Explanation:

Three limittaions of computer are:

It requires reqular power supply to operate .

It needs instructions to perform a task.

It cannot memorize and recall.

8 0
3 years ago
QUESTION 1 Someone may choose to own a car instead of leasing because: AThe car can be resold later to make some money back. BTh
Lorico [155]
This is not a computers and technology question but ill still answer it and the answer to this that i think it is ...A ...but i may be wrong
3 0
3 years ago
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