我的bool函数不断返回true，但我不确定为什么

Question

Im doing an excercise sheet to get an understanding of functions and I am currently working on the following question.

Write function prototypes for each of the following:

1. A function HasValue that may be passed a reference to an array, the size of the array and a search value. The function should return true if the search value exists in the array

In my code I have sent the contents of the array, the array size and the value to be searched in the array to the bool function.

In the function I compared the value to each element of the array using a for loop.

I then created a variable count in the function that will be incremented if the value matches any element in the array.

I then used an if else statment to return true if count is greater than 0 and false if count is equal to 0. The problem is however that the function is only returning true thus the output will always be "this number appears in the array"

Logically these steps seem correct to me but obviously there is a flaw somewhere that I cant see. I presume its just I do not have a decent understanding of Bool functions yet but if someone could explain where and why I'm going wrong it would be greatly appreciated in my learning process to understanding functions and c++.

#include <iostream>
#include <iomanip>
#include "stdafx.h"

using namespace std;

bool HasValue(int Array[], int size, int Value);

int main()
{
    int value;
    int Array[10]{ 3,5,6,8,9,1,2,14,12,43 };

    cout << "enter value you wish to search for in array " << endl;
    cin >> value;

    HasValue(Array, 10 , value);

    if (true)
        cout << "This number appears in the array " << endl;
    else
        cout << "This number does not appear in the array " << endl;

    return 0;
}

bool HasValue(int Array[], int size, int Value)
{
    int count = 0;

    for (int i = 0; i < size; i++)
    {
       if (Value == Array[i])
       {
           count++;
       }    
    }

    if (count > 0)
    {
        return true;
    }
    else
        return false;
}

Answer 1

You test code is the problem

HasValue(Array, 10 , value);
if (true)
    cout << "This number appears in the array " << endl;
else
    cout << "This number does not appear in the array " << endl;

This ignores the return value of HasValue and always prints "This number appears in the array" .

Answer 2

HasValue(Array, 10 , value);

This line of code executes the function but ignores the returned value. When a function returns a value, you need to assign it to a variable:

bool result = HasValue(Array, 10 , value);

Then if (true) does not have any reference to the returned value. The true inside the if will cause the first cout to always print. You will never see the output from the else . But once you have the return value in a variable, you can use it in the if :

if(result)

You can reduce this all to one line of code, if you want:

if(HasValue(Array, 10 , value))

Now the if statement will directly test the return value from HasValue() . In this particular case, combining the code into a single line seems reasonable. You must be careful doing this, though. When you combine too much into a single line, the code becomes more difficult to debug. You will need to find a balance between readability and convenience as you continue learning how to program.