Beginner C++ small question?

Hey!


I wrote this code for the question: Write a function called findReading. It should be a Reading structure reference variable as its parameter. The function should ask the user to enter values for each member of the structure. My code:


struct TempScale


{


double fahrenheit;


double centigrade;


};


struct Reading


{


int windSpeed;


double humidity;


tempScale temperature;


};





void findReading(struct Reading)


{


cout %26lt;%26lt; "Enter the following values:\n";


cout %26lt;%26lt; "Wind speed: ";


cin %26gt;%26gt; Reading.windSpeed;


cout %26lt;%26lt; "Humidity: ";


cin %26gt;%26gt; Reading.humidity;


cout %26lt;%26lt; "Fahrenheit temperature: ";


cin %26gt;%26gt; TempScale.temperature.fahrenheit;


cout %26lt;%26lt; "Centigrade temperature: ";


cin %26gt;%26gt; TempScale.temperature.centigrade;


}


}





Can u help me fix it or change it? Thanks

Beginner C++ small question?
ur parameter should be





void findReading(Reading input_readings)





input_readings is just a variable name, which is how you are going to refer to the Reading object in your function. Structs are types. So u can use them as types, just like





void some_function(int x), you need void findReadings(Reading name_of_variable);





so if you use the name input_readings, your input it going to look like this





the %26amp; passes it the memory address of the variable so you can directly work on it





void findReading( Reading%26amp; input_reading)


{


cout %26lt;%26lt; "Enter the following values:\n";


cout %26lt;%26lt; "Wind speed: ";


cin %26gt;%26gt; input_reading.windSpeed;





ect. change all of them





also, TempScale is an object inside of readings, so it would look like





cin %26gt;%26gt; input_reading.temperature.Fahrenheit


cout %26lt;%26lt; "Centigrade temperature: ";


cin %26gt;%26gt; input_reading.temperature.centigra...


}











and u have an extra } at the end








hope that helps ya.
Reply:Your problem is the declaration of the function findReading. You write:


void findReading(struct Reading)


{





What you need is:


void findReading(struct Reading %26amp; value)


{


...


cin %26gt;%26gt; value.windSpeed;


...


cin %26gt;%26gt; value.humidity;


// etc





Now, there is a big difference when you use the '%26amp;' in the function signature.





void findReading(struct Reading value)


then you call:


fincReading(reading);


In this case, the function receives a copy of the caller's parameter. Any changes are local to the function are not reflected to the caller.





void findReading(struct Reading %26amp; value)


then you call:


fincReading(reading);


Here, the function instead of receiving a copy, gets a reference. A reference is an object that says "I'm really over there." (it's a hidden pointer). Any changes you make to the reference are made to the original. When you call the function, the parameter you pass can be modified by the function.
Reply:struct TempScale


{


double fahrenheit;


double centigrade;


};





struct Reading


{


int windSpeed;


double humidity;


tempScale temperature;


};





struct Reading is a structure, you need to declare a variable of type struct Reading





In C++ you can define a function with just the type to pass in but this parameter is ignored.


Avoids the unused parameter warning.


void f(int) {}


Call this function as follows: f(0);


The function 'f' doesn't use the value passed in.


To use the value, just change the declaration to


void f(int f)


{if (f) return;} %26lt;%26lt;- Here i use the value.





void findReading(struct Reading %26amp;r)


{


cout %26lt;%26lt; "Enter the following values:\n";


cout %26lt;%26lt; "Wind speed: ";


cin %26gt;%26gt; r.windSpeed;


cout %26lt;%26lt; "Humidity: ";


cin %26gt;%26gt; r.humidity;


tempScale t;


cout %26lt;%26lt; "Fahrenheit temperature: ";


cin %26gt;%26gt; t.fahrenheit;


// You can calculate degree Celsius


t.centigrade = (t.fahrenheit - 32.0) * 5.0 / 9.0;


r.temperature = t;


}