Menu
So I was writing this simple program to calculate the day of any date using the Gaussian algorithm found here.
#include <iostream> using namespace std; //Using the Gaussian algorithm int dayofweek(int date, int month, int year ){ int d=date; if (month==1||month==2) {int y=((year-1)%100);int c=(year-1)/100;} else {int y=year%100;int c=year/100;} int m=(month+9)%12+1; int product=(d+(2.6*m-0.2)+y+y/4+c/4-2*c); return product%7; } int main(){ cout<<dayofweek(19,1,2054); return 0; } It's a very simple program and what's even more puzzling is the output.
:In function dayofweek(int, int, int)’: :19: warning: unused variable ‘y’ :19: warning: unused variable ‘c’ :21: warning: unused variable ‘y’ :21: warning: unused variable ‘c’ :23: error: ‘y’ was not declared in this scope :25: error: ‘c’ was not declared in this scope It says that my variable is unused but then says that it isn't declared? Could anyone please tell me whats wrong.
806
To resolve this error, a first method that is helpful would be declaring the function prototype before the main() method. So, we have used the function prototype before the main method in the updated code. When we have compiled the code, it throws no exceptions and runs properly.
In case of getting the Serial1 was not declared in this scope error, chances are your Arduino has no Serial1. Assuming that you use Arduino Uno, you need to comment the Serial1 and un-comment the SoftwareSerial part in order to solve the problem.
When you declare a program element such as a class, function, or variable, its name can only be "seen" and used in certain parts of your program. The context in which a name is visible is called its scope. For example, if you declare a variable x within a function, x is only visible within that function body.
The error 'was not declared in this scope' generally occurs when a variable or function is not accessible to its call. A global variable in Arduino is the variable that is declared outside any function, including the setup() and loop() functions.
The scope of a variable is always the block it is inside. For example if you do something like
if(...) { int y = 5; //y is created } //y leaves scope, since the block ends. else { int y = 8; //y is created } //y leaves scope, since the block ends. cout << y << endl; //Gives error since y is not defined. The solution is to define y outside of the if blocks
int y; //y is created if(...) { y = 5; } else { y = 8; } cout << y << endl; //Ok In your program you have to move the definition of y and c out of the if blocks into the higher scope. Your Function then would look like this:
//Using the Gaussian algorithm int dayofweek(int date, int month, int year ) { int y, c; int d=date; if (month==1||month==2) { y=((year-1)%100); c=(year-1)/100; } else { y=year%100; c=year/100; } int m=(month+9)%12+1; int product=(d+(2.6*m-0.2)+y+y/4+c/4-2*c); return product%7; } If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
