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dont understand BMI-Body Mass Index code beginner "Programming 1"

Body Mass Index program console window version since im learning with Programming 1 course its essentially before object oriented stuff.
i cant seem to understand what im doing wrong as it says
Error CS0019 Operator '^' cannot be applied to operands of type 'double' and 'double
and i have no clue what im doing wrong and are completely lost! :(
namespace BMI_o_BMR
{
internal class Program
{
static void Main(string[] args)
{
double weight, height, bmi, bmr;
string gender; // Female f = kvinna, male m = man
Console.WriteLine("Vi ska räkna ut ditt BMI och BMR");
Console.WriteLine("Write your height");
weight = Convert.ToDouble(Console.ReadLine());
Console.WriteLine("Write Your weight");
height = Convert.ToDouble(Console.ReadLine());
Console.WriteLine("Write your age");
int age = Convert.ToInt32(Console.ReadLine());
if (height >= 0.50 && height <= 2.20 && weight >= 10 && weight <= 250 && age >= 18 && age <= 70)
{
bmi = (weight * 1.3) / height ^ 2.5;
Console.WriteLine(bmi);
}
Console.WriteLine("choose your Gender M for Male or F for Female");
}
}
}

In C# ^ is either a boolean logical XOR or bitwise logical XOR (C# operators and expressions).
Use Math.Pow:
Returns a specified number raised to the specified power.
bmi = (weight * 1.3) / Math.Pow(height, 2.5);

C# use of unassigned local variable on three variables

I am trying to create a code that allows me to calculate the price of two products however there are some changes throughout the code, (I want to aplly discounts according to the volume that will be bought. However when I make counts with three variables it show use of unassigned local variable...
static void Main(string[] args)
{
double peso_morango, peso_maca, total;
double totalmorango, totalmaca, final_real;
const double percentual = 0.1;
Console.WriteLine("Insira a quantidade de morangos, em kg: ");
peso_morango = Convert.ToInt32(Console.ReadLine());
Console.WriteLine("Insira a quantidade de maça, em kg:");
peso_maca = Convert.ToInt32(Console.ReadLine());
if (peso_morango <= 5)
{
totalmorango = 2.50 * peso_morango;
}
else if (peso_morango > 5)
{
totalmorango = 2.20 * peso_morango;
}
else if (peso_morango > 8)
{
totalmorango = 1.62 * peso_morango;
}
else if (peso_maca <= 5)
{
totalmaca = 1.62 * peso_maca;
}
else if (peso_maca > 5)
{
totalmaca = 2.20 * peso_maca;
}
else if (peso_maca > 8)
{
totalmaca = 1.35 * peso_maca;
}
else if (peso_maca + peso_morango > 8)
{
total = totalmorango + totalmaca;
final_real = (total) - (percentual * total);
Console.WriteLine(" O valor final é de {0} euros", final_real);
}
else if (totalmaca + totalmorango > 25)
{
total = totalmorango + totalmaca;
final_real = (total) - (percentual * total);
Console.WriteLine(" O valor final é de {0} euros", final_real);
}
else if (peso_maca + peso_morango < 8)
{
Console.WriteLine(" O valor final é de {0} euros", final_real);
}
else if (totalmorango + totalmaca < 25)
{
Console.WriteLine(" O valor final é de {0} euros", final_real);
}
Console.ReadKey();
}

I understand this is not a solution to your problem, but the only part of the code which will run is:
static void Main(string[] args)
{
double peso_morango, peso_maca, total;
double totalmorango, totalmaca, final_real;
const double percentual = 0.1;
Console.WriteLine("Insira a quantidade de morangos, em kg: ");
peso_morango = Convert.ToInt32(Console.ReadLine());
Console.WriteLine("Insira a quantidade de maça, em kg:");
peso_maca = Convert.ToInt32(Console.ReadLine());
if (peso_morango <= 5)
{
totalmorango = 2.50 * peso_morango;
}
else if (peso_morango > 5)
{
totalmorango = 2.20 * peso_morango;
}
Console.ReadKey();
}
This is because EITHER peso_morango <= 5 OR peso_morango > 5 will be true.
I don't think "else if" does what you expect it to do.

To make such mistake means that you're a beginner in programming and you'll probably do with some explanation to what is happening.
We gonna use int for this example.
When you declare variable like this int x; that is basically say hey i need some memory allocate some for me, How much of memory depend on the int
Now if you ever done some C++ or any unmanaged programming languages (Means the compiler doesn't do half of the work for you), you would understand that
int x; in memory will actually be holding some values from previous operations we don't know about, So you'll have to explicitly clean it from that garbage.
In C# that is basically the same, however here we have the compiler complaining to help you out, since you want some memory it needs to be fresh so you'll have to zero it out int x = 0;
So you'll have to zero it out or assign it a value you want to start with, you can use negative values too if you want to keep the zero, depending on the Valuetype int or uint etc....

totalmorango and totalmaca are not initialized, they are just declared. You need to initialize your variables with proper Initial values first before you use it.
By looking into your code, you are using them in addition operation.
assuming if any if elseif are not satisfy (where you assign value to them), you need them to be 0. (also it is unit value for addition operation)
replace your line
double totalmorango, totalmaca, final_real;
with
double totalmorango = 0, totalmaca = 0, final_real = 0;
Update
however whatever error you are facing will be solved by above steps, but as mentioned by others too. your checking ,
else if (peso_morango > 5)
is not appropriate if you want to check if value of peso_morango is greater than 5 but also if you want that to be less or equal to 8.
you should be checking else if (peso_morango > 5 && peso_morango <= 8)
And also you should be having separate condition checking for peso_maca . like,
if (peso_maca <= 5)
{
totalmaca = 1.62 * peso_maca;
}
else if (peso_maca > 5 && peso_maca <= 8)
{
totalmaca = 2.20 * peso_maca;
}
else// no need to check else part "peso_maca > 8" as it always be true
{
totalmaca = 1.35 * peso_maca;
}
and same way, separate condition checking for later part too.
By doing this, you will not even need to do,
double totalmorango = 0, totalmaca = 0, final_real = 0;
as suggested in first part of answer.

x++ operator for playing gif animations: How do it work together in this example? [duplicate]

I have next code
int a,b,c;
b=1;
c=36;
a=b%c;
What does "%" operator mean?

It is the modulo (or modulus) operator:
The modulus operator (%) computes the remainder after dividing its first operand by its second.
For example:
class Program
{
static void Main()
{
Console.WriteLine(5 % 2); // int
Console.WriteLine(-5 % 2); // int
Console.WriteLine(5.0 % 2.2); // double
Console.WriteLine(5.0m % 2.2m); // decimal
Console.WriteLine(-5.2 % 2.0); // double
}
}
Sample output:
1
-1
0.6
0.6
-1.2
Note that the result of the % operator is equal to x – (x / y) * y and that if y is zero, a DivideByZeroException is thrown.
If x and y are non-integer values x % y is computed as x – n * y, where n is the largest possible integer that is less than or equal to x / y (more details in the C# 4.0 Specification in section 7.8.3 Remainder operator).
For further details and examples you might want to have a look at the corresponding Wikipedia article:
Modulo operation (on Wikipedia)

% is the remainder operator in many C-inspired languages.
3 % 2 == 1
789 % 10 = 9
It's a bit tricky with negative numbers. In e.g. Java and C#, the result has the same sign as the dividend:
-1 % 2 == -1
In e.g. C++ this is implementation defined.
See also
Wikipedia/Modulo operation
References
MSDN/C# Language Reference/% operator

That is the Modulo operator. It will give you the remainder of a division operation.

It's the modulus operator. That is, 2 % 2 == 0, 4 % 4 % 2 == 0 (2, 4 are divisible by 2 with 0 remainder), 5 % 2 == 1 (2 goes into 5 with 1 as remainder.)

It is the modulo operator. i.e. it the remainder after division 1 % 36 == 1 (0 remainder 1)

That is the modulo operator, which finds the remainder of division of one number by another.
So in this case a will be the remainder of b divided by c.

It's is modulus, but you example is not a good use of it. It gives you the remainder when two integers are divided.
e.g. a = 7 % 3 will return 1, becuase 7 divided by 3 is 2 with 1 left over.

It is modulus operator
using System;
class Test
{
static void Main()
{
int a = 2;
int b = 6;
int c = 12;
int d = 5;
Console.WriteLine(b % a);
Console.WriteLine(c % d);
Console.Read();
}
}
Output:
0
2

is basic operator available in almost every language and generally known as modulo operator.
it gives remainder as result.

Okay well I did know this till just trying on a calculator and playing around so basically:
5 % 2.2 = 0.6 is like saying on a calculator 5/2.2 = 2.27 then you multiply that .27 times the 2.27 and you round and you get 0.6. Hope this helps, it helped me =]

Nobody here has provided any examples of exactly how an equation can return different results, such as comparing 37/6 to 37%6, and before some of you get upset thinking that you did, pause for a moment and think about it for a minute, according to Dirk Vollmar in here the int x % int y parses as (x - (x / y) * y) which seems fairly straightforward at first glance, but not all Math is performed in the same order.
Since not every equation has it's proper brackets, some Schools will teach that the Equation is to be parsed as ((x - (x / y)) * y) whilst other Schools teach (x - ((x / y) * y)).
Now I experimented with my example (37/6 & 37%6) and figured out which selection was intended (it's (x - ((x / y) * y))), and I even displayed a nicely built if Loop (even though I forgot every End of Line Semicolon) to simulate the Divide Equation at the most Fundamental Scale, as that was in fact my point, the Equation is similar, yet Fundamentally different.
Here's what I can remember from my deleted Post (this took me around an Hour to Type from my Phone, indents are Double Spaced)
using System;
class Test
{
static void Main()
{
float exact0 = (37 / 6); //6.1666e∞
float exact1 = (37 % 6); //1
float exact2 = (37 - (37 / 6) * 6); //0
float exact3 = ((37 - (37 / 6)) * 6); //0
float exact4 = (37 - ((37 / 6) * 6)); //185
int a = 37;
int b = 6;
int c = 0;
int d = a;
int e = b;
string Answer0 = "";
string Answer1 = "";
string Answer2 = "";
string Answer0Alt = "";
string Answer1Alt = "";
string Answer2Alt = "";
Console.WriteLine("37/6: " + exact0);
Console.WriteLine("37%6: " + exact1);
Console.WriteLine("(37 - (37 / 6) * 6): " + exact2);
Console.WriteLine("((37 - (37 / 6)) * 6): " + exact3);
Console.WriteLine("(37 - ((37 / 6) * 6)): " + exact4);
Console.WriteLine("a: " + a + ", b: " + b + ", c: " + c + ", d: " + d + ", e: " + e);
Console.WriteLine("Answer0: " + Answer0);
Console.WriteLine("Answer0Alt: " + Answer0Alt);
Console.WriteLine("Answer1: " + Answer1);
Console.WriteLine("Answer0Alt: " + Answer1Alt);
Console.WriteLine("Answer2: " + Answer2);
Console.WriteLine("Answer2Alt: " + Answer2Alt);
Console.WriteLine("Init Complete, starting Math...");
Loop
{
if (a !< b) {
a - b;
c +1;}
if else (a = b) {
a - b;
c +1;}
else
{
String Answer0 = c + "." + a; //6.1
//this is = to 37/6 in the fact that it equals 6.1 ((6*6=36)+1=37) or 6 remainder 1,
//which according to my Calculator App is technically correct once you Round Down the .666e∞
//which has been stated as the default behavior of the C# / Operand
//for completion sake I'll include the alternative answer for Round Up also
String Answer0Alt = c + "." + (a + 1); //6.2
Console.WriteLine("Division Complete, Continuing...");
Break
}
}
string Answer1 = ((d - (Answer0)) * e); //185.4
string Answer1Alt = ((d - (Answer0Alt​)) * e); // 184.8
string Answer2 = (d - ((Answer0) * e)); //0.4
string Answer2Alt = (d - ((Answer0Alt​) * e)); //-0.2
Console.WriteLine("Math Complete, Summarizing...");
Console.WriteLine("37/6: " + exact0);
Console.WriteLine("37%6: " + exact1);
Console.WriteLine("(37 - (37 / 6) * 6): " + exact2);
Console.WriteLine("((37 - (37 / 6)) * 6): " + exact3);
Console.WriteLine("(37 - ((37 / 6) * 6)): " + exact4);
Console.WriteLine("Answer0: " + Answer0);
Console.WriteLine("Answer0Alt: " + Answer0Alt);
Console.WriteLine("Answer1: " + Answer1);
Console.WriteLine("Answer0Alt: " + Answer1Alt);
Console.WriteLine("Answer2: " + Answer2);
Console.WriteLine("Answer2Alt: " + Answer2Alt);
Console.Read();
}
}
This also CLEARLY demonstrated how an outcome can be different for the exact same Equation.

C# value is always 0 console application

I have no idea why my BMI value is always = to 0. I am programming noob what am i missing? other than that is my if statement alright? what am i missing?
static void Main(string[] args) {
double WeightKg = 0.0, HeightCm = 0.0, Weightlbs = 0.0, WeightOz = 0.0, BMI = 0.0, Feet = 0.0, Inches = 0.0;
int BMIOption;
string AnotherConversion;
string BMIMenu = ("Which Measurement You Want to use to enter the weight and height?"
+ "\n1)Enter 1 for Metric"
+ "\n2)Enter 2 for British Imperial:");
Console.Write(BMIMenu);
BMIOption = int.Parse(Console.ReadLine());
if (BMIOption == 1) {
Console.Write("\nPlease Enter your Weight in Kilogram (kg):");
WeightKg = int.Parse(Console.ReadLine());
Console.Write("\nPlease Enter your Height in in centimetres (cm):");
HeightCm = int.Parse(Console.ReadLine());
BMI = WeightKg / (HeightCm * HeightCm);
if (BMI >= 35) {
Console.WriteLine("\nYour BMI is {0:C},Severe Obesity", BMI);
} else if (BMI >= 30) {
Console.WriteLine("\nYour BMI is {0:C},Obese", BMI);
} else if (BMI >= 25) {
Console.WriteLine("\nYour BMI is {0:C},OverWeight", BMI);
} else if (BMI >= 18.5) {
Console.WriteLine("\nYour BMI is {0:C},Healthy BodyWeight", BMI);
} else if (BMI <= 18.5) {
Console.WriteLine("\nYour BMI is {0:C},UnderWeight", BMI);
}//End if
Console.Write("\nWould you like to make an another conversion? \n\n(Enter Y to make an another conversion/Enter any other key to exit):");
Console.ReadKey();

BMI is calculated with meters, not centimeters. So you need to convert the HeightCm to HeightM. If you don't do this, you get really small number, that is then printed as 0.
double HeightM = HeightCm / 100.0;
BMI = WeightKg / (HeightM * HeightM);
Also, when parsing, use double.Parse instead of int.Parse. The way it is right now, you will only parse the number without the decimal part.

In this line:
BMI = WeightKg / (HeightCm * HeightCm);
The result ends up being very small (less than 1). Take an example where WeightKg is 55 and HeightCm is 165. The result is around 0.002.
When you display it as {0:C}, it's being displayed as zero. Use {0:G} to see the actual value.

even so your formula is wrong, as Martin says u must replace the int.Parse with double AND
BMI = WeightKg / ((HeightCm/100) * (HeightCm/100));
use this for calculation it is HEIGHT IN METERS :)

What does the '%' operator mean?

I have next code
int a,b,c;
b=1;
c=36;
a=b%c;
What does "%" operator mean?

It is the modulo (or modulus) operator:
The modulus operator (%) computes the remainder after dividing its first operand by its second.
For example:
class Program
{
static void Main()
{
Console.WriteLine(5 % 2); // int
Console.WriteLine(-5 % 2); // int
Console.WriteLine(5.0 % 2.2); // double
Console.WriteLine(5.0m % 2.2m); // decimal
Console.WriteLine(-5.2 % 2.0); // double
}
}
Sample output:
1
-1
0.6
0.6
-1.2
Note that the result of the % operator is equal to x – (x / y) * y and that if y is zero, a DivideByZeroException is thrown.
If x and y are non-integer values x % y is computed as x – n * y, where n is the largest possible integer that is less than or equal to x / y (more details in the C# 4.0 Specification in section 7.8.3 Remainder operator).
For further details and examples you might want to have a look at the corresponding Wikipedia article:
Modulo operation (on Wikipedia)

% is the remainder operator in many C-inspired languages.
3 % 2 == 1
789 % 10 = 9
It's a bit tricky with negative numbers. In e.g. Java and C#, the result has the same sign as the dividend:
-1 % 2 == -1
In e.g. C++ this is implementation defined.
See also
Wikipedia/Modulo operation
References
MSDN/C# Language Reference/% operator

That is the Modulo operator. It will give you the remainder of a division operation.

It's the modulus operator. That is, 2 % 2 == 0, 4 % 4 % 2 == 0 (2, 4 are divisible by 2 with 0 remainder), 5 % 2 == 1 (2 goes into 5 with 1 as remainder.)

It is the modulo operator. i.e. it the remainder after division 1 % 36 == 1 (0 remainder 1)

That is the modulo operator, which finds the remainder of division of one number by another.
So in this case a will be the remainder of b divided by c.

It's is modulus, but you example is not a good use of it. It gives you the remainder when two integers are divided.
e.g. a = 7 % 3 will return 1, becuase 7 divided by 3 is 2 with 1 left over.

It is modulus operator
using System;
class Test
{
static void Main()
{
int a = 2;
int b = 6;
int c = 12;
int d = 5;
Console.WriteLine(b % a);
Console.WriteLine(c % d);
Console.Read();
}
}
Output:
0
2

is basic operator available in almost every language and generally known as modulo operator.
it gives remainder as result.

Okay well I did know this till just trying on a calculator and playing around so basically:
5 % 2.2 = 0.6 is like saying on a calculator 5/2.2 = 2.27 then you multiply that .27 times the 2.27 and you round and you get 0.6. Hope this helps, it helped me =]

Nobody here has provided any examples of exactly how an equation can return different results, such as comparing 37/6 to 37%6, and before some of you get upset thinking that you did, pause for a moment and think about it for a minute, according to Dirk Vollmar in here the int x % int y parses as (x - (x / y) * y) which seems fairly straightforward at first glance, but not all Math is performed in the same order.
Since not every equation has it's proper brackets, some Schools will teach that the Equation is to be parsed as ((x - (x / y)) * y) whilst other Schools teach (x - ((x / y) * y)).
Now I experimented with my example (37/6 & 37%6) and figured out which selection was intended (it's (x - ((x / y) * y))), and I even displayed a nicely built if Loop (even though I forgot every End of Line Semicolon) to simulate the Divide Equation at the most Fundamental Scale, as that was in fact my point, the Equation is similar, yet Fundamentally different.
Here's what I can remember from my deleted Post (this took me around an Hour to Type from my Phone, indents are Double Spaced)
using System;
class Test
{
static void Main()
{
float exact0 = (37 / 6); //6.1666e∞
float exact1 = (37 % 6); //1
float exact2 = (37 - (37 / 6) * 6); //0
float exact3 = ((37 - (37 / 6)) * 6); //0
float exact4 = (37 - ((37 / 6) * 6)); //185
int a = 37;
int b = 6;
int c = 0;
int d = a;
int e = b;
string Answer0 = "";
string Answer1 = "";
string Answer2 = "";
string Answer0Alt = "";
string Answer1Alt = "";
string Answer2Alt = "";
Console.WriteLine("37/6: " + exact0);
Console.WriteLine("37%6: " + exact1);
Console.WriteLine("(37 - (37 / 6) * 6): " + exact2);
Console.WriteLine("((37 - (37 / 6)) * 6): " + exact3);
Console.WriteLine("(37 - ((37 / 6) * 6)): " + exact4);
Console.WriteLine("a: " + a + ", b: " + b + ", c: " + c + ", d: " + d + ", e: " + e);
Console.WriteLine("Answer0: " + Answer0);
Console.WriteLine("Answer0Alt: " + Answer0Alt);
Console.WriteLine("Answer1: " + Answer1);
Console.WriteLine("Answer0Alt: " + Answer1Alt);
Console.WriteLine("Answer2: " + Answer2);
Console.WriteLine("Answer2Alt: " + Answer2Alt);
Console.WriteLine("Init Complete, starting Math...");
Loop
{
if (a !< b) {
a - b;
c +1;}
if else (a = b) {
a - b;
c +1;}
else
{
String Answer0 = c + "." + a; //6.1
//this is = to 37/6 in the fact that it equals 6.1 ((6*6=36)+1=37) or 6 remainder 1,
//which according to my Calculator App is technically correct once you Round Down the .666e∞
//which has been stated as the default behavior of the C# / Operand
//for completion sake I'll include the alternative answer for Round Up also
String Answer0Alt = c + "." + (a + 1); //6.2
Console.WriteLine("Division Complete, Continuing...");
Break
}
}
string Answer1 = ((d - (Answer0)) * e); //185.4
string Answer1Alt = ((d - (Answer0Alt​)) * e); // 184.8
string Answer2 = (d - ((Answer0) * e)); //0.4
string Answer2Alt = (d - ((Answer0Alt​) * e)); //-0.2
Console.WriteLine("Math Complete, Summarizing...");
Console.WriteLine("37/6: " + exact0);
Console.WriteLine("37%6: " + exact1);
Console.WriteLine("(37 - (37 / 6) * 6): " + exact2);
Console.WriteLine("((37 - (37 / 6)) * 6): " + exact3);
Console.WriteLine("(37 - ((37 / 6) * 6)): " + exact4);
Console.WriteLine("Answer0: " + Answer0);
Console.WriteLine("Answer0Alt: " + Answer0Alt);
Console.WriteLine("Answer1: " + Answer1);
Console.WriteLine("Answer0Alt: " + Answer1Alt);
Console.WriteLine("Answer2: " + Answer2);
Console.WriteLine("Answer2Alt: " + Answer2Alt);
Console.Read();
}
}
This also CLEARLY demonstrated how an outcome can be different for the exact same Equation.