I can't set my reference numbers's format to “byte” or “short” in console project?












3















Actually this is my first question on this site and i hope it's right way to ask this type question.



I'm just learning C# from scratch in Youtube.



I created simple console project that has multiply function as you can see below, but I can't set my reference numbers (number1, number2) to "short" or "byte". But I can set them "long", "decimal" "int" etc.



I did little search and I have theory that this is about bits and bytes counts of each parameter but can't really comprehend the concept.



May anyone explain this error that I'm encountering with simple language any chance? Thanks for any explanation :)



using System;

namespace HelloWorld
{
class Program
{
static void Main(string args)
{
try
{
Start:
int number1;
int number2;

Console.Write("enter number to multiply:");
number1 = Convert.ToInt32(Console.ReadLine());
Console.Write("enter another number:");
number2 = Convert.ToInt32(Console.ReadLine());
Console.WriteLine("the result is:" + number1 * number2 + "(" + number1 + "X" + number2 + ")");
Console.WriteLine();
goto Start;
}
catch (Exception)
{
Console.WriteLine("hey silly! that's not even a number:)");
}
}
}


}










share|improve this question

























  • I am not sure how to answer your question because I couldn't clearly understand it. can you include the error you got when you tried to set your number1 to byte or short? Moreover, just a hint: Don't use goto, just put your whole code in a loop instead.

    – Paul Karam
    Nov 23 '18 at 14:22













  • Thanks for advice. I will check for loop. If I set them byte or short, Visual Studio can't compile them and says "there were build errors."

    – Carla Köhler
    Nov 23 '18 at 14:31











  • What errors are you getting when declaring those variables as short or byte?

    – Alex Leo
    Nov 23 '18 at 14:53


















3















Actually this is my first question on this site and i hope it's right way to ask this type question.



I'm just learning C# from scratch in Youtube.



I created simple console project that has multiply function as you can see below, but I can't set my reference numbers (number1, number2) to "short" or "byte". But I can set them "long", "decimal" "int" etc.



I did little search and I have theory that this is about bits and bytes counts of each parameter but can't really comprehend the concept.



May anyone explain this error that I'm encountering with simple language any chance? Thanks for any explanation :)



using System;

namespace HelloWorld
{
class Program
{
static void Main(string args)
{
try
{
Start:
int number1;
int number2;

Console.Write("enter number to multiply:");
number1 = Convert.ToInt32(Console.ReadLine());
Console.Write("enter another number:");
number2 = Convert.ToInt32(Console.ReadLine());
Console.WriteLine("the result is:" + number1 * number2 + "(" + number1 + "X" + number2 + ")");
Console.WriteLine();
goto Start;
}
catch (Exception)
{
Console.WriteLine("hey silly! that's not even a number:)");
}
}
}


}










share|improve this question

























  • I am not sure how to answer your question because I couldn't clearly understand it. can you include the error you got when you tried to set your number1 to byte or short? Moreover, just a hint: Don't use goto, just put your whole code in a loop instead.

    – Paul Karam
    Nov 23 '18 at 14:22













  • Thanks for advice. I will check for loop. If I set them byte or short, Visual Studio can't compile them and says "there were build errors."

    – Carla Köhler
    Nov 23 '18 at 14:31











  • What errors are you getting when declaring those variables as short or byte?

    – Alex Leo
    Nov 23 '18 at 14:53
















3












3








3








Actually this is my first question on this site and i hope it's right way to ask this type question.



I'm just learning C# from scratch in Youtube.



I created simple console project that has multiply function as you can see below, but I can't set my reference numbers (number1, number2) to "short" or "byte". But I can set them "long", "decimal" "int" etc.



I did little search and I have theory that this is about bits and bytes counts of each parameter but can't really comprehend the concept.



May anyone explain this error that I'm encountering with simple language any chance? Thanks for any explanation :)



using System;

namespace HelloWorld
{
class Program
{
static void Main(string args)
{
try
{
Start:
int number1;
int number2;

Console.Write("enter number to multiply:");
number1 = Convert.ToInt32(Console.ReadLine());
Console.Write("enter another number:");
number2 = Convert.ToInt32(Console.ReadLine());
Console.WriteLine("the result is:" + number1 * number2 + "(" + number1 + "X" + number2 + ")");
Console.WriteLine();
goto Start;
}
catch (Exception)
{
Console.WriteLine("hey silly! that's not even a number:)");
}
}
}


}










share|improve this question
















Actually this is my first question on this site and i hope it's right way to ask this type question.



I'm just learning C# from scratch in Youtube.



I created simple console project that has multiply function as you can see below, but I can't set my reference numbers (number1, number2) to "short" or "byte". But I can set them "long", "decimal" "int" etc.



I did little search and I have theory that this is about bits and bytes counts of each parameter but can't really comprehend the concept.



May anyone explain this error that I'm encountering with simple language any chance? Thanks for any explanation :)



using System;

namespace HelloWorld
{
class Program
{
static void Main(string args)
{
try
{
Start:
int number1;
int number2;

Console.Write("enter number to multiply:");
number1 = Convert.ToInt32(Console.ReadLine());
Console.Write("enter another number:");
number2 = Convert.ToInt32(Console.ReadLine());
Console.WriteLine("the result is:" + number1 * number2 + "(" + number1 + "X" + number2 + ")");
Console.WriteLine();
goto Start;
}
catch (Exception)
{
Console.WriteLine("hey silly! that's not even a number:)");
}
}
}


}







c# console






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 23 '18 at 14:20









fubo

29.6k966104




29.6k966104










asked Nov 23 '18 at 14:19









Carla KöhlerCarla Köhler

183




183













  • I am not sure how to answer your question because I couldn't clearly understand it. can you include the error you got when you tried to set your number1 to byte or short? Moreover, just a hint: Don't use goto, just put your whole code in a loop instead.

    – Paul Karam
    Nov 23 '18 at 14:22













  • Thanks for advice. I will check for loop. If I set them byte or short, Visual Studio can't compile them and says "there were build errors."

    – Carla Köhler
    Nov 23 '18 at 14:31











  • What errors are you getting when declaring those variables as short or byte?

    – Alex Leo
    Nov 23 '18 at 14:53





















  • I am not sure how to answer your question because I couldn't clearly understand it. can you include the error you got when you tried to set your number1 to byte or short? Moreover, just a hint: Don't use goto, just put your whole code in a loop instead.

    – Paul Karam
    Nov 23 '18 at 14:22













  • Thanks for advice. I will check for loop. If I set them byte or short, Visual Studio can't compile them and says "there were build errors."

    – Carla Köhler
    Nov 23 '18 at 14:31











  • What errors are you getting when declaring those variables as short or byte?

    – Alex Leo
    Nov 23 '18 at 14:53



















I am not sure how to answer your question because I couldn't clearly understand it. can you include the error you got when you tried to set your number1 to byte or short? Moreover, just a hint: Don't use goto, just put your whole code in a loop instead.

– Paul Karam
Nov 23 '18 at 14:22







I am not sure how to answer your question because I couldn't clearly understand it. can you include the error you got when you tried to set your number1 to byte or short? Moreover, just a hint: Don't use goto, just put your whole code in a loop instead.

– Paul Karam
Nov 23 '18 at 14:22















Thanks for advice. I will check for loop. If I set them byte or short, Visual Studio can't compile them and says "there were build errors."

– Carla Köhler
Nov 23 '18 at 14:31





Thanks for advice. I will check for loop. If I set them byte or short, Visual Studio can't compile them and says "there were build errors."

– Carla Köhler
Nov 23 '18 at 14:31













What errors are you getting when declaring those variables as short or byte?

– Alex Leo
Nov 23 '18 at 14:53







What errors are you getting when declaring those variables as short or byte?

– Alex Leo
Nov 23 '18 at 14:53














2 Answers
2






active

oldest

votes


















1














You are on the right track; it's about implicit conversion of types and strong typing.



Not to overload you with all the details, you can make some adjustments for



short:



note the ToInt16



short number1;
//etc
number1 = Convert.ToInt16(Console.ReadLine());


byte:



note: the ToByte



byte number1;
//etc
number1 = Convert.ToByte(Console.ReadLine());


long




left as an exercise






Crucial is that, although all the types consists of bits, and are basically numbers, C# want to be sure you mean what you are saying and enforces you to use the correct type.






So, why does it work for decimal etc.?

Is because C# thinks it's valid to implicitly convert them, and therefor the compiler is doing it for you.



In the link you can see that for int (aka Int32) the folowing implicit conversions are predefined:



//from  |  to
//int | long, float, double, or decimal


In the linked table you can also see that the other way round, from byte to int is allowed.






The big lesson: C# is strong typed: if you say it's an Int16 (aka short), you must use it only as an Int16 (and not an Int32)

Happy coding ;-)






share|improve this answer


























  • Thanks Stefan! That link helped me a lot to understand my struggle and ToInt16 solved it :)

    – Carla Köhler
    Nov 23 '18 at 17:28



















1














if I understand your problem - in order to "set" a variable to a given type you will need to declare it as such before compiling (before the program run).



So if you want to work with long numbers you will need to declare:



long number1;
long number2;


Your conversions should reflect it:



number1 = Convert.ToInt64(Console.ReadLine());


For decimal



decimal number1;
decimal number2;


Conversion e.g. decimal.parse(number1).



For byte



byte number1;
byte number2;


Again your conversion must enforced per each type.



Just a note. Using a goto is something that I don't approve neither suggest. Learn to use a loop. If those videos are using the goto, I would advise to learn from a better source.



Hope it helps.






share|improve this answer



















  • 1





    Thanks Alex! I solved with using ToInt16 and stopped using "goto" :)

    – Carla Köhler
    Nov 23 '18 at 17:29











  • Good stuff. Happy coding :-)

    – Alex Leo
    Nov 24 '18 at 18:07











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














You are on the right track; it's about implicit conversion of types and strong typing.



Not to overload you with all the details, you can make some adjustments for



short:



note the ToInt16



short number1;
//etc
number1 = Convert.ToInt16(Console.ReadLine());


byte:



note: the ToByte



byte number1;
//etc
number1 = Convert.ToByte(Console.ReadLine());


long




left as an exercise






Crucial is that, although all the types consists of bits, and are basically numbers, C# want to be sure you mean what you are saying and enforces you to use the correct type.






So, why does it work for decimal etc.?

Is because C# thinks it's valid to implicitly convert them, and therefor the compiler is doing it for you.



In the link you can see that for int (aka Int32) the folowing implicit conversions are predefined:



//from  |  to
//int | long, float, double, or decimal


In the linked table you can also see that the other way round, from byte to int is allowed.






The big lesson: C# is strong typed: if you say it's an Int16 (aka short), you must use it only as an Int16 (and not an Int32)

Happy coding ;-)






share|improve this answer


























  • Thanks Stefan! That link helped me a lot to understand my struggle and ToInt16 solved it :)

    – Carla Köhler
    Nov 23 '18 at 17:28
















1














You are on the right track; it's about implicit conversion of types and strong typing.



Not to overload you with all the details, you can make some adjustments for



short:



note the ToInt16



short number1;
//etc
number1 = Convert.ToInt16(Console.ReadLine());


byte:



note: the ToByte



byte number1;
//etc
number1 = Convert.ToByte(Console.ReadLine());


long




left as an exercise






Crucial is that, although all the types consists of bits, and are basically numbers, C# want to be sure you mean what you are saying and enforces you to use the correct type.






So, why does it work for decimal etc.?

Is because C# thinks it's valid to implicitly convert them, and therefor the compiler is doing it for you.



In the link you can see that for int (aka Int32) the folowing implicit conversions are predefined:



//from  |  to
//int | long, float, double, or decimal


In the linked table you can also see that the other way round, from byte to int is allowed.






The big lesson: C# is strong typed: if you say it's an Int16 (aka short), you must use it only as an Int16 (and not an Int32)

Happy coding ;-)






share|improve this answer


























  • Thanks Stefan! That link helped me a lot to understand my struggle and ToInt16 solved it :)

    – Carla Köhler
    Nov 23 '18 at 17:28














1












1








1







You are on the right track; it's about implicit conversion of types and strong typing.



Not to overload you with all the details, you can make some adjustments for



short:



note the ToInt16



short number1;
//etc
number1 = Convert.ToInt16(Console.ReadLine());


byte:



note: the ToByte



byte number1;
//etc
number1 = Convert.ToByte(Console.ReadLine());


long




left as an exercise






Crucial is that, although all the types consists of bits, and are basically numbers, C# want to be sure you mean what you are saying and enforces you to use the correct type.






So, why does it work for decimal etc.?

Is because C# thinks it's valid to implicitly convert them, and therefor the compiler is doing it for you.



In the link you can see that for int (aka Int32) the folowing implicit conversions are predefined:



//from  |  to
//int | long, float, double, or decimal


In the linked table you can also see that the other way round, from byte to int is allowed.






The big lesson: C# is strong typed: if you say it's an Int16 (aka short), you must use it only as an Int16 (and not an Int32)

Happy coding ;-)






share|improve this answer















You are on the right track; it's about implicit conversion of types and strong typing.



Not to overload you with all the details, you can make some adjustments for



short:



note the ToInt16



short number1;
//etc
number1 = Convert.ToInt16(Console.ReadLine());


byte:



note: the ToByte



byte number1;
//etc
number1 = Convert.ToByte(Console.ReadLine());


long




left as an exercise






Crucial is that, although all the types consists of bits, and are basically numbers, C# want to be sure you mean what you are saying and enforces you to use the correct type.






So, why does it work for decimal etc.?

Is because C# thinks it's valid to implicitly convert them, and therefor the compiler is doing it for you.



In the link you can see that for int (aka Int32) the folowing implicit conversions are predefined:



//from  |  to
//int | long, float, double, or decimal


In the linked table you can also see that the other way round, from byte to int is allowed.






The big lesson: C# is strong typed: if you say it's an Int16 (aka short), you must use it only as an Int16 (and not an Int32)

Happy coding ;-)







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 23 '18 at 14:50

























answered Nov 23 '18 at 14:28









StefanStefan

8,35373760




8,35373760













  • Thanks Stefan! That link helped me a lot to understand my struggle and ToInt16 solved it :)

    – Carla Köhler
    Nov 23 '18 at 17:28



















  • Thanks Stefan! That link helped me a lot to understand my struggle and ToInt16 solved it :)

    – Carla Köhler
    Nov 23 '18 at 17:28

















Thanks Stefan! That link helped me a lot to understand my struggle and ToInt16 solved it :)

– Carla Köhler
Nov 23 '18 at 17:28





Thanks Stefan! That link helped me a lot to understand my struggle and ToInt16 solved it :)

– Carla Köhler
Nov 23 '18 at 17:28













1














if I understand your problem - in order to "set" a variable to a given type you will need to declare it as such before compiling (before the program run).



So if you want to work with long numbers you will need to declare:



long number1;
long number2;


Your conversions should reflect it:



number1 = Convert.ToInt64(Console.ReadLine());


For decimal



decimal number1;
decimal number2;


Conversion e.g. decimal.parse(number1).



For byte



byte number1;
byte number2;


Again your conversion must enforced per each type.



Just a note. Using a goto is something that I don't approve neither suggest. Learn to use a loop. If those videos are using the goto, I would advise to learn from a better source.



Hope it helps.






share|improve this answer



















  • 1





    Thanks Alex! I solved with using ToInt16 and stopped using "goto" :)

    – Carla Köhler
    Nov 23 '18 at 17:29











  • Good stuff. Happy coding :-)

    – Alex Leo
    Nov 24 '18 at 18:07
















1














if I understand your problem - in order to "set" a variable to a given type you will need to declare it as such before compiling (before the program run).



So if you want to work with long numbers you will need to declare:



long number1;
long number2;


Your conversions should reflect it:



number1 = Convert.ToInt64(Console.ReadLine());


For decimal



decimal number1;
decimal number2;


Conversion e.g. decimal.parse(number1).



For byte



byte number1;
byte number2;


Again your conversion must enforced per each type.



Just a note. Using a goto is something that I don't approve neither suggest. Learn to use a loop. If those videos are using the goto, I would advise to learn from a better source.



Hope it helps.






share|improve this answer



















  • 1





    Thanks Alex! I solved with using ToInt16 and stopped using "goto" :)

    – Carla Köhler
    Nov 23 '18 at 17:29











  • Good stuff. Happy coding :-)

    – Alex Leo
    Nov 24 '18 at 18:07














1












1








1







if I understand your problem - in order to "set" a variable to a given type you will need to declare it as such before compiling (before the program run).



So if you want to work with long numbers you will need to declare:



long number1;
long number2;


Your conversions should reflect it:



number1 = Convert.ToInt64(Console.ReadLine());


For decimal



decimal number1;
decimal number2;


Conversion e.g. decimal.parse(number1).



For byte



byte number1;
byte number2;


Again your conversion must enforced per each type.



Just a note. Using a goto is something that I don't approve neither suggest. Learn to use a loop. If those videos are using the goto, I would advise to learn from a better source.



Hope it helps.






share|improve this answer













if I understand your problem - in order to "set" a variable to a given type you will need to declare it as such before compiling (before the program run).



So if you want to work with long numbers you will need to declare:



long number1;
long number2;


Your conversions should reflect it:



number1 = Convert.ToInt64(Console.ReadLine());


For decimal



decimal number1;
decimal number2;


Conversion e.g. decimal.parse(number1).



For byte



byte number1;
byte number2;


Again your conversion must enforced per each type.



Just a note. Using a goto is something that I don't approve neither suggest. Learn to use a loop. If those videos are using the goto, I would advise to learn from a better source.



Hope it helps.







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 23 '18 at 14:40









Alex LeoAlex Leo

461211




461211








  • 1





    Thanks Alex! I solved with using ToInt16 and stopped using "goto" :)

    – Carla Köhler
    Nov 23 '18 at 17:29











  • Good stuff. Happy coding :-)

    – Alex Leo
    Nov 24 '18 at 18:07














  • 1





    Thanks Alex! I solved with using ToInt16 and stopped using "goto" :)

    – Carla Köhler
    Nov 23 '18 at 17:29











  • Good stuff. Happy coding :-)

    – Alex Leo
    Nov 24 '18 at 18:07








1




1





Thanks Alex! I solved with using ToInt16 and stopped using "goto" :)

– Carla Köhler
Nov 23 '18 at 17:29





Thanks Alex! I solved with using ToInt16 and stopped using "goto" :)

– Carla Köhler
Nov 23 '18 at 17:29













Good stuff. Happy coding :-)

– Alex Leo
Nov 24 '18 at 18:07





Good stuff. Happy coding :-)

– Alex Leo
Nov 24 '18 at 18:07


















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