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Why a var of the kind double can store a var of the kind int?
Using C#
int x;
double y;
x = 5;
y = 2*x;
Console.WriteLine(x);
Console.WriteLine(y);
Console.ReadLine();
1 answer
7
Let’s start calling things by the right name, then let’s call x and y variable, because var is a C# command and using that word can confuse what you’re talking about.
Then at the end I’d be saying
a variable of type
doublecan store a variable of typeint
Then I’ll correct the term once again, it’s meant to say:
a variable of type
doublecan store a value of typeint
Variable is one thing, value is another.
It turns out that a value of the type is not being stored int in y. The type of value stored there will always be of the type double because C# is a static typing language and the variable can only have a value of the type it was declared. Well, there is the question of null, but let’s leave it there so as not to confuse because in this case can not have this value.
So what happened? There was one cast implicit. Had a multiplication of two values of type int (2 and x) and the result of that which is also int is automatically converted to a value of the type double. Then effectively a value reconstruction operation in another format is performed (since int and double sane internally incompatible), even if it doesn’t seem to have it.
Automatic conversion is possible only in cases that have no data loss, and as whole value of int is a valid value in double the conversion is placed by the compiler.
Congratulations on your curiosity, just get used to using correct terms to communicate properly.
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Oops man, thank you for being contributing to my learning. This issue of data loss I’ve come to see in the course I’m taking. But, to be more clear, what would be this data loss ?
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1
Let’s take the opposite, you want to store 1.4 which is
doublein a variable of typeint, then you have to save 1 since you can only store decimal part in this type, and where did 0.4 go? There is a loss. You’re allowed to do this, but only manually, you have to say that you want the loss, the compiler doesn’t do it for you. If solved, you can accept the answer, see how it is in [tour]. -
So in this case I can make an explicit casting ? type double x = 1.4; int y = (int) x;
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Can yes, you said you want the loss.
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All right, thanks man.
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@Fluss you read the [tour]? understood what you should do?
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