C＃简单划分问题

Question

I have this:

 double result = 60 / 23;

In my program, the result is 2, but correct is 2,608695652173913. Where is problem?

Answer 1

60 and 23 are integer literals so you are doing integer division and then assigning to a double. The result of the integer division is 2.

Try

double result = 60.0 / 23.0;

Or equivalently

double result = 60d / 23d;

Where the d suffix informs the complier that you meant to write a double literal.

Answer 2

You can use any of the following all will give 2.60869565217391 :

 double result = 60 / 23d;  
 double result = 60d / 23;  
 double result = 60d/ 23d;  
 double result = 60.0 / 23.0;   

But

double result = 60 / 23;  //give 2

Explanation:

if any of the number is double it will give a double

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EDIT:

Documentation

The evaluation of the expression is performed according to the following rules:

• If one of the floating-point types is double, the expression evaluates to double (or bool in the case of relational or Boolean expressions).

• If there is no double type in the expression, it evaluates to float (or bool in the case of relational or Boolean expressions).

Answer 3

It will work

double result = (double)60 / (double) 23;

Or equivalently

double result = (double)60 /  23;

Answer 4

（双）60/23

Answer 5

Haven't used C# for a while, but you are dividing two integers, which as far as I remember makes the result an integer as well.

You can force your number literals to be doubles by adding the letter "d", likes this:

double result = 60d / 23d;

Answer 6

double result = 60.0 / 23.0;

Answer 7

It is best practice to correctly decorate numerals for their appropriate type. This avoids not only the bug you are experiencing, but makes the code more readable and maintainable.

double x = 100d;
single x = 100f;
decimal x = 100m;

Answer 8

convert the dividend and divisor into double values, so that result is double
double res= 60d/23d;

Answer 9

To add to what has been said so far... 60/23 is an operation on two constants. The compiler recognizes the result as a constant and pre-computes the answer. Since the operation is on two integers, the compiler uses an integer result The integer operation of 60/23 has a result of 2; so the compiler effective creates the following code:

double result = 2;

As has been pointed out already, you need to tell the compiler not to use integers, changing one or both of the operands to non-integer will get the compiler to use a floating-point constant.