为什么我不能将大值传递给Int32？

Question

I have a number: 94,800,620,800

Float is 4-byte data-type. Int32 is also 4-byte data-type.

float f = 94800620800; // ok
Int32 t = 94800620800; // error

Please explain this problem. Why I get a error when using Int32. Why I can use this number for float data-type because both of them are 4-byte data-type. Thanks.

Answer 1

Because the number you are trying to assign is larger than the largest possible value for a number of type Int32 , which happens to be 2,147,483,647. To note, the maximum value for a Single is 3.402823 × 10 ³⁸ .

Answer 2

The max value for Int32 is 2,147,483,647 - which is less than 94,800,620,800.

A float can take a value in the following range: ±1.5 × 10−45 to ±3.4 × 1038

Also, check out this SO question - what the difference between the float and integer data type when the size is same in java? . It's a Java question, but the concept is the same and there's a detailed explanation of the difference, even though they're the same size.

Answer 3

Because that number is too big for a 4 byte int. Scalar values like Int32 have a minimum and maximum limit (which are -2 ³¹ and 2 ³¹ - 1 in this case, respectively), and you simply can't store a value outside this range.

Floating point numbers are stored totally differently, so you won't get compiler errors with huge values, only possible precision problems later, during runtime.

Answer 4

Because of those types internal representation.

float uses something like i,d ^ n where i is the integral part, d is the decimal part and n is the exponent (of course, this happens in base 2).

In this way, you can store bigger numbers in a float , but it won't be accurate as a, say, Int32 in storing integral numbers. If you try to convert

float f = 94800620800; 

to an integral type large enough to store its value, it may not be the same as the initial 94800620800.

Answer 5

Maybe you should read the error message? ;)

Error   Integral constant is too large

the maximum valiue of a 32 bit int is 2,147,483,647

the float on the other hand works because it stores a mantissa and exponent, rather than just a single number, so it can cope with a much bigger range at the expense of possibly losing precision

try printing out your float and you will get 94800620000 instead of 94800620800 as the lower bits are lost

Answer 6

Integers types are exact representations, while floating point numbers are a combination of significant digits and exponent.

The floating point wiki page is explaining how this works.

Answer 7

Int32 has a max value of 2,147,483,647. Your value is much higher.

Take a look at system.int32.maxvalue

Answer 8

提供的常量表达式值不在int数据类型的范围内。

Answer 9

The range of Int32 goes from − 2,147,483,648 to 2,147,483,647. Your variable is way out of range.