printf浮点变量和常量之间有什么区别？

Question

Here's my code:

float x = 21.195;
printf("%.2f\n", x);
printf("%.2f\n", 21.195);

I would expect both print statements to have identical output, but instead, the first prints 21.19 , and the second prints 21.20 .

Could someone explain why the output is different?

Answer 1

The values are different. The first is a float , which is typically 4 bytes. The second is a double , which is typically 8 bytes.

The rules for rounding are based on the third digit after the decimal place. So, in one case, the value is something like 21.19499997 and the other 21.1950000000001, or something like that. (These are made up to illustrate the issue with rounding and imprecise numeric formats.)

Answer 2

By default 21.195 is a double.

If you want a float, write :

21.195F

or

(float)21.195

Regards

Answer 3

When a floating point variable is defined in C, by default it is set to double . So x is set to float because you mentioned it explicitly , else 21.195 is considered to be double.

Now, as mentioned above, float is usually about 4 bytes and double is about 8 bytes . So a float value has 24 significant bits with 7 digits of precision and double has 53 significant bits with 15 to 16 digits of precision .

A roundoff function %.2f works to round off the number correct to 2 decimal places and checks the third digit after the decimal point for rounding off. So 21.195 in float expands to 21.19499998 and is then reduced to 21.19 after %.2f an 21.195 in double expands to 21.1950000000000001 and hence reduces to 21.20.

Hope it helps!