显式/隐式类型转换c ++

Question

I have a line of code

double i = 1 + (long)1.5* 5.0f

My question is what is the conversion order and the result? Been searching for examples like this, but to no avail. Any good guides out there that may help me understand it?

Answer 1

My question is what is the conversion order and the result?

The cast is applied to 1.5 , giving a long with value 1 .

That's converted to float for multiplication with 5.0f , giving a float with value 5.0f .

1 is converted to float for addition with that value, giving a float with value 6.0f .

Finally, that's promoted to double (retaining the value 6.0 ) to assign to i .

This assumes a non-crazy floating point format that can represent small integers exactly; otherwise, there may be rounding errors.

If you wanted to cast the result of the multiplication, then use parentheses to control the operator precedence:

double i = 1 + (long)(1.5* 5.0f);  // = 8.0

or use a C++-style cast, which forces the use of parentheses:

double i = 1 + static_cast<long>(1.5* 5.0f)

Any good guides out there that may help me understand it?

Here's one: http://en.cppreference.com/w/cpp/language/operator_precedence . Note that the type cast has a higher precedence than multiplication, which is in turn higher than addition (3 vs. 5 vs. 6).

Answer 2

从这个表中可以看出，强制转换操作符的优先级高于乘法，但请遵循建议使用括号。

Answer 3

This precedence table should tell you everything you need to know.

1. Casting: 1.5 is cast to a long
2. Multiplication: 1.5 * 5.0f , which casts this product as a float
3. Addition: 1 + ( ((long) 1.5) * 5.0f)
4. Assignment: i = 1 + ((long) 1.5 * 5.0f)

Answer 4

If you're not sure what the precedence of the casting operator is then rewrite the expression (in your head)

(long)1.5 * 5.0

to

5.0 * (long)1.5

Here its pretty obvious what has precedence and its the same with the first version