Because Convert.ToInt32 rounds: Return Value: rounded to the nearest 32-bit signed integer. If value is halfway between two whole numbers, the even number is returned; that is, 4.5 is converted to 4, and 5.5 is converted to 6. …while the cast truncates: When you convert from a double or float value to an integral type, … Read more
Use to_string() (running example here): let x: u32 = 10; let s: String = x.to_string(); println!(“{}”, s); You’re right; to_str() was renamed to to_string() before Rust 1.0 was released for consistency because an allocated string is now called String. If you need to pass a string slice somewhere, you need to obtain a &str reference … Read more
The first two are constant expressions, the last one isn’t. The C# specification allows an implicit conversion from int to short for constants, but not for other expressions. This is a reasonable rule, since for constants the compiler can ensure that the value fits into the target type, but it can’t for normal expressions. This … Read more
In [16]: df = DataFrame(np.arange(10).reshape(5,2),columns=list(‘AB’)) In [17]: df Out[17]: A B 0 0 1 1 2 3 2 4 5 3 6 7 4 8 9 In [18]: df.dtypes Out[18]: A int64 B int64 dtype: object Convert a series In [19]: df[‘A’].apply(str) Out[19]: 0 0 1 2 2 4 3 6 4 8 Name: A, … Read more
Assuming you’re on at least 3.2, there’s a built in for this: int.from_bytes( bytes, byteorder, *, signed=False ) … The argument bytes must either be a bytes-like object or an iterable producing bytes. The byteorder argument determines the byte order used to represent the integer. If byteorder is “big”, the most significant byte is at … Read more
When you select data from a MySQL database using PHP the datatype will always be converted to a string. You can convert it back to an integer using the following code: $id = (int) $row[‘userid’]; Or by using the function intval(): $id = intval($row[‘userid’]);
There’s nothing wrong with that statement; you’re just multiplying 4 numbers and assigning it to an int, there just happens to be an overflow. This is different than assigning a single literal, which would be bounds-checked at compile-time. It is the out-of-bounds literal that causes the error, not the assignment: System.out.println(2147483648); // error System.out.println(2147483647 + … Read more
int a = 1; char b = (char) a; System.out.println(b); will print out the char with Unicode code point 1 (start-of-heading char, which isn’t printable; see this table: C0 Controls and Basic Latin, same as ASCII) int a=”1″; char b = (char) a; System.out.println(b); will print out the char with Unicode code point 49 (one … Read more
Just do (int)myLongValue. It’ll do exactly what you want (discarding MSBs and taking LSBs) in unchecked context (which is the compiler default). It’ll throw OverflowException in checked context if the value doesn’t fit in an int: int myIntValue = unchecked((int)myLongValue);
Math.ceil() is the correct function to call. I’m guessing a is an int, which would make a / 100 perform integer arithmetic. Try Math.ceil(a / 100.0) instead. int a = 142; System.out.println(a / 100); System.out.println(Math.ceil(a / 100)); System.out.println(a / 100.0); System.out.println(Math.ceil(a / 100.0)); System.out.println((int) Math.ceil(a / 100.0)); Outputs: 1 1.0 1.42 2.0 2 See http://ideone.com/yhT0l