Java：長時間投射到int

Question

我們假設變量start和stop都是長型變量。

int diff = (int) start-stop; 

此代碼產生了錯誤Type mismatch: cannot convert from long to int，而

int diff = (int) (start-stop); 

運行正常。爲什麼？

Answer 1

因爲在第一，(int)適用於start，不表達，如：

int diff = ((int) start) - stop; // <== Your first expression clarified 

然後就從所得int減去long，而且表達的結果是一個long。 （如果已經工作，它可能將不得不給予你一個不正確的結果，如果start曾經有過大在int存儲值的副作用。）

通過先做減法和獲得long結果，然後將其轉換爲int，您可以將其存儲在您的int變量中。 （大概你知道它適合你的應用邏輯。）

Answer 2

因爲Java不允許隱含縮小轉換。一個將在第一種情況下是需要的，因爲它是等效於：

int diff = ((long)(int)start) - stop; 

Answer 3

int diff = (int) start - stop; 

等於

int diff = (int) (long) start - (long) stop; // we substract long types 

可以簡化爲

int diff = (int) start - (long) stop; // (int)(long) x means (int) x 

等於

int diff = ((int) start) - (long) stop; // casting is applied only to start 

等於

int diff = ((long)(int) start) - (long) stop; // compiler make types equal 

可以簡化爲

int diff = (long) start - (long) stop; // we can substract now 

等於

int diff = (long) startMinusStop; // because long - long => long 

在這裏我們是可讀的錯誤Type mismatch: cannot convert from long to int。

Answer 4

在您的第一個聲明中，您只會將變量start轉換爲整數。所以結果是int - long這是一個很長的，不適合整數。它會像int = (int - long)和java不會允許。