問題描述
我一直在試圖找出原因，但我找不到.有人可以幫幫我嗎?
I've been trying to find out the reason, but I couldn't. Can anybody help me?
看下面的例子.
float f = 125.32f;
System.out.println("value of f = " + f);
double d = (double) 125.32f;
System.out.println("value of d = " + d);
這是輸出:
value of f = 125.32
value of d = 125.31999969482422
推薦答案
float 的值在轉換為 double 時不會改變.顯示的數字有所不同，因為需要更多數字來區分 double 值與其相鄰值，即 Java 文檔要求.那是 toString 的文檔，從 println 的文檔中引用(通過幾個鏈接).
The value of a float does not change when converted to a double. There is a difference in the displayed numerals because more digits are required to distinguish a double value from its neighbors, which is required by the Java documentation. That is the documentation for toString, which is referred (through several links) from the documentation for println.
125.32f 的確切值是 125.31999969482421875.兩個相鄰的 float 值是 125.3199920654296875 和 125.32000732421875.觀察到 125.32 比任何一個鄰居都更接近 125.31999969482421875.因此，通過顯示125.32"，Java 顯示了足夠的數字，以便從十進制數字轉換回 float 再現了傳遞給 println<的 float 的值/代碼>.
The exact value for 125.32f is 125.31999969482421875. The two neighboring float values are 125.3199920654296875 and 125.32000732421875. Observe that 125.32 is closer to 125.31999969482421875 than to either of the neighbors. Therefore, by displaying "125.32", Java has displayed enough digits so that conversion back from the decimal numeral to float reproduces the value of the float passed to println.
在兩個相鄰<代碼>雙的125.3199996948242的 1875 是125.3199996948242的 045391452847979962825775146484375 并125.3199996948242的 329608547152020037174224853515625 即可.值結果觀察到 125.32 更接近后一個鄰居而不是原始值 (125.31999969482421875).因此，打印125.32"不包含足夠的數字來區分原始值.Java 必須打印更多的數字，以確保從顯示的數字轉換回 double 再現傳遞給 println 的 double 的值.
The two neighboring double values of 125.31999969482421875 are 125.3199996948242045391452847979962825775146484375 and 125.3199996948242329608547152020037174224853515625.
Observe that 125.32 is closer to the latter neighbor than to the original value (125.31999969482421875). Therefore, printing "125.32" does not contain enough digits to distinguish the original value. Java must print more digits in order to ensure that a conversion from the displayed numeral back to double reproduces the value of the double passed to println.
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