问题描述
所以我有这样的：
char cr = "9783815820865".charAt(0);
System.out.println(cr); //prints out 9
如果我这样做：
int cr = "9783815820865".charAt(0);
System.out.println(cr); //prints out 57
我理解char和int之间的转换不是简单地从'9'到 9 。我的问题是现在我只需要保持9作为int值，而不是57.如何获取值9而不是57作为 int 类型？
I understand that the conversion between char and int is not simply from '9' to 9. My problem is right now I simply need to keep the 9 as the int value, not 57. How to get the value 9 instead of 57 as a int type?
推荐答案
您可以尝试：
int cr = "9783815820865".charAt(0) - '0';
charAt（0） c $ c>'9'（作为 char ），它是一个数字类型。从这个值，我们只是减去'0'的值，它再次是数值，并且在' '字符。
charAt(0) will return '9' (as a char), which is a numeric type. From this value we'll just subtract the value of '0', which is again numeric and is exactly nine entries behind the entry of the '9' character in the ASCII table.
因此，在幕后，减法将使用'9'和'0'，这意味着将计算 57 - 48 。
So, behind the scenes, the the subtraction will work with the ASCII codes of '9' and '0', which means that 57 - 48 will be calculated.
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