当我将其设置为变量时，它会更改值。 有谁知道为什么？

Question

int temp = a1.charAt(num);
System.out.println(a1.charAt(num));
System.out.println(temp);

(a1 is a string that contains 6 8 1 5...)

(num is 0)

the output is:

6

54

Answer 1

(int)'6' == 54
'6' == (char)54
'6' == (char)0x36

cf. ASCII

System.out.println has different overloads for parameters of type int (prints the numeric/ordinal representation of the value) and for parameters of type char (prints the character representation of the ordinal value).

Answer 2

It is because the .chatAt() method returns a char and you are storing it in an integer variable. Hence, it is stored as it's ASCII value.

6 in ASCII is 54.

Link - https://www.cs.cmu.edu/~pattis/15-1XX/common/handouts/ascii.html

Answer 3

ASCII value of 6 is 54.

When you a1.charAt(0) it prints character 6 from the string = 6815.

But when you assign char 6 to int, then it prints its equivalent ASCII value, which is 54.