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[英](java) convert a decimal number to binary without using parseint
[英]binary to decimal converter without using parseint or arrays
使字符串索引超出范圍,但我不明白為什么我經歷了大約 50 次
import java.util.Scanner;
public class binary {
public static void main(String[] args) {
System.out.println("Enter the first binary number");
Scanner keyboard = new Scanner(System.in);
String num1 = keyboard.next();
//System.out.println("Enter the second binary number");
//String num2 = keyboard.next();
int total = 0;
for(int i = num1.length(); i>0;i--) {
if(num1.charAt(i) == 1) {
total += 2*i;
}
}
if(num1.charAt(3) == 1) {
total -= 1;
}
System.out.println(total);
}
}
這是您嘗試做的事情的完整解決方案,包括一組測試:
class binary {
private static int binaryToInt(String binary) {
int total = 0;
for (int i = 0 ; i < binary.length(); i++) {
total *= 2;
if (binary.charAt(i) == '1')
total += 1;
}
return total;
}
private static void test(String binary, int expected) {
int n = binaryToInt(binary);
String rightWrong = "right";
if (n != expected) {
rightWrong = String.format("WRONG! (should be %d)", expected);
System.out.printf("%s -> %d is %s\n", binary, n, rightWrong);
}
public static void main(String[] args) {
test("0", 0);
test("1", 1);
test("10", 2);
test("100", 4);
test("111", 7);
test("0000111", 7);
test("1010101010", 682);
test("1111111111", 1023);
System.out.println("");
// test sanity check
System.out.println("This last test should fail (we are just testing the test method itself here)...");
test("1010101010", 0);
}
}
結果:
0 -> 0 is right
1 -> 1 is right
10 -> 2 is right
100 -> 4 is right
111 -> 7 is right
0000111 -> 7 is right
1010101010 -> 682 is right
1111111111 -> 1023 is right
This last test should fail (we are just testing the test method itself here)...
1010101010 -> 682 is WRONG! (should be 0)
您的代碼中的一個重要問題尚未在評論或早期答案中得到解決。 請注意這一行與代碼中的行:
if (binary.charAt(i) == '1')
您正在測試數值1
,這永遠不會是true
,因為您從charAt()
取回一個字符,而不是一個數字。
雖然length()
計算元素的數量,但它們的索引從 0 開始。對於“1111”字符串,最后一個字符的索引為 3,而不是 4,因此.length()-1
。 您需要將 for 語句更改為for(int i = num1.length()-1; i>=0;i--)
(還要注意條件更改)或將 charAt 語句更改為if(num1.charAt(i-1) == '1')
。
另外,根據您要執行的操作,我假設對於total += 2*i
您實際上需要像total += Math.pow(2, i-length())
類的東西,具體取決於您首先決定對i
做什么.
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