string ="abc"+“d”+“e”
时间: 2024-05-21 16:11:02 浏览: 8
Invalid syntax. A string must be enclosed in quotes, and if you want to concatenate multiple strings, you must use the + operator between them. Here are some examples:
- s = "abc" + "d" + "e" -> s will be "abcde"
- s = "abc" "d" "e" -> s will be "abcd" "e" (two separate string literals)
- s = "abc" "d" + "e" -> s will be "abcd" + "e" -> s will be "abcde"
相关问题
java String str="abc" 创建了几个对象 ;java String s = a+b+c+d+e 创建了几个对象
第一个问题:在Java中,创建了2个对象。一个是字符串常量池中的"abc",另一个是在堆中的String对象,它引用了字符串常量池中的"abc"。
第二个问题:在Java中,创建了5个对象。在编译时,"a"、"b"、"c"、"d"和"e"都被当作字符串常量处理,会被放入字符串常量池中。然后,在运行时,会创建4个String对象。首先,"a"和"b"会被拼接成一个新的String对象,然后再将其与"c"拼接成另一个新的String对象,以此类推,直到拼接完成。因此,总共创建5个对象。
JAVA语言用1, 2, 3...9 这九个数字组成一个数学公式,满足:ABC + DEF = GHI,每个数字只能出现一次,编写程序输出所有的组合。
下面是用Java语言编写的程序,可以输出所有符合要求的组合:
```java
public class Main {
public static void main(String[] args) {
for(int a = 1; a <= 9; a++) {
for(int b = 1; b <= 9; b++) {
if(b == a) continue;
for(int c = 1; c <= 9; c++) {
if(c == a || c == b) continue;
for(int d = 1; d <= 9; d++) {
if(d == a || d == b || d == c) continue;
for(int e = 1; e <= 9; e++) {
if(e == a || e == b || e == c || e == d) continue;
for(int f = 1; f <= 9; f++) {
if(f == a || f == b || f == c || f == d || f == e) continue;
for(int g = 1; g <= 9; g++) {
if(g == a || g == b || g == c || g == d || g == e || g == f) continue;
for(int h = 1; h <= 9; h++) {
if(h == a || h == b || h == c || h == d || h == e || h == f || h == g) continue;
int abc = a * 100 + b * 10 + c;
int def = d * 100 + e * 10 + f;
int ghi = g * 100 + h * 10 + i;
if(abc + def == ghi) {
System.out.println(abc + " + " + def + " = " + ghi);
}
}
}
}
}
}
}
}
}
}
}
```
这个程序使用了6层嵌套的循环来枚举所有可能的组合,并且通过判断每个数字是否已经被使用过来避免重复。最后,将符合要求的组合输出到控制台上。