我正在尝试计算一个算术表达式，它以字符串形式输入（例如，( 5+4*5-1/8 )，它将给出结果3）。我输入一个表达式并将其转换为数组。第一;结果将从第一个元素开始，它将在循环中更改。但问题是运营商优先。如何在循环中使用运算符presedence？这是我的代码：
import java.util.Scanner;
public class HesapMakinesi {
private char value[];
private int count;
private Scanner str = new Scanner(System.in);
private String process;
HesapMakinesi() {
System.out.print("Enter the process ");
process = str.next();
//System.out.println(islem);
Initializer(process);
}
private void Initializer(String process) {
count = process.toCharArray().length;
value = new char [count];
int i;
System.arraycopy(process.toCharArray(), 0, value, 0, count);
//System.out.println(value);
if(value[0]=='-' || value[0]=='+' || value[0]=='/' || value[0]=='*' || // A process cannot start with an operator
value[count-1]=='-' || value[count-1]=='+' || value[count-1]=='/' || value[count-1]=='*') {
System.out.println("You have entered a wrong process.Please enter again!!!");
System.out.print("Enter the process: ");
process = str.next();
Initializer(process);
}
for(i=0; i<count; i++) { // A process cannot include a character except operators
if( value[i]!='+' && value[i]!='-' && value[i]!='*' && value[i]!='/' && value[i]!='(' && value[i]!=')' && !Character.isDigit(value[i]) ) {
System.out.println("You have entered a wrong process.Please enter again!!!");
System.out.print("Enter the process: ");
process = str.next();
Initializer(process);
}
}
for(i=0; i<count-1; i++) { // A process cannot have operators sequantially
if( !Character.isDigit(value[i]) && !Character.isDigit(value[i+1]) ) {
if( (value[i] == '+' && value[i+1] == '+' ) || (value[i] == '+' && value[i+1] == '-' ) || (value[i] == '+' && value[i+1] == '*' ) ||
(value[i] == '+' && value[i+1] == '/' ) ) {
System.out.println("You have entered a wrong process.Please enter again!!!");
System.out.print("Enter the process: ");
process = str.next();
Initializer(process);
}
else if( (value[i] == '-' && value[i+1] == '+' ) || (value[i] == '-' && value[i+1] == '-' ) || (value[i] == '-' && value[i+1] == '*' ) ||
(value[i] == '-' && value[i+1] == '/' ) ) {
System.out.println("You have entered a wrong process.Please enter again!!!");
System.out.print("Enter the process: ");
process = str.next();
Initializer(process);
}
else if( (value[i] == '*' && value[i+1] == '+' ) || (value[i] == '*' && value[i+1] == '-' ) || (value[i] == '*' && value[i+1] == '*' ) ||
(value[i] == '*' && value[i+1] == '/' ) ) {
System.out.println("You have entered a wrong process.Please enter again!!!");
System.out.print("Enter the process: ");
process = str.next();
Initializer(process);
}
else if( (value[i] == '/' && value[i+1] == '+' ) || (value[i] == '/' && value[i+1] == '-' ) || (value[i] == '/' && value[i+1] == '*' ) ||
(value[i] == '/' && value[i+1] == '/' ) ) {
System.out.println("You have entered a wrong process.Please enter again!!!");
System.out.print("Enter the process: ");
process = str.next();
Initializer(process);
}
}
}
//sCount();
}
/*private void Count(){
double result,temp;
int i;
for(i=0; i<count; i++) {
if( value[i]!= )
}
}*/
}
答案 0 :(得分：1)
这不是你怎么做的。您需要在评估表达式之前解析表达式。我建议你阅读Shunting-yard algorithm。
答案 1 :(得分：0)
继续我的评论...如果你正在处理一个简单的表达式，你只能有数字和符号+ - / *那么你可以采用一种简单的方法：
在你的例子中，你最终会得到这样的东西：