I am basically trying to figure out a technique in which I can scan a string from left to right and it will give me the integer of it.
Starting from right to left is quiet easy as shown in the code below.
I am just struggling on finding a way to add each integer to make it the correct value it is supposed to be.
Output:
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y
enter an expression
2567
value = 7652
I am trying to get the output to be in the correct order.
e = String input // string needing to be converted to double
p = 0 // position
// If a substring of e whose rightmost character is at position p
// represents a number (possibly with a decimal point, possibly negative),
// return the numerical value of the substring as a double value and
// re-position p just to the left of the substring.
private double formNum() {
double total = 0.0;
int count = 0;
int flag = 0;
double mult = 1;
char c, d;
do {
c = e.charAt(p); // examine the current character in the string (from right to left)
if (c == '.') {
flag = 1; // set a flag to remember that the character is a decimal point
} else {
// if the current character is a digit, convert to its
// int value and include it in the value corresponding to the string.
if ((c >= '0') && (c <= '9')) {
total = total + mult * (c - '0');
mult = mult * 10.0;
if (flag == 0) {
count++; // count the number of digits to the right of a possible decimal point
}
} else {
if (c == '(' && e.charAt(p + 1) == '-') {
total = -total; // an underscore character represents a negative value
}
}
}
p++; // Prepare to move to the next character to the left.
// This is a private non-static method because it changes the member variable p
d = '?';
if (p <= e.length() - 1) {
d = e.charAt(p); // the next character to the left
}
} while ((p <= e.length() - 1) && (((d <= '9') && (d >= '0')) || (d == '_') || (d == '.')));// check for a valid
// character
if (flag == 1) {
total = total / Math.pow(10, count * 1.0); // compute the value taking into account
// the number of decimal places
}
return total;
}
question from:
https://stackoverflow.com/questions/66057000/compute-total-from-scanning-string-from-left-to-right-instead-of-right-to-left 与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…