[英]evaluate expression binary tree c++
我正在嘗試學習這種對表達式進行評估的二叉樹的實現。 我無法運行它並查看輸出。 我將如何獲得3 *(7 + 1)/ 4 +(17-5),結果為18。這是鏈接http://math.hws.edu/eck/cs225/s03/binary_trees/
class ExpNode {
// Represents a node of any type in an expression tree.
// This is an "abstract" class, since it contains an undefined
// function, value(), that must be defined in subclasses.
// The word "virtual" says that the defintion can change
// in a subclass. The "= 0" says that this function has
// no definition in this class.
public:
virtual double value() = 0; // Return the value of this node.
}; // end class ExpNode
class ConstNode : public ExpNode {
// Represents a node that holds a number. (The
// ": public ExpNode" says that this class is
// a subclass of ExpNode.)
double number; // The number in the node.
public:
ConstNode( double val ) {
// Constructor. Create a node to hold val.
number = val;
}
double value() {
// The value is just the number that the node holds.
return number;
}
}; // end class ConstNode
class BinOpNode : public ExpNode {
// Represents a node that holds an operator.
char op; // The operator.
ExpNode *left; // The left operand.
ExpNode *right; // The right operand.
public:
BinOpNode( char op, ExpNode *left, ExpNode *right ) {
// Constructor. Create a node to hold the given data.
this->op = op;
this->left = left;
this->right = right;
}
double value() {
// To get the value, compute the value of the left and
// right operands, and combine them with the operator.
double leftVal = left->value();
double rightVal = right->value();
switch ( op ) {
case '+': return leftVal + rightVal;
case '-': return leftVal - rightVal;
case '*': return leftVal * rightVal;
case '/': return leftVal / rightVal;
}
}
}; // end class BinOpNode
這是我嘗試做的主要功能:
int main() {
BinOpNode *opnode;
opnode = new BinOpNode;
opnode->value()=5;
ExpNode *expnode;
expnode = opnode;
expnode->value();
return 0;
}
它不編譯,這是錯誤
15:58:27 **** Incremental Build of configuration Debug for project ExpNode ****
Info: Internal Builder is used for build
g++ -O0 -g3 -Wall -c -fmessage-length=0 -o "src\\ExpNode.o" "..\\src\\ExpNode.cpp"
..\src\ExpNode.cpp: In function 'int main()':
..\src\ExpNode.cpp:60:15: error: no matching function for call to 'BinOpNode::BinOpNode()'
..\src\ExpNode.cpp:36:2: note: candidates are: BinOpNode::BinOpNode(char, ExpNode*, ExpNode*)
..\src\ExpNode.cpp:30:33: note: BinOpNode::BinOpNode(const BinOpNode&)
..\src\ExpNode.cpp:61:18: error: lvalue required as left operand of assignment
15:58:28 Build Finished (took 405ms)
沒有一個類具有默認構造函數。
value
返回評估表達式的結果,並且在構造表達式時,需要將表達式的必要部分作為其參數傳遞。
(尚不清楚如何期望將值5分配給二進制表達式。)
您需要從葉子(將是常量)到根部構建一棵樹。
例如,這是表達式5 + 3
:
ConstNode five(5);
ConstNode three(3);
BinOpNode fiveplusthree('+', &five, &three);
std::cout << fiveplusthree.value(); // Should print 8
我認為問題出在您的main()函數的邏輯中。
根據給定類的定義,首先應為表達式中的每個數字創建一個類型為ConstNode
的對象。 然后,您應該為表達式中的每個運算符創建BinOpNode
。
順便說一下,該表達式的值為18,而不是82!
像這樣:
//3*(7+1)/4+(17-5) = 18
int main()
{
BinOpNode *a, *b;
a = new BinOpNode('+', new ConstNode(7), new ConstNode(1));
a = new BinOpNode('*', new ConstNode(3), a);
a = new BinOpNode('/', a, new ConstNode(4));
b = new BinOpNode('-', new ConstNode(17), new ConstNode(5));
b = new BinOpNode('+', a, b);
cout << b->value();
}
PS:當ExpNode
的構造函數中需要BinOpNode
的對象時,我們可以傳遞ConstNode
類的對象,因為ConstNode
繼承自ExpNode
抽象基類。
在C ++中,默認構造函數的工作有趣。
如果您未定義任何構造函數,則會為您生成一個默認構造函數:
class A {};
int main()
{
A a; // perfectly fine
但是,如果定義任何其他構造函數,這些生成的構造函數就會消失:
class A
{
A(int) { ... }
};
int main()
{
A a; // ERROR!
}
在這種情況下,默認構造函數不存在,因為您定義了一個構造函數,並且編譯器沒有為您生成一個。
這是您的問題,因為在main
,您有以下一行:
opnode = new BinOpNode;
運行默認的BinOpNode
構造BinOpNode
。 查看您的BinOpNode
構造函數:
BinOpNode( char op, ExpNode *left, ExpNode *right )
嘿,這不是默認的構造函數!
您有兩個選擇:向該類添加默認構造函數:
BinOpNode() { ... }
或在調用new
時使用參數:
opnode = new BinOpNode(op, left, right);
祝好運!
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