So I was tinkering with || and && operators with return. I developed an understanding by comparing int as for example return 1 || 0; return 0 || 0; that the program returns 1 or 0 in case of an int function and true or false in case of bool functions after comparing values.
Now I was writing a code to find path sum to given value using implementation of trees.
bool HasPathSum(ds::sn::BNode<int>* root, int value)
{
if (!root && !value)
{
return true;
}
else if(root)
{
return HasPathSum(root->left, value - root->data)
|| HasPathSum(root->right, value - root->data);
}
return false;
}
I managed to write it after massive tinkering. For recursion I was doing
HasPathSum(root->left, value - root->data);
return HasPathSum(root->right, value - root->data);
To explore all the paths, but it wasn't quite all right because if the path sum was in left, It did go to true but eventually returned false. I figured I will have to write the recursive definition in a single line for it to work and wrote || just as an experiment and it blew my mind that it actually worked. I wrote std::cout << value << " "; at the top of if statement to see what is happening and it does what its supposed do and stops just as a path has been found.
This is all that I've tried but I am unable too wrap my head around how the recursive function with || works or will work if given &&.
If someone experienced can provide an explanation! :)
operator || and operator && does short circuit, so might doesn't evaluate second operand depending of the first one.
bool b = f() || g(); is mostly equivalent to
bool b = f();
if (b == false) b = g();
bool b = f() && g(); is mostly equivalent to
bool b = f();
if (b == true) b = g();
Related
I have written the following infix to postfix program but it's not working.
My program takes input but doesn't show any result. Can anyone help find the problem in my program.
And also it would be a great help if you tell if my Algorithm for converting infix to postfix is correct or not.
using namespace std;
class Stack
{
private:
int top;
char s[mx];
public:
Stack()
{
top=-1;
}
void push(char c)
{
if(!stackFull())
s[++top]=c;
}
void pop()
{
if(!stackEmpty())
top--;
else cout<<"Stack is empty"<<endl;
}
char topShow()
{
if(!stackEmpty())
return s[top];
}
bool stackEmpty()
{
if(top==-1)
return 1;
else return 0;
}
bool stackFull()
{
if(top == (mx-1))
return 1;
else return 0;
}
};
class Expression
{
private:
char entry2;
int precedence;
char infix[mx];
char postfix[mx];
public:
int prec(char symbol)
{
switch(symbol)
{
case '(':return 0; break;
case '-':return 1; break;
case '+':return 2; break;
case '*':return 3; break;
case '/':return 4; break;
}
}
void Read()
{
cout<<"Enter the infix expression: ";cin>>infix;
for(int i=0;infix[i]!='\0';i++)
{
convertToPostfix(infix[i]);
}
}
void ShowResult()
{
cout<<"Postfix expression"<<endl;
for(int j=0;postfix[j]!='\0';j++)
{
cout<<postfix[j];
}
}
void convertToPostfix(char c)
{
int p=0;
Stack myStack;
precedence=prec(c);
entry2=myStack.topShow();
if(isdigit(c))
{
postfix[++p]=c;
}
if(precedence>prec(entry2))
{
myStack.push(c);
}
if(precedence<prec(entry2))
{
switch(c)
{
case '(': myStack.push(c); break;
case ')': while(myStack.topShow()!= '(')
{
postfix[++p]=myStack.topShow();
myStack.pop();
};myStack.pop();break;
case '+':
case '-':
case '*':
case '/': while(prec(myStack.topShow())>=precedence)
{
postfix[++p]=myStack.topShow();
myStack.pop();
};break;
}
}
}
};
int main()
{
Expression myExp;
myExp.Read();
myExp.ShowResult();
return 0;
}
Here are some issues I found:
Boolean Functions Return true or false
Match return types with return values. The numbers 1 and 0 are not Boolean values.
Precedence table
Add and subtract have same precedence.
Multiply and divide have same precedence.
Multiply and divide have higher precedence than add and subtract.
Stack disappears
Since the stack is declared as a local variable in the function, it will be created fresh when entering the function and destroyed before exiting the function.
Solution: move it to the class as a class member or declare it as static.
Multiple statements per line are not more efficient
Blank lines and newlines do not affect performance, and add negligible time to the build.
However, they make your program more readable which helps when inspecting or debugging. Use them.
And similarly with space before and after operators.
Build the habit now rather than correcting when you get a job.
Call function once and store the value
You call prec(entry2) twice, which is a waste of time. Call it once and save the value in a variable. Similarly with stack.TopShow().
Use std::vector not an array
The std::vector will grow as necessary and reduce the chance of buffer overflow.
With an array, you must check that your indices are always within range. Also, array capacities don't change; you have to declare a new instance and copy the data over.
The variable mx is not declared
The compiler should catch this one. You use mx as the capacity for an array and comparing for full. However, it is never declared, defined nor initialized. Prefer std::vector and you won't have to deal with these issues.
Input is not validated
You input a letter, but don't validate it.
Try these characters: space, #, #, A, B, etc.
Missing default for switch
Crank up your compiler warnings to maximum.
Your switch statements need defaults.
What precedence do numeric characters ('0'..'9') have?
(You check the precedence of numeric characters.)
Check all paths through your functions and program.
Using a debugger (see below) or pen and paper, check your program flow through you functions. Include boundary values and values not within the bounds.
Case statements: break or return
You don't need a break after a return statement. Think about it. Can the program continue executing at the line after a return statement?
Use a debugger or print statements
You can print variables at different points in your program. This is an ancient technique when debuggers are not available.
Learn to use a debugger. Most IDEs come with them. You can single step each statement and print out variable values. Very, very, useful.
class infixToPostfix{
public static void postfix(String str){
Stack<Character> stk = new Stack<Character>();
for(Character c : str.toCharArray()){
// If operands appears just print it
if(c >= 'A' && c <= 'Z' || c >= 'a' && c <= 'z'){
System.out.print(c);
}else{
// Open paranthesis push is
if(c == '('){
stk.push(c);
//Close paranthesis pop until close paranthesis
}else if( c == ')'){
while(stk.peek() != '(')
System.out.print(stk.pop());
stk.pop();
// check the precedence of operator with the top of stack
}else if(c == '+' || c == '-'){
if(!stk.isEmpty()){
char top = stk.peek();
if(top == '*' || top == '/' || top == '+' || top == '-'){
System.out.print(stk.pop());
}
}
stk.push(c);
}else{
if(!stk.isEmpty()){
char top = stk.peek();
if(top == '/' || top == '*'){
System.out.print(stk.pop());
}
}
stk.push(c);
}
}
}
//Print all the remaining operands
while(!stk.isEmpty()) System.out.print(stk.pop());
System.out.println();
}
public static void main(String args[]){
String str = "A+B-(c+d*Z+t)/e";
postfix(str);
}
}
using stack and map u can solve the problem
1) create a map having operator as key and some integer to set priority. operator with same precedence will have same value something like:
map<char,int>oprMap;
oprMap['^'] = 3;
oprMap['*'] = 2;
oprMap['/'] = 2;
oprMap['+'] = 1;
oprMap['-'] = 1;
2) iterate through given expression call these checks
1) if current element
i) is operand add it to result
ii) not operand do following check
a. while not (stack is empty and element is open bracket and found operator with higher precedence.
add top of the stack to the result and pop()
b. push current element to stack
iii) if open brackets push to stack
iv) if closed brackets pop until get closed bracket and add it to result
3) while stack is not empty pop() and add top element to the result.
{
stack<char>S;
for (int i = 0; i < n; i++) {
if(isOperand(exps[i])) {
res = res + exps[i];
} else if(isOperator(exps[i])){
while(!(S.empty() && isOpenParanthesis(S.top()) && isHeigherPrecedence(S.top(),exps[i])){
res = res+S.top();
S.pop();
}
S.push(exps[i]);
} else if(isOpenParanthesis(exps[i])) {
S.push(exps[i]);
} else if(isClosingParanthesis(exps[i])) {
while(!S.empty() && !isOpenParanthesis(S.top())) {
res = res+S.top();
S.pop();
}
S.pop();
}
}
while(!S.empty()) {
res = res + S.top();
S.pop();
}
}
}
#include<bits/stdc++.h>
using namespace std;
// This isHigher function checks the priority of character a over b.
bool isHigher(char a,char b)
{
if(a=='+' || a=='-')
return false;
else if((a=='*' && b=='*') || (a=='*' && b=='/') || (a=='/' && b=='*') ||
(a=='/' && b == '/')|| (a=='^' && b=='^')||(a=='*' && b=='^') || (a=='/' &&
b=='^'))
return false;
return true;
}
int main(){
string s;
cin>>s;
s = s + ")";
//Vector postfix contains the postfix expression.
vector<char>postfix;
stack<char>mid;
mid.push('(');
for(int i=0;i<s.length();i++)
{
if(s[i] == '(')
mid.push(s[i]);
else if(s[i] == '+' || s[i] == '^' || s[i] == '-' || s[i] == '*'||
s[i] == '/')
{
if(mid.top() == '(')
mid.push(s[i]);
else {
if(isHigher(s[i],mid.top()))
mid.push(s[i]);
else
{
while(mid.top()!='(')
{
if(!isHigher(s[i],mid.top()))
{
postfix.push_back(mid.top());
mid.pop();
}
else
break;
}
mid.push(s[i]);
}
}
}
else if(s[i] == ')')
{
while(mid.top() != '(')
{
postfix.push_back(mid.top());
mid.pop();
}
mid.pop();
}
else
postfix.push_back(s[i]);
}
for(int i=0;i<postfix.size();i++)
cout<<postfix[i];
return 0;
}
This code that I have is not working as it should.
for (int i = 0; i <= 140; i++)
{
if (OneLine_Array.GetAt(i) == "Pass" || "Fail" || "Warn" || "Active")
{
OneLine_State.Add(OneLine_Array.GetAt(i));
}
}
It will work though if i have it as
for (int i = 0; i <= 140; i++)
{
if ((OneLine_Array.GetAt(i) == "Pass") || (OneLine_Array.GetAt(i) == "Fail") || (OneLine_Array.GetAt(i) == "Warn") || (OneLine_Array.GetAt(i) == "Active"))
{
OneLine_State.Add(OneLine_Array.GetAt(i));
}
}
I was wondering is there a shorter way of doing this rather than replicating the same line of code over and over again?
thanks.
You can store the result of GetAt before you get to the condition, then you don't need to evaluate it multiple times:
auto x = OneLine_Array.GetAt(i);
if (x == "Pass" || x == "Fail" || x == "Warn" || x == "Active")
For a sufficiently short variable name (but please call it something nicer than x), this will be shorter than your second example.
The reason your original condition doesn't work is because each of "Fail", "Warn", and "Active" are true, so the logical OR will also be true. They are true because they are of array type which can be converted to a pointer to their first elements (char*). A non-null pointer is converted to true.
you can't do this ( OneLine_Array.GetAt(i) == "Pass" || "Fail" || "Warn" || "Active") ==> this is not possible in C++
it will be better to save the string of the OneLine_Array.GetAt(i) in variable and make the "=="
Since you mention MFC C++. With a better performance, I'd suggest using CMap (http://msdn.microsoft.com/en-us/library/s897094z.aspx) to store your strings. It does speed up the comparision.
Although I prefer what #sftrabbit answered, you could write a little helper for this.
template<typename A, typename B>
bool equals_one_of(const A& a, const B& b)
{
return a == b;
}
template<typename A, typename B, typename... Args>
bool equals_one_of(const A& a, const B& b, const Args&... args)
{
return equals_one_of(a, b) || equals_one_of(a, args...);
}
int f()
{
if (equals_one_of(OneLine_Array.GetAt(i), "Pass", "Fail", "Warn", "Active"))
{
// ....
}
}
struct Something {
union {
float k;
int n;
};
bool isFloat;
bool operator==(const Something& mS)
{
if(isFloat != mS.isFloat) return false;
if(isFloat && mS.k == k) return true;
if(!isFloat && mS.n == n) return true;
}
};
My implementation of Something::operator== seems rather expensive and convoluted. Is this the only way to check equality in classes with union types?
Or is there a better way that avoids branches/checking additional variables?
bool operator==(const Something& mS)
{
if (isFloat != mS.isFloat)
{
return false;
}
else if (isFloat)
{
return mS.k == k;
}
else
{
return mS.n == n;
}
}
Clear and debuggable with the minimum number of checks. You want to have a constructor and/or set methods to ensure isFloat is correct at all times.
You can remove one redundant check, and perhaps enhance readability slightly, by replacing the last two lines with
if(isFloat != mS.isFloat) return false; // As you have
return isFloat ? mS.k == k : mS.n == n;
(or the equivalent if construct, as in Sean Perry's answer) but the compiler will probably do just as good a job of optimising your version.
There's no way to avoid a runtime check that the types match. You might consider a ready-made discriminated union type like Boost.Variant; it won't be any more efficient, but it might be easier and less error-prone to use.
return (isFloat && mS.isFloat && k==mS.k) || (!isFloat && !mS.isFloat && n==mS.n);
I do not think that you can escape checking all the conditions. So the question can be how to write them more simpler and expressively.
I would write them the following way
bool operator==( const Something &mS ) const
{
return ( ( isFloat == mS.isFloat ) && ( isFloat ? k == mS.k : n == mS.n ) );
}
I have written the following function to check of two b-trees are similar or not. But I am not getting the desired output.
int bt_similar(btree *b1, btree *b2)
{
int m=1;
if ((b1->data==b2->data&&(((b1->lchild==NULL&&b2->lchild==NULL)||(b1->lchild!=NULL&&b2->lchild!=NULL))&&((b1->rchild==NULL&&b2->rchild==NULL)||(b1->rchild!=NULL&&b2->rchild!=NULL))))&&(m))
{
if (b1->lchild!=NULL)
m=bt_similar(b1->lchild,b2->lchild);
if (b1->rchild!=NULL)
m=bt_similar(b1->rchild,b2->rchild);
if (m)
return 1;
}
return 0;
}
Firstly, you could benefit a lot from breaking up your code, in particular that big condition in the if statement on the fourth line is nasty. Why not just handle the null case in the method itself? eg:
if (b1 == null && b2 != null)
return 0;
if (b2 == null && b1 != null)
return 0;
if (b1 == null && b2 == null)
return 1;
As far as I can tell, the actual problem with your code is where you overwrite the value of m at line 9, without taking the previous value into consideration. If the left-hand-side of the tree is not similar but the right is, the return will be 1. Instead of:
if (b1->lchild!=NULL)
m=bt_similar(b1->lchild,b2->lchild);
if (b1->rchild!=NULL)
m=bt_similar(b1->rchild,b2->rchild);
You should have:
if (b1->lchild!=NULL)
m = bt_similar(b1->lchild,b2->lchild);
if (b1->rchild!=NULL)
m = m && bt_similar(b1->rchild,b2->rchild);
you should make sure that if b1 == NULL or b2 == NULL...
bool compare(struct node* b1, struct node* b2) {
// 1. both empty -> true
if (b1==NULL && b2==NULL) return(true);
// 2. both non-empty -> compare them
else if (b1!=NULL && b2!=NULL) {
return(
b1->data == b2->data &&
compare(b1->lchild, b2->lchild) &&
compare(b1->rchild, b2->rchild));
}
// 3. one empty, one not -> false
else return false;
}
and, b-tree is not short for binary tree.
can this be done somehow?
if((a || b) == 0) return 1;
return 0;
so its like...if a OR b equals zero, then...but it is not working for me.
my real code is:
bool Circle2::contains(Line2 l) {
if((p1.distanceFrom(l.p1) || p1.distanceFrom(l.p2)) <= r) {
return 1;
}
return 0;
}
You need to write the full expression:
(a==0)||(b==0)
And in the second code:
if((p1.distanceFrom(l.p1)<= r) || (p1.distanceFrom(l.p2)<=r) )
return 1;
If you do ((a || b) == 0) this means "Is the logical or of a and b equal to 0. And that's not what you want here.
And as a side note: the if (BooleanExpression)return true; else return false pattern can be shortened to return BooleanExpression;
You have to specify the condition separately each time:
if (a == 0) || (b == 0))
bla bla;
When you do
if ((a || b) == 0)
bla bla;
it has a different meaning: (a || b) means "if either a or b is non-zero (ie. true), then the result of this expression is true".
So when you do (a||b) == 0, you are checking if the result of the previously explained expression is equal to zero (or false).
The C++ language specifies that the operands of || ("or") be boolean expressions.
If p1.distanceFrom(l.p1) is not boolean (that is, if distanceFrom returns int, or double, or some numeric class type), the compiler will attempt to convert it to boolean.
For built in numeric type, the conversion is: non-zero converts to true, zero converts to false. If the type of p1.distanceFrom(l.p1) is of class type Foo, the compiler will call one (and only one) user defined conversion, e.g., Foo::operator bool(), to convert the expression's value to bool.
I think you really want something like this:
bool Circle2::contains(Line2 l) {
if((p1.distanceFrom(l.p1) <= r) || (p1.distanceFrom(l.p2) <= r)) return 1;
return 0;
}
Fun with templates:
template <typename T>
struct or_t
{
or_t(const T& a, const T& b) : value1(a), value2(b)
{
}
bool operator==(const T& c)
{
return value1 == c || value2 == c;
}
private:
const T& value1;
const T& value2;
};
template <typename T>
or_t<T> or(const T& a, const T& b)
{
return or_t<T>(a, b);
}
In use:
int main(int argc, char** argv)
{
int a = 7;
int b = 9;
if (or(a, b) == 7)
{
}
return 0;
}
It performs the same comparison you would normally do, though, but at your convenience.
If you have lot of that code, you may consider a helping method:
bool distanceLE (Point p1, Point p2, double threshold) {
return (p1.distanceFrom (p2) <= threshold)
}
bool Circle2::contains (Line2 l) {
return distanceLE (p1, l.p1, r) && distanceLE (p1, l.p2, r);
}
If you sometimes have <, sometimes <=, >, >= and so on, maybe you should pass the operator too, in form of a function.
In some cases your intentions by writing this:
if ((a || b) == 0) return 1;
return 0;
could be expressed with an bitwise-or:
if ((a | b) == 0) return 1;
return 0;
and simplified to
return ! (a | b);
But read up on bitwise operations and test it carefully. I use them rarely and especially I didn't use C++ for some time.
Note, that you inverted the meaning between your examples 1 and 2, returning true and false in the opposite way.
And bitwise less-equal doesn't make any sense, of course. :)
C++ doesn't support any construct like that. Use if (a == 0 || b == 0).
Your condition should be (a == 0 || b == 0) or (p1.distanceFrom(l.p1) <= r || p1.distanceFrom(l.p2)) <= r)
C++ isn't that smart. You have to do each comparison manually.
bool Circle2::contains(Line2 l) {
if((p1.distanceFrom(l.p1) <= r) || (p1.distanceFrom(l.p2) <= r)) return 1;
return 0;
}