This is a piece of code I found in my textbook for using recursion to evaluate prefix expressions. I'm having trouble understanding this code and the process in which it goes through.
char *a; int i;
int eval()
{ int x = 0;
while (a[i] == ' ') i++;
if (a[i] == '+')
{ i++; return eval() + eval(); }
if (a[i] == '*')
{ i++; return eval() * eval(); }
while ((a[i] >= '0') && (a[i] <= '9'))
x = 10*x + (a[i++] - '0');
return x;
}
I guess I'm confused primarily with the return statements and how it eventually leads to solving a prefix expression. Thanks in advance!
The best way to understand recursive examples is to work through an example :
char* a = "+11 4"
first off, i is initialized to 0 because there is no default initializer. i is also global, so updates to it will affect all calls of eval().
i = 0, a[i] = '+'
there are no leading spaces, so the first while loop condition fails. The first if statement succeeds, i is incremented to 1 and eval() + eval() is executed. We'll evaluate these one at a time, and then come back after we have our results.
i = 1, a[1] = '1'
Again, no leading spaces, so the first while loop fails. The first and second if statements fail. In the last while loop, '1' is between 0 and 9(based on ascii value), so x becomes 0 + a[1] - '0', or 0 + 1 = 1. Important here is that i is incremented after a[i] is read, then i is incremented. The next iteration of the while loop adds to x. Here x = 10 * 1 + a[2] - '0', or 10 + 1 = 11. With the correct value of x, we can exit eval() and return the result of the first operand, again here 11.
i = 2, a[2] = '4'
As in the previous step, the only statement executed in this call of eval() is the last while loop. x = 0 + a[2] - '0', or 0 + 4 = 4. So we return 4.
At this point the control flow returns back to the original call to eval(), and now we have both values for the operands. We simply perform the addition to get 11 + 4 = 15, then return the result.
Every time eval() is called, it computes the value of the immediate next expression starting at position i, and returns that value.
Within eval:
The first while loop is just to ignore all the spaces.
Then there are 3 cases:
(a) Evaluate expressions starting with a + (i.e. An expression of the form A+B which is "+ A B" in prefix
(b) Evaluate expressions starting with a * (i.e. A*B = "* A B")
(c) Evaluate integer values (i.e. Any consecutive sequence of digits)
The while loop at the end takes care of case (c).
The code for case (a) is similar to that for case (b). Think about case (a):
If we encounter a + sign, it means we need to add the next two "things" we find in the sequence. The "things" might be numbers, or may themselves be expressions to be evaluated (such as X+Y or X*Y).
In order to get what these "things" are, the function eval() is called with an updated value of i. Each call to eval() will fetch the value of the immediate next expression, and update position i.
Thus, 2 successive calls to eval() obtain the values of the 2 following expressions.
We then apply the + operator to the 2 values, and return the result.
It will help to work through an example such as "+ * 2 3 * 4 5", which is prefix notation for (2*3)+(4*5).
So this piece of code can only eat +, *, spaces and numbers. It is supposed to eat one command which can be one of:
- + <op1> <op2>
- * <op1> <op2>
<number>
It gets a pointer to a string, and a reading position which is incremented as the program goes along that string.
char *a; int i;
int eval()
{ int x = 0;
while (a[i] == ' ') i++; // it eats all spaces
if (a[i] == '+')
/* if the program encounters '+', two operands are expected next.
The reading position i already points just before the place
from which you have to start reading the next operand
(which is what first eval() call will do).
After the first eval() is finished,
the reading position is moved to the begin of the second operand,
which will be read during the second eval() call. */
{ i++; return eval() + eval(); }
if (a[i] == '*') // exactly the same, but for '*' operation.
{ i++; return eval() * eval(); }
while ((a[i] >= '0') && (a[i] <= '9')) // here it eats all digit until something else is encountered.
x = 10*x + (a[i++] - '0'); // every time the new digit is read, it multiplies the previously obtained number by 10 and adds the new digit.
return x;
// base case: returning the number. Note that the reading position already moved past it.
}
The example you are given uses a couple of global variables. They persist outside of the function's scope and must be initialized before calling the function.
i should be initialized to 0 so that you start at the beginning of the string, and the prefix expression is the string in a.
the operator is your prefix and so should be your first non-blank character, if you start with a number (string of numbers) you are done, that is the result.
example: a = " + 15 450"
eval() finds '+' at i = 1
calls eval()
which finds '1' at i = 3 and then '5'
calculates x = 1 x 10 + 5
returns 15
calls eval()
which finds '4' at i = 6 and then '5' and then '0'
calclulates x = ((4 x 10) + 5) x 10) + 0
returns 450
calculates the '+' operator of 15 and 450
returns 465
The returns are either a value found or the result of an operator and the succeeding results found. So recursively, the function successively looks through the input string and performs the operations until either the string ends or an invalid character is found.
Rather than breaking up code into chunks and so on, i'll try and just explain the concept it as simple as possible.
The eval function always skips spaces so that it points to either a number character ('0'->'9'), an addition ('+') or a multiply ('*') at the current place in the expression string.
If it encounters a number, it proceeds to continue to eat the number digits, until it reaches a non-number digit returning the total result in integer format.
If it encounters operator ('+' and '*') it requires two integers, so eval calls itself twice to get the next two numbers from the expression string and returns that result as an integer.
One hair in the soup may be evaluation order, cf. https://www.securecoding.cert.org/confluence/display/seccode/EXP10-C.+Do+not+depend+on+the+order+of+evaluation+of+subexpressions+or+the+order+in+which+side+effects+take+place.
It is not specified which eval in "eval() + eval()" is, well, evaluated first. That's ok for commutative operators but will fail for - or /, because eval() as a side effect advances the global position counter so that the (in time) second eval gets the (in space) second expression. But that may well be the (in space) first eval.
I think the fix is easy; assign to a temp and compute with that:
if (a[i] == '-')
{ i++; int tmp = eval(); return tmp - eval(); }
Related
I'm having a hard time understanding why using a increment operator in a for loop in C++ has a different result than doing 'variable' + 1. The variable in the second case simply isn't remembered after each iteration of the loop:
Take the following code:
#include <iostream>
int main(){
int a{0};
for (int i = 0; i < 5; i++){
std::cout << ++a;
}
return 0;
}
It outputs as expected:
12345
However, if I instead replace ++a with a + 1:
#include <iostream>
int main(){
int a{0};
for (int i = 0; i < 5; i++){
std::cout << a + 1;
}
return 0;
}
I get:
11111
Even if I make the 'a' variable static:
static int a{0};
It still outputs 11111.
Why doesn't 'a + 1' preserve its value after every loop? What would I need to do if I wanted to preserve its value after every iteration without using ++ (for instance, with another operation like a * 2)?
Why doesn't 'a + 1' preserve its value after every loop?
a + 1 is an expression that doesn't assign anything to a. That is a is not affected in any way. This means when you do just a + 1, the value of a is still 0(the old value).
On the other hand ++a has the same effect as:
v---------->assignment done here implicitly
a = a+1
In the above expression ++a, the value of is incremented by 1 so that the new value of a is now 1.
++a is equivalent to the assignment a= a + 1 (or a+= 1). The expression a + 1 alone does not modify a.
C++ will allow you to write std::cout << (a= a + 1);.
a++ is not the same thing as a+1 but as a=a+1.
What's the difference?
Well:
a+1 contains the value, which is one higher than the value of a.
a=a+1 means that, in top of this, this value gets assigned to the variable a, which means something like "replace a by its value plus one".
This gives following possibilities (I also show the difference between a++ and ++a):
int a=3;
cout<<a+1; // output : 4. Value of 'a' equals 3.
cout<<a+1; // output : 4. Value of 'a' equals 3.
cout<<a+1; // output : 4. Value of 'a' equals 3.
int a=3;
cout<<a++; // output : 3. Value of 'a' becomes 4 after having shown it on screen.
cout<<a++; // output : 4. Value of 'a' becomes 5 after having shown it on screen.
cout<<a++; // output : 5. Value of 'a' becomes 6 after having shown it on screen.
int a=3;
cout<<++a; // output : 4. Value of 'a' becomes 4 before showning it on screen.
cout<<++a; // output : 5. Value of 'a' becomes 5 before showning it on screen.
cout<<++a; // output : 6. Value of 'a' becomes 6 before showning it on screen.
Conceptually, the built-in arithmetic operators like + and * evaluate their operands, perform the respective operation with the thusly obtained values, and create a temporary object that contains the result.
The operands, one of which is your a, are only read from, not written to. That is more obvious when you consider that + is commutative, that is, its operands can be swapped without affecting the result: a+1 is by definition the same as 1+a, and clearly you cannot write anything back to the immediate value 1.
Generally, as you certainly know, = is used in C and C++ to write a value to a variable. C, and by means of ancestry also C++, provide shortcuts to write back the result of certain operations to one of their operands: it's the combination of the operator and the assignment =, for example a += 1. Like all assignments, this one is an expression (that is, it has a value) which has the value of the variable after the operation and assignment. Like all other expressions, you can use it as a subexpression, even though that's not very common: cout << (a += 1); would achieve the desired effect (the parentheses are necessary because assignment has about the lowest operator precedence). Because incrementing (and decrementing) by one is so very common, C and C++ have a shortcut for the shortcut: ++a is the same as a+=1, so that one would typically write cout << ++a; (as a side effect obviating the need for parentheses).
I can't understand how to count number of 1's in binary representation.
I have my code, and I hope someone can explain it for me.
Code:
int count (int x)
{
int nr=0;
while(x != 0)
{
nr+=x%2;
x/=2;
}
return nr;
}
Why while ? For example if i have 1011, it wouldn't stop at 0?
Why nr += x%2 ?
Why x/=2 ?!
First:
nr += x % 2;
Imagine x in binary:
...1001101
The Modulo operator returns the remainder from a / b.
Now the last bit of x is either a 0, in which case 2 will always go into x with 0 remainder, or a 1, in which case it returns a 1.
As you can see x % 2 will return (if the last bit is a one) a one, thus incrementing nr by one, or not, in which case nr is unchanged.
x /= 2;
This divides x by two, and because it is a integer, drops the remainder. What this means is is the binary was
....10
It will find out how many times 2 would go into it, in this case 1. It effectively drops the last digit of the binary number because in base 2 (binary) the number of times 2 goes into a number is just the same as 'shifting' everything down a space (This is a poor explanation, please ask if you need elaboration). This effectively 'iterates' through the binary number, allowing the line about to check the next bit.
This will iterate until the binary is just 1 and then half that, drop the remainder and x will equal 0,
while (x != 0)
in which case exit the loop, you have checked every bit.
Also:
'count`is possibly not the most descriptive name for a function, consider naming it something more descriptive of its purpose.
nr will always be a integer greater or equal to zero, so you should probably have the return type unsigned int
int count (int x)
{
int nr=0;
while(x != 0)
{
nr+=x%2;
x/=2;
}
return nr;
}
This program basically gives the numbers of set bits in a given integer.
For instance, lets start with the example integer 11 ( binary representation - 1011).
First flow will enter the while loop and check for the number, if it is equal to zero.
while(11 != 0)
Since 11 is not equal to zero it enter the while loop and nr is assigned the value 1 (11%2 = 1).nr += 11%2;
Then it executes the second line inside the loop (x = x/2). This line of code assigns the value 5 (11/2 = 5 ) to x.
Once done with the body of the while loop, it then again checks if x ie 5 is equal to zero.
while( 5 != 0).
Since it is not the case,the flow goes inside the while loop for the second time and nr is assigned the value 2 ( 1+ 5%2).
After that the value of x is divided by 2 (x/2, 5/2 = 2 )and it assigns 2 to x.
Similarly in the next loop, while (2 != 0 ), nr adds (2 + 2%2), since 2%2 is 0, value of nr remains 2 and value of x is decreased to 1 (2/2) in the next line.
1 is not eqaul to 0 so it enters the while loop for the third time.
In the third execution of the while loop nr value is increased to 3 (2 + 1%2).
After that value of x is reduced to 0 ( x = 1/2 which is 0).
Since it fails the check (while x != 0), the flow comes out of the loop.
At the end the value of nr (Which is the number of bits set in a given integer) is returned to the calling function.
Best way to understand the flow of a program is executing the program through a debugger. I strongly suggest you to execute the program once through a debugger.It will help you to understand the flow completely.
I am asking the user to input an expression which will be evaluated in postfix notation. The beginning of the expression is the variable name where the answer of the evaluated expression will be stored. Ex: A 4 5 * 6 + 2 * 1 – 6 / 4 2 + 3 * * = where A is the variable name and the equal sign means the answer to the expression will be stored in the variable A. The OUT A statement means that the number stored in the variable A will be printed out.
What I need help with is that when I input the second expression, I do not get the right answer. For example, my first expression A 4 5 * 6 + 2 * 1 – 6 / 4 2 + 3 * * = will evaluate to 153 and then when I input my second expression B A 10 * 35.50 + =, it has to evaluate to 1565.5, but it doesn't. It evaluates to 35.5. I cannot figure out why I am getting the wrong answer. Also, I need help with the OUT statement.
else if (isalpha(expr1[i]))
{
stackIt.push(mapVars1[expr1[i]]);
}
Will place the variable, or zero if the variable has not been set, onto the stack.
else if (isalpha(expr1[i]))
{
map<char, double>::iterator found = mapVars1.find(expr1[i]);
if (found != mapVars1.end())
{
stackIt.push(found->second);
}
else
{
// error message and exit loop
}
}
Is probably better.
Other suggestions:
Compilers are pretty sharp these days, but you may get a bit out of char cur = expr1[i]; and then using cur (or suitably descriptive variable name) in place of the remaining expr1[i]s in the loop.
Consider using isdigit instead of expr1[i] >= '0' && expr1[i] <= '9'
Test your code for expressions with multiple spaces in a row or a space after an operator. It looks like you will re-add the last number you parsed.
Test for input like 123a456. You might not like the result.
If spaces after each token in the expression are specified in the expression protocol, placing your input string into a stringstream will allow you to remove a great deal of your parsing code.
stringstream in(expr1);
string token;
while (in >> token)
{
if (token == "+" || token == "-'" || ...)
{
// operator code
}
else if (token == "=")
{
// equals code
}
else if (mapVars1.find(token) != mapVars1.end())
{
// push variable
}
else if (token.length() > 0)
{
char * endp;
double val = strtod(token.c_str(), &endp);
if (*endp == '\0')
{
// push val
}
}
}
To use previous symbol names in subsequent expressions add this to the if statements in your parsing loop:
else if (expr1[i] >= 'A' && expr1[i] <= 'Z')
{
stackIt.push(mapVars1[expr[i]]);
}
Also you need to pass mapVars by reference to accumulate its contents across Eval calls:
void Eval(string expr1, map<char, double> & mapVars1)
For the output (or any) other command I would recommend parsing the command token that's at the front of the string first. Then call different evaluators based on the command string. You are trying to check for OUT right now after you have already tried to evaluate the string as an arithmetic assignment command. You need to make that choice first.
while ( (i=t-i%10 ? i/10 : !printf("%d\n",j)) || (i=++j<0?-j:j)<101 );
I came across this on codegolf
Please explain the usage of ? and : and why is there no statement following the while loop? As in why is there a ; after the parenthesis.
There is a boolean operation going on inside the parentheses of the while loop:
while (boolean);
Since the ternary operator is a boolean operator, it's perfectly legal.
So what's this doing? Looks like modular arithmetic, printing going on over a range up to 101.
I'll agree that it's cryptic and obscure. It looks more like a code obfuscation runner up. But it appears to be compilable and runnable. Did you try it? What did it do?
The ?: is a ternary operator.
An expression of form <A> ? <B> : <C> evaluates to:
If <A> is true, then it evaluates to <B>
If <A> is false, then it evaluates to <C>
The ; after the while loop indicates an empty instruction. It is equivalent to writing
while (<condition>) {}
The code you posted seems like being obfuscated.
Please explain the usage of ? and :
That's the conditional operator. a ? b : c evaluates a and converts it to a boolean value. Then it evaluates b if its true, or c if its false, and the overall value of the expression is the result of evaluating b or c.
So the first sub-expression:
assigns t-i%10 to i. The result of that expression is the new value of i.
if i is not zero, the result of the expression is i/10
otherwise, print j, and the result of the expression is zero (since printf returns a non-zero count of characters printed, which ! converts to zero).
Then the second sub-expression, after ||, is only evaluated if the result of the first expression was zero. I'll leave you to figure out what that does.
why is there no statement following the while loop?
There's an empty statement, ;, so the loop body does nothing. All the action happens in the side effects of the conditional expression. This is a common technique when the purpose of the code is to baffle the reader; but please don't do this sort of thing when writing code that anyone you care about might need to maintain.
This is the Conditional Operator (also called ternary operator).
It is a one-line syntax to do the same as if (?) condition doA else (:) doB;
In your example:
(i=t-i%10 ? i/10 : !printf("%d\n",j)
Is equivalent to
if (i=t-i%10)
i/10;
else
!printf("%d\n",j);
?: is the short hand notation for if then else
(i=t-i%10 ? i/10 : !printf("%d\n",j)<br>
equals to
if( i= t-i%10 )
then { i/10 }
else { !printf("%d\n",j) }
Your while loop will run when the statement before the || is true OR the statement after the || is true.
notice that your code does not make any sense.
while ( (i=t-i%10 ? i/10 : !printf("%d\n",j)) || (i=++j<0?-j:j)<101 );
in the most human-readable i can do it for u, it's equivalent to:
while (i < 101)
{
i = (t - i) % 10;
if (i > 0)
{
i = i / 10;
}
else
{
printf("%d\n",j);
}
i = ++j;
if (i < 0)
{
i = i - j;
}
else
{
i = j;
}
}
Greetings.
I am the proud perpetrator of that code. Here goes the full version:
main()
{
int t=getchar()-48,i=100,j=-i;
while ((i=t-i%10?i/10:!printf("%d\n",j)) || (i=++j<0?-j:j)<101 );
}
It is my submission to a programming challenge or "code golf" where you are asked to create the tinniest program that would accept a digit as a parameter and print all the numbers in the range -100 to 100 that include the given digit. Using strings or regular expressions is forbidden.
Here's the link to the challenge.
The point is that it is doing all the work into a single statement that evaluates to a boolean. In fact, this is the result of merging two different while loops into a single one. It is equivalent to the following code:
main()
{
int i,t=getchar()-'0',j=-100;
do
{
i = j<0? -j : j;
do
{
if (t == i%10)
{
printf("%d\n",j);
break;
}
}
while(i/=10);
}
while (j++<100);
}
Now lets dissect that loop a little.
First, the initialisation.
int t=getchar()-48,i=100,j=-i;
A character will be read from the standard input. You are supposed to type a number between 0 and 9. 48 is the value for the zero character ('0'), so t will end up holding an integer between 0 and 9.
i and j will be 100 and -100. j will be run from -100 to 100 (inclusive) and i will always hold the absolute value of j.
Now the loop:
while ((i=t-i%10?i/10:!printf("%d\n",j)) || (i=++j<0?-j:j)<101 );
Let's read it as
while ( A || B ) /* do nothing */ ;
with A equals to (i=t-i%10?i/10:!printf("%d\n",j)) and B equals to (i=++j<0?-j:j)<101
The point is that A is evaluated as a boolean. If true, B won't be evaluated at all and the loop will execute again. If false, B will be evaluated and in turn, if B is true we'll repeat again and once B is false, the loop will be exited.
So A is the inner loop and B the outer loop. Let's dissect them
(i=t-i%10?i/10:!printf("%d\n",j))
It's a ternary operator in the form i = CONDITION? X : Y; It means that first CONDITION will be evaluated. If true, i will be set to the value of X; otherwise i will be set to Y.
Here CONDITION (t-i%10) can be read as t - (i%10). This will evaluate to true if i modulo 10 is different than t, and false if i%10 and t are the same value.
If different, it's equivalent to i = i / 10;
If same, the operation will be i = !printf("%d\n",j)
If you think about it hard enough, you'll see that it's just a loop that checks if any of the decimal digits in the integer in i is equal to t.
The loop will keep going until exhausting all digits of i (i/10 will be zero) or the printf statement is run. Printf returns the number of digits printed, which should always be more than zero, so !printf(...) shall always evaluate to false, also terminating the loop.
Now for the B part (outer loop), it will just increment j until it reaches 101, and set i to the absolute value of j in the way.
Hope I made any sense.
Yes, I found this thread by searching for my code in google because I couldn't find the challenge post.
gooday programers. I have to design a C++ program that reads a sequence of positive integer values that ends with zero and find the length of the longest increasing subsequence in the given sequence. For example, for the following
sequence of integer numbers:
1 2 3 4 5 2 3 4 1 2 5 6 8 9 1 2 3 0
the program should return 6
i have written my code which seems correct but for some reason is always returning zero, could someone please help me with this problem.
Here is my code:
#include <iostream>
using namespace std;
int main()
{
int x = 1; // note x is initialised as one so it can enter the while loop
int y = 0;
int n = 0;
while (x != 0) // users can enter a zero at end of input to say they have entered all their numbers
{
cout << "Enter sequence of numbers(0 to end): ";
cin >> x;
if (x == (y + 1)) // <<<<< i think for some reason this if statement if never happening
{
n = n + 1;
y = x;
}
else
{
n = 0;
}
}
cout << "longest sequence is: " << n << endl;
return 0;
}
In your program, you have made some assumptions, you need to validate them first.
That the subsequence always starts at 1
That the subsequence always increments by 1
If those are correct assumptions, then here are some tweaks
Move the cout outside of the loop
The canonical way in C++ of testing whether an input operation from a stream has worked, is simply test the stream in operation, i.e. if (cin >> x) {...}
Given the above, you can re-write your while loop to read in x and test that x != 0
If both above conditions hold, enter the loop
Now given the above assumptions, your first check is correct, however in the event the check fails, remember that the new subsequence starts at the current input number (value x), so there is no sense is setting n to 0.
Either way, y must always be current value of x.
If you make the above logic changes to your code, it should work.
In the last loop, your n=0 is execute before x != 0 is check, so it'll always return n = 0. This should work.
if(x == 0) {
break;
} else if (x > y ) {
...
} else {
...
}
You also need to reset your y variable when you come to the end of a sequence.
If you just want a list of increasing numbers, then your "if" condition is only testing that x is equal to one more than y. Change the condition to:
if (x > y) {
and you should have more luck.
You always return 0, because the last number that you read and process is 0 and, of course, never make x == (y + 1) comes true, so the last statement that its always executed before exiting the loop its n=0
Hope helps!
this is wrong logically:
if (x == (y + 1)) // <<<<< i think for some reason this if statement if never happening
{
Should be
if(x >= (y+1))
{
I think that there are more than one problem, the first and most important that you might have not understood the problem correctly. By the common definition of longest increasing subsequence, the result to that input would not be 6 but rather 8.
The problem is much more complex than the simple loop you are trying to implement and it is usually tackled with Dynamic Programming techniques.
On your particular code, you are trying to count in the if the length of the sequence for which each element is exactly the successor of the last read element. But if the next element is not in the sequence you reset the length to 0 (else { n = 0; }), which is what is giving your result. You should be keeping a max value that never gets reset back to 0, something like adding in the if block: max = std::max( max, n ); (or in pure C: max = (n > max? n : max );. Then the result will be that max value.