Question: What is y after the following switch statement is executed? Rewrite the code using an if statement.
y = 3; x = 3;
switch (x + 3)
{
case 6: y = 1;
default: y += 1;
}
I'm at the beginning of the C++ hike. I don't know what do with this. It's not working in C++ Visual Studio 2013. I've put it in as is and nothing happens.
I use:
y = 3; x = 3;
switch (x + 3)
{
case 6: y = 1;
default: y += 1;
}
return 0;
}
And nothing happens. I have both the answers but I have no clue how to get them...
y is 2
if (x + 3 == 6)
y = 1;
y += 1;
I'm strictly supposed to be using
#include <iostream>
using namespace std;
int main()
{
return 0;
}
switch basics in C/C++:
switch expression is evaluated once, at the beginning. Then the 1st matching case is executed
after the case clause, the execution falls through to the next clause unless there's a break. Since this typically isn't what was indended, fall-throughs are normally documented with a comment or (if multiple cases refer to the same code) by stacking them together (case 2: case 3: <code>).
Related
Is there a base function in Rcpp that:
Fills entirely by a single value if size of a vector is 1.
Fills the other vector completely if same length.
Fills with an NA value if neither Vector are the same length nor a vector is of size 1.
I've written the above criteria as a function below using a NumericVector as an example. If there isn't a base function in Rcpp that performs said operations there should be a way to template the function so that given any type of vector (e.g. numeric, character and so on) the above logic would be able to be executed.
// [[Rcpp::export]]
NumericVector cppvectorize(NumericVector x,NumericVector y) {
NumericVector y_out(y.size());
if(x.size() == 1) {
for(int i = 0; i < y_out.size(); i++) {
y_out[i] = x[0];
}
} else if(x.size() == y_out.size()) {
for(int i = 0; i < y_out.size(); i++) {
y_out[i] = x[i];
}
} else {
for(int i = 0; i < y_out.size(); i++) {
y_out[i] = NA_REAL;
}
}
return y_out;
}
Unfortunately, the closest you will come to such a function is one of the rep variants that Rcpp supports. However, none of the variants match the desired output. Therefore, the only option is to really implement a templated version of your desired function.
To create the templated function, we will first create a routing function that handles the dispatch of SEXP objects. The rationale behind the routing function is SEXP objects are able to be retrieved from and surfaced into R using Rcpp Attributes whereas a templated version is not. As a result, we need to specify the SEXTYPE (used as RTYPE) dispatches that are possible. The TYPEOF() macro retrieves the coded number. Using a switch statement, we can dispatch this number into the appropriate cases.
After dispatching, we arrive at the templated function. The templated function makes use of the base Vector class of Rcpp to simplify the data flow. From here, the notable novelty will be the use of ::traits::get_na<RTYPE>() to dynamically retrieve the appropriate NA value and fill it.
With the plan in place, let's look at the code:
#include <Rcpp.h>
using namespace Rcpp;
// ---- Templated Function
template <int RTYPE>
Vector<RTYPE> vec_helper(const Vector<RTYPE>& x, const Vector<RTYPE>& y) {
Vector<RTYPE> y_out(y.size());
if(x.size() == 1){
y_out.fill(x[0]);
} else if (x.size() == y.size()) {
y_out = x;
} else {
y_out.fill(::traits::get_na<RTYPE>());
}
return y_out;
}
// ---- Dispatch function
// [[Rcpp::export]]
SEXP cppvectorize(SEXP x, SEXP y) {
switch (TYPEOF(x)) {
case INTSXP: return vec_helper<INTSXP>(x, y);
case REALSXP: return vec_helper<REALSXP>(x, y);
case STRSXP: return vec_helper<STRSXP>(x, y);
default: Rcpp::stop("SEXP Type Not Supported.");
}
// Need to return a value even though this will never be triggered
// to quiet the compiler.
return R_NilValue;
}
Sample Tests
Here we conduct a few sample tests on each of the supported data
# Case 1: x == 1
x = 1:5
y = 2
cppvectorize(x, y)
## [1] NA
# Case 2: x == y
x = letters[1:5]
y = letters[6:10]
cppvectorize(x, y)
## [1] "a" "b" "c" "d" "e"
# Case 3: x != y && x > 1
x = 1.5
y = 2.5:6.5
cppvectorize(x, y)
## [1] 1.5 1.5 1.5 1.5 1.5
ı want to calculate determinant of matrix with thread but i have a error "term does not eveluate to a function taking 0 arguments" ı want to solve big matrix with thread and parsing matrix,what can ı do
int determinant(int f[1000][1000], int x)
{
int pr, c[1000], d = 0, b[1000][1000], j, p, q, t;
if (x == 2)
{
d = 0;
d = (f[1][1] * f[2][2]) - (f[1][2] * f[2][1]);
return(d);
}
else
{
for (j = 1; j <= x; j++)
{
int r = 1, s = 1;
for (p = 1; p <= x; p++)
{
for (q = 1; q <= x; q++)
{
if (p != 1 && q != j)
{
b[r][s] = f[p][q];
s++;
if (s > x - 1)
{
r++;
s = 1;
}
}
}
}
for (t = 1, pr = 1; t <= (1 + j); t++)
pr = (-1)*pr;
c[j] = pr*determinant(b, x - 1);
}
for (j = 1, d = 0; j <= x; j++)
{
d = d + (f[1][j] * c[j]);
}
return(d);
}
}
int main()
{
srand(time_t(NULL));
int i, j;
printf("\n\nEnter order of matrix : ");
scanf_s("%d", &m);
printf("\nEnter the elements of matrix\n");
for (i = 1; i <= m; i++)
{
for (j = 1; j <= m; j++)
{
a[i][j] = rand() % 10;
}
}
thread t(determinant(a, m));
t.join();
printf("\n Determinant of Matrix A is %d .", determinant(a, m));
}
The immediate problem is that here: thread t(determinant(a, m)); you pass the result of calling determinant(a, m) as the function to execute, and zero arguments to call that function with - but an int is not a function or other callable object, which is what the error you got complains about.
std::thread's constructor takes the function to run and the arguments to supply separately, so you would need to call std::thread(determinant, a, m).
Now we have another problem, std::thread doesn't provide a way to retrieve the return value, and so you calculate it again here: printf("\n Determinant of Matrix A is %d .", determinant(a, m));.
To fix this, we can use std::async from the <future> header, which will manage the thread handling for us, and lets us retrieve the result later:
auto result = std::async(std::launch::async, determinant, a, m);
int det = result.get()
This will run determinant(a,m) on a new thread, and return a std::future<int> into which the return value may eventually be placed.
We can then try to retrieve that value with std::future::get(), which will block until the value can be retrieved (or until an exception occurs in the thread).
In this example, we still execute determinant in a pretty serial fashion, since we delegate the work to a thread, then wait for that thread to finish its work before continuing.
However we are now free to store the future, and defer calling std::future::get() until we actually need the value, potentially much later in your program.
There are a few other problems in the rest of your code:
all your array indexing is off by one (array indices run from 0 to N-1 in C and C++)
a few of the variables you're using don't exist (like a and m)
C-arrays are passed by pointer, so if you ever change the code not to block on the thread right there, the array will go out of scope and your thread may read garbage from the dangling pointer. If you use a proper container like std::array or std::vector, you can pass it by value so your thread will own the data to operate on for its entire lifetime.
Here's a problem:
Given string A and a substring B, remove the first occurence of substring B in string A till it is possible to do so. Note that removing a substring, can further create a new same substring. Ex. removing 'hell' from 'hehelllloworld' once would yield 'helloworld' which after removing once more would become 'oworld', the desired string.
Write a program for the above for input constraints of length 10^6 for A, and length 100 for B.
This question was asked to me in an interview, I gave them a simple algorithm to solve it that was to do exactly what the statement was and remove it iteratievly(to decresae over head calls), I later came to know there's a better solution for it that's much faster what would it be ? I've thought of a few optimizations but it's still not as fast as the fastest soln for the problem(acc. the company), so can anyone tell me of a faster way to solve the problem ?
P.S> I know of stackoverflow rules and that having code is better, but for this problem, I don't think that having code would be in any way beneficial...
Your approach has a pretty bad complexity. In a very bad case the string a will be aaaaaaaaabbbbbbbbb, and the string b will be ab, in which case you will need O(|a|) searches, each taking O(|a| + |b|) (assuming using some sophisticated search algorithm), resulting in a total complexity of O(|a|^2 + |a| * |b|), which with their constraints is years.
For their constraints a good complexity to aim for would be O(|a| * |b|), which is around 100 million operations, will finish in subsecond. Here's one way to approach it. For each position i in the string a let's compute the largest length n_i, such that the a[i - n_i : i] = b[0 : n_i] (in other words, the longest suffix of a at that position which is a prefix of b). We can compute it in O(|a| + |b|) by using Knuth-Morris-Pratt algorithm.
After we have n_i computed, finding the first occurrence of b in a is just a matter of finding the first n_i that is equal to |b|. This will be the right end of one of the occurrences of b in a.
Finally, we will need to modify Knuth-Morris-Pratt slightly. We will be logically removing occurrences of b as soon as we compute an n_i that is equal to |b|. To account for the fact that some letters were removed from a we will rely on the fact that Knuth-Morris-Pratt only relies on the last value of n_i (and those computed for b), and the current letter of a, so we just need a fast way of retrieving the last value of n_i after we logically remove an occurrence of b. That can be done with a deque, that stores all the valid values of n_i. Each value will be pushed into the deque once, and popped from it once, so that complexity of maintaining it is O(|a|), while the complexity of the Knuth-Morris-Pratt is O(|a| + |b|), resulting in O(|a| + |b|) total complexity.
Here's a C++ implementation. It could have some off-by-one errors, but it works on your sample, and it flies for the worst case that I described at the beginning.
#include <deque>
#include <string>
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main() {
string a, b;
cin >> a >> b;
size_t blen = b.size();
// make a = b$a
a = b + "$" + a;
vector<size_t> n(a.size()); // array for knuth-morris-pratt
vector<bool> removals(a.size()); // positions of right ends at which we remove `b`s
deque<size_t> lastN;
n[0] = 0;
// For the first blen + 1 iterations just do vanilla knuth-morris-pratt
for (size_t i = 1; i < blen + 1; ++ i) {
size_t z = n[i - 1];
while (z && a[i] != a[z]) {
z = n[z - 1];
}
if (a[i] != a[z]) n[i] = 0;
else n[i] = z + 1;
lastN.push_back(n[i]);
}
// For the remaining iterations some characters could have been logically
// removed from `a`, so use lastN to get last value of n instaed
// of actually getting it from `n[i - 1]`
for (size_t i = blen + 1; i < a.size(); ++ i) {
size_t z = lastN.back();
while (z && a[i] != a[z]) {
z = n[z - 1];
}
if (a[i] != a[z]) n[i] = 0;
else n[i] = z + 1;
if (n[i] == blen) // found a match
{
removals[i] = true;
// kill last |b| - 1 `n_i`s
for (size_t j = 0; j < blen - 1; ++ j) {
lastN.pop_back();
}
}
else {
lastN.push_back(n[i]);
}
}
string ret;
size_t toRemove = 0;
for (size_t pos = a.size() - 1; a[pos] != '$'; -- pos) {
if (removals[pos]) toRemove += blen;
if (toRemove) -- toRemove;
else ret.push_back(a[pos]);
}
reverse(ret.begin(), ret.end());
cout << ret << endl;
return 0;
}
[in] hehelllloworld
[in] hell
[out] oworld
[in] abababc
[in] ababc
[out] ab
[in] caaaaa ... aaaaaabbbbbb ... bbbbc
[in] ab
[out] cc
When context action such as convert if to switch is invoked, all my comments are lost. Can this be prevented?
e.g.:
int a = 1;
int i;
if (a == 1)
//comment I want to keep 1
i = 1;
else if (a == 2)
//comment I want to keep 2
i = 2;
else
i = 3;
This is a known issue - here it is on YouTrack (and here's a duplicate).
I'm writing a Conway's life game for school. In the program I am having trouble with the arrays taking the values I am assigning them. At one point in the program they print out the value assigned to them (1) yet at the end of the program when I need to print the array to show the iterations of the game it shows an incredibly low number. The other trouble was I was encountering difficulties when putting in a loop that would ask if it wants you to run another iteration. So I removed it until the previous errors were fixed.
Im writing this with C++
#include <stdio.h>
int main (void)
{
int currentarray [12][12];
int futurearray [12][12];
char c;
char check = 'y';
int neighbors = 0;
int x = 0; // row
int y = 0; //column
printf("Birth an organism will be born in each empty location that has exactly three neighbors.\n");
printf("Death an organism with four or more organisms as neighbors will die from overcrowding.\n");
printf("An organism with fewer than two neighbors will die from loneliness.\n");
printf("Survival an organism with two or three neighbors will survive to the next generation.\n");
printf( "To create life input x, y coordinates.\n");
while ( check == 'y' )
{
printf("Enter x coordinate.\n");
scanf("%d", &x ); while((c = getchar()) != '\n' && c != EOF);
printf("Enter y coordinate.\n");
scanf("%d", &y ); while((c = getchar()) != '\n' && c != EOF);
currentarray [x][y] = 1;
printf ("%d\n", currentarray[x][y]);
printf( "Do you wish to enter more input? y/n.\n");
scanf("%c", &check); while((c = getchar()) != '\n' && c != EOF);
}
// Note - Need to add a printf statement showing the array before changes are made after input added.
// check for neighbors
while(check == 'y')
{
for(y = 0; y <= 12; y++)
{
for(x = 0; x <= 12; x++)
{
//Begin counting number of neighbors:
if(currentarray[x-1][y-1] == 1) neighbors += 1;
if(currentarray[x-1][y] == 1) neighbors += 1;
if(currentarray[x-1][y+1] == 1) neighbors += 1;
if(currentarray[x][y-1] == 1) neighbors += 1;
if(currentarray[x][y+1] == 1) neighbors += 1;
if(currentarray[x+1][y-1] == 1) neighbors += 1;
if(currentarray[x+1][y] == 1) neighbors += 1;
if(currentarray[x+1][y+1] == 1) neighbors += 1;
//Apply rules to the cell:
if(currentarray[x][y] == 1 && neighbors < 2)
futurearray[x][y] = 0;
else if(currentarray[x][y] == 1 && neighbors > 3)
futurearray[x][y] = 0;
else if(currentarray[x][y] == 1 && (neighbors == 2 || neighbors == 3))
futurearray[x][y] = 1;
else if(currentarray[x][y] == 0 && neighbors == 3)
futurearray[x][y] = 1;
}
}
}
// Set the current array to the future and change the future to 0
{
for(y = 0; y < 12; y++)
{
for(x = 0; x < 12; x++)
{
//Begin the process
currentarray [x][y] = futurearray [x][y];
futurearray [x][y] = 0;
}
}
}
{
for(y = 0; y < 12; y++)
{
for(x = 0; x < 12; x++)
{
//print the current life board
printf("%d ", currentarray[x][y]);
}
}
}
// Have gone through one iteration of Life
//Ask to do another iteration
printf("Do you wish to continue y/n?\n");
scanf("%c", &check); while((c = getchar()) != '\n' && c != EOF);
return 0;
}
You are defining your arrays as [12][12].
In your generation loop you walk from i = 0 to i <= 12, which is 13 steps instead of the 12 of the array. Additionally you are trying to access x-1 and y-1, which can be as low as -1. Again not inside your array.
Sometimes you get semi-useful values from within your array, but on some borders you are just accessing random data.
Try to correct your border.
You forgot to set neighbors to 0 before counting them.
Since this is C++ (not C), you might as well declare neighbors inside the loop body. Makes these kinds of issues easier to spot, too.
Also, is it me, or is that while loop never going to finish? Your braces are a mess, in general, as is your indentation. You could do yourself and us a favour by cleaning those up.
Obviously agree with all the above suggestions. One nice trick you might want to implement with Life is to create an extra border around your area. So if the user wants a 12x12 grid (and you should allow width/height to be specified and allocate memory dynamically) internally you hold a 14x14 grid corresponding to a border around the actual grid. Before running the calculation copy the top row to the bottom border, bottom row to the top border etc. Now you can run the main algorithm on the inner 12x12 grid without worrying about edge cases. This will enable your patterns to re-appear on the other side if they fall off the edge.
You're also forgetting to set the values of both arrays to zero. This will take care of the ridiculous number issue you're having. you can do that by copying this for loop:
for(y = 0; y < 12; y++)
{
for(x = 0; x < 12; x++)
{
//Begin the process
currentarray [x][y] = futurearray [x][y];
futurearray [x][y] = 0;
}
}
and pasting it before the while loop but instead of setting currentarray[x][y] = futurearray[x][y], set it to 0. Also, if the coordinates are viewable locations instead of array co-ordinates, you'll want to change this:
printf ("%d\n", currentarray[x][y]);
to this:
printf ("%d\n", currentarray[x-1][y-1]);
I would also recommend putting a printf with a newline (\n) after each row has been printed and a tab (\t) after each item so that the formatting looks cleaner.