I'm trying to set html content dynamically to document node in the main thread.
QWebElement dynamicContent = ui->webView->page()->mainFrame()->documentElement().findFirst("div#dynamicContent");
if (QWebFrame *frame = dynamicContent.webFrame())
{
param = "<html><body></body></html>";
frame->setHtml(param);
}
These lines are executed normally, but after that i get read access violation with call stack
0 QWebPluginDatabase::searchPaths QtWebKitd4 0x1009deca
1 QWebPluginDatabase::searchPaths QtWebKitd4 0x10749451
2 QWebPluginDatabase::searchPaths QtWebKitd4 0x10749434
3 QWebPluginDatabase::searchPaths QtWebKitd4 0x10749347
4 QWebPluginDatabase::searchPaths QtWebKitd4 0x10748469
5 QWebPluginDatabase::searchPaths QtWebKitd4 0x1074856f
6 QWebPluginDatabase::searchPaths QtWebKitd4 0x107100d6
7 QWebPluginDatabase::searchPaths QtWebKitd4 0x107244ce
8 QWebPluginDatabase::searchPaths QtWebKitd4 0x1074cd58
9 QWebPluginDatabase::searchPaths QtWebKitd4 0x109f9462
10 QWebPluginDatabase::searchPaths QtWebKitd4 0x109fa78e
11 QWebPluginDatabase::searchPaths QtWebKitd4 0x10725e17
12 QWebPluginDatabase::searchPaths QtWebKitd4 0x10725b25
13 QWebPluginDatabase::searchPaths QtWebKitd4 0x1074c180
14 QWebPluginDatabase::searchPaths QtWebKitd4 0x1074cf2e
15 QWebPluginDatabase::searchPaths QtWebKitd4 0x109f9462
16 QWebPluginDatabase::searchPaths QtWebKitd4 0x109fcdb2
17 QWebPluginDatabase::searchPaths QtWebKitd4 0x1074ca2e
18 QWebPluginDatabase::searchPaths QtWebKitd4 0x10721800
19 QWebPluginDatabase::searchPaths QtWebKitd4 0x10721383
20 QWebPluginDatabase::searchPaths QtWebKitd4 0x107253c3
21 QWebPluginDatabase::searchPaths QtWebKitd4 0x10720293
22 QWebPluginDatabase::searchPaths QtWebKitd4 0x10750cba
23 QWebPluginDatabase::searchPaths QtWebKitd4 0x10751ab9
24 QWebPluginDatabase::searchPaths QtWebKitd4 0x108639e4
25 QWebPluginDatabase::searchPaths QtWebKitd4 0x10863916
26 QWebPluginDatabase::searchPaths QtWebKitd4 0x109ecb72
27 QTemporaryFile::tr QtCored4 0x671ec7fa
28 QPictureIO::init QtGuid4 0x65071a2e
29 QPictureIO::init QtGuid4 0x6506f6aa
30 QTemporaryFile::tr QtCored4 0x671ceb81
31 QTemporaryFile::tr QtCored4 0x671d3d29
32 QTemporaryFile::tr QtCored4 0x67214812
33 QPictureIO::init QtGuid4 0x65071a2e
34 QPictureIO::init QtGuid4 0x6506f6aa
35 QTemporaryFile::tr QtCored4 0x671ceb81
36 QTemporaryFile::tr QtCored4 0x671d3d29
37 QTemporaryFile::tr QtCored4 0x671cfadb
38 QTemporaryFile::tr QtCored4 0x672129dd
39 InternalCallWinProc USER32 0x74d66238
40 UserCallWinProcCheckWow USER32 0x74d668ea
41 DispatchMessageWorker USER32 0x74d67d31
42 DispatchMessageW USER32 0x74d67dfa
43 QTemporaryFile::tr QtCored4 0x672139f6
44 QPictureIO::init QtGuid4 0x6512c4ce
45 QTemporaryFile::tr QtCored4 0x671cc68e
46 QTemporaryFile::tr QtCored4 0x671cc7c0
47 QTemporaryFile::tr QtCored4 0x671cf0fd
48 QPictureIO::init QtGuid4 0x6506f398
49 main main.cpp 28 0x401ec9
Any googling or stackoverflowing of problem did not succeed. Had anyone else the same issue? What is the proper usage of QWebFrame::setHtml?
Thank you
[Solved] This issue happens while calling QWebFrame::setHtml not from main thread.
Have solved this issue, and answer is DO NOT use setHtml method from thread, that differs from main.
Related
I have two sheets in an Excel workbook. I need the formula which creates tables by using vlookup.
I have 10 columns and 10 rows like this
1 2 3 4 5 6 7 8 9 10
2
3
4
5
6
7
8
9
10
I have tried to use Vlookup with sum but not get the actual results.
The expected result should be like this
1 2 3 4 5 6 7 8 9 10
2 4 6 8 10 12 14 16 18 20
3 6 9 12 15 18 21 24 27 30
4 8 12 16 20 24 28 32 36 40
5 10 15 20 25 30 35 40 45 50
6 12 18 24 30 36 42 48 54 60
7 14 21 28 35 42 49 56 63 70
8 16 24 32 42 48 56 64 72 80
9 18 27 36 49 54 63 72 81 90
10 20 30 40 50 60 70 80 90 100
I have a QTextStream which contain:
Line1: 3 5 7 17 19 23 25
Line2: 3 5 7 17 19 23 26
Line3: 3 5 8 17 19 23 27
Line4: 3 5 9 17 21 35 37
Line5: 3 5 10 17 21 35 38
Line6: 3 5 11 17 21 35 39
Line7: 3 5 12 17 21 36 37
Line8: 3 5 13 17 21 37 38
Line9: 3 5 15 17 21 36 39
Line10: 3 5 16 17 21 37 38
I need to create regular expression to select only lines which contain numbers 3, 17 and 37.
Line4, 7, 8 and 10.
How to setup pattern?
Assuming they are ordered, you might use \b3\b.*\b17\b.*\b37\b.
So in C++, with raw string R"(\b3\b.*\b17\b.*\b37\b)".
Hi so I have created the nested list/matrix:
q)m:((1 2 3);(4 5 6);(7 8 9))
q)m
1 2 3
4 5 6
7 8 9
I have also identified the middle column in the list:
q)a:m[0;1],m[1;1],m[2;1]
I now want to replace the middle row (4 5 6) with a to finish with m looking like:
q)m
1 2 3
2 5 8
7 8 9
You've already seen you can index into the matrix with syntax like m[0;1], where 0 refers to the first level of nesting and 1 refers to the second level
KDB also allows you to assign to an index of a list in a similar way e.g.
q)l:1 2 3 4
q)l[1]:20
q)l
1 20 3 4
So you can use something similar in this example:
q)m[1]:a
q)m
1 2 3
2 5 8
7 8 9
As an aside, KDB also allows you to leave out an index, in which case it will take all items from the corresponding level of nesting, e.g.
q)m[0] /first level of nesting i.e. first row
1 2 3
q)m[;0] /second level of nesting i.e. first column
1 4 7
Hope that helps
Jonathon McMurray
AquaQ Analytics
You want to generalise for larger matrices (which must also be square) so your answer needs two parts:
how to construct a
how to insert it
for row/col x where x<count m.
The general expression you want is simply m[x;]:m[;x], because m[x;] denotes row x and m[;x] denotes column x.
See Q for Mortals 3.11.3 Two- and Three-Dimensional Matrices
You can make this a function of the index and the matrix:
q)show m:5 5#1_til 26
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25
q){y[x;]:y[;x];:y}[3;m]
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
4 9 14 19 24
21 22 23 24 25
Just adding another approach for you.
q)m:8 cut til 64
q)0 0+\:til 8
0 1 2 3 4 5 6 7
0 1 2 3 4 5 6 7
q)(m)./:flip 0 0+\:til 8
0 9 18 27 36 45 54 63
q)#[m;4;:;(m)./:flip 0 0+\:til 8]
0 1 2 3 4 5 6 7
8 9 10 11 12 13 14 15
16 17 18 19 20 21 22 23
24 25 26 27 28 29 30 31
0 9 18 27 36 45 54 63
40 41 42 43 44 45 46 47
48 49 50 51 52 53 54 55
56 57 58 59 60 61 62 63
q)
For fun, here it is in a function which takes the length&width of the matrix and replaces the 'middle' row with the diagonal values
q){n:x*x;m:x cut til n;#[m;x div 2;:;](m)./:flip 0 0+\:til x}8
0 1 2 3 4 5 6 7
8 9 10 11 12 13 14 15
16 17 18 19 20 21 22 23
24 25 26 27 28 29 30 31
0 9 18 27 36 45 54 63
40 41 42 43 44 45 46 47
48 49 50 51 52 53 54 55
56 57 58 59 60 61 62 63
q){n:x*x;m:x cut til n;#[m;x div 2;:;](m)./:flip 0 0+\:til x}5
0 1 2 3 4
5 6 7 8 9
0 6 12 18 24
15 16 17 18 19
20 21 22 23 24
q){n:x*x;m:x cut til n;#[m;x div 2;:;](m)./:flip 0 0+\:til x}4
0 1 2 3
4 5 6 7
0 5 10 15
12 13 14 15
q)
q)#[((1 2 3);(4 5 6);(7 8 9));1;:;(2;5;8)]
1 2 3
2 5 8
7 8 9
Indexing in q can be straight forward and I believe a intermediate can be omitted:
q)m:((1 2 3);(4 5 6);(7 8 9))
q)m[1]:m[;1]
q)m
1 2 3
2 5 8
7 8 9
I know that this question has been asked but i cant understand the problem in my code. I know that we have to use cost in desending order for minimum cost and i did same but still gives wrong output.
A board composed of m×n wooden squares and asks him to find the minimum cost of breaking the board back down into individual 1×1 pieces. To break the board down, Bob must make cuts along its horizontal and vertical lines.
To reduce the board to squares, xn−1 vertical cuts must be made at locations x1,x2,…,xn−2,xn−1 and ym−1 horizontal cuts must be made at locations y1,y2,…,ym−2,ym−1. Each cut along some xi (or yj) has a cost, cxi (or cyj). If a cut of cost c passes through n already-cut segments, the total cost of the cut is n×c.
The cost of cutting the whole board down into 1×1 squares is the sum of the cost of each successive cut. Recall that the cost of a cut is multiplied by the number of already-cut segments it crosses through, so each cut is increasingly expensive.
Input Format
The first line contains a single integer, T, denoting the number of test cases. The subsequent 3T lines describe each test case in 3 lines.
For each test case, the first line has two positive space-separated integers, m and n, detailing the respective height (y) and width (x) of the board.
The second line has m−1 space-separated integers listing the cost, cyj, of cutting a segment of the board at each respective location from y1,y2,…,ym−2,ym−1.
The third line has n−1 space-separated integers listing the cost, cxi, of cutting a segment of the board at each respective location from x1,x2,…,xn−2,xn−1.
Note: If we were to superimpose the m×n board on a 2D graph, x0, xn, y0, and yn would all be edges of the board and thus not valid cut lines.
Constraints
1≤T≤20
2≤m,n≤1000000
,0≤cxi,cyj≤1000000000
Output Format
For each of the T test cases, find the minimum cost (MinimumCost) of cutting the board into 1×1 squares and print the value of MinimumCost % (1000000000+7).
#include <iostream>
#include <limits>
using namespace std;
int main() {
int t,ch=0;
long int pos,m,n,h=1,l=1;
long long int cost=0,*x,*y,temp;
cin>>t;
while(t>0)
{cin>>m>>n;
cost=0;
x = new long long int[n-1];
y = new long long int[m-1];
for (long i=0;i<m-1;i++)
cin>>y[i];
for(long i=0;i<n-1;i++)
cin>>x[i];
h=1;
l=1;
while((h!=m)|(l!=n))
{ch=0;
temp=0;
for (long i=0;i<m-1;i++)
if (temp<y[i])
{temp=y[i];
pos=i;
}
for(long i=0;i<n-1;i++)
if (temp<x[i])
{temp=x[i];
pos=i;
ch=1;
}
cost=cost+temp*(ch==0?l:h);
if (ch==0)
{y[pos]=-1;
h++;}
else
{x[pos]=-1;
l++;
}
}
cout<<cost%1000000007;
t--;
}
return 0;
}
Test case that gives wrong output:
Input:
5
52 30
2 30 79 47 4 56 47 67 25 30 75 58 47 54 66 61 6 64 28 41 75 36 1 92 42 61 35 56 12 86 84 14 68 63 13 72 19 60 39 96 43 14 55 42 21 73 3 27 37 84 68
64 72 21 56 14 35 44 71 47 82 7 14 50 71 79 23 42 92 14 39 35 81 46 29 2 19 84 81 57
23 43
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23
60 76
7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7
30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30
58 40
71 58 61 51 33 3 43 48 94 30 29 40 59 83 12 43 64 69 64 65 42 57 40 72 64 98 98 47 56 6 85 79 65 46 30 98 49 25 98 96 7 27 88 66 10 0 62 26 69 78 92 64 87 84 88 51 35
87 50 91 45 35 22 62 81 53 61 83 30 59 31 38 39 19 56 1 20 70 28 41 48 72 57 35 56 46 39 91 85 41 34 30 77 57 93 10
47 94
6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6
17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17
Output
51028
1912
33638
91124
27525
This line could be a contributing factor:
while((h!=m)|(l!=n))
The | operator is a binary arithmetic operator.
Maybe you were looking for ||, which is the logical OR operator:
while((h!=m) || (l!=n))
For example, I have a matrix M of size 10x10 and I have a column matrix ind of length 5
How can I assign A(ind,:) to a new matrix B in C++ with OpenCV?
Below is how I do in Matlab:
A = [ 41 8 33 36 22 14 38 43 18 4
46 49 2 2 20 34 13 13 42 3
7 48 43 14 39 33 26 41 30 27
46 25 47 3 40 9 35 13 28 39
32 41 34 5 10 6 45 47 46 47
5 8 38 42 25 25 48 18 15 7
14 22 38 35 23 48 28 10 38 29
28 46 20 16 33 18 7 13 38 24
48 40 33 48 36 30 8 31 20 1
49 48 9 2 38 12 13 24 29 17]
ind = [2; 8; 4; 6; 2]
B = A(ind, :);
B = [ 46 49 2 2 20 34 13 13 42 3
28 46 20 16 33 18 7 13 38 24
46 25 47 3 40 9 35 13 28 39
5 8 38 42 25 25 48 18 15 7
46 49 2 2 20 34 13 13 42 3]
Can anyone tell me how to do this in C++ with OpenCV without using for loop
There is no direct way to extract a random ordering of rows/columns without iterating in some way. The simplest method is to extract rows and push them into the target matrix one by one. Given you have your matrix A declared and its data set:
cv::Mat B;
B.push_back(A(cv::Range(2,3),cv::Range::all()));
B.push_back(A(cv::Range(8,9),cv::Range::all()));
B.push_back(A(cv::Range(4,5),cv::Range::all()));
B.push_back(A(cv::Range(6,7),cv::Range::all()));
B.push_back(A(cv::Range(2,3),cv::Range::all()));
should do what you want. This uses the overloaded operator()(cv::rowRange, cv::colRange) to extract the selected rows.
I don't think this is possible without using for loop but the fastest way of doing this is by using memcpy. You can see the complete code here