pthread_t resources are not released. You should call pthread_detach or pthread_join, otherwise the pthread_t value remains valid consumes resources and I can guess that happens in this case. 0 solved Process memory examination
This should do the trick: #include <stdio.h> #include <string.h> int extract_line (char* line, char* buffer, char* ctag) { char line_buffer[255] = {0}; char* tagStart; char* tagEnd; if( strlen(line) < (sizeof(line_buffer)/sizeof(char)) ) { strcpy(line_buffer,line); } else { printf(“Line size is too big.\n”); return 1; } tagStart = strstr(line_buffer,ctag); if(tagStart != NULL) { tagEnd = strstr(tagStart+1,ctag); if(tagEnd … Read more
The std::cin method from the iostream library returns true when it can read an input and false when it tries to read but there’s nothing more to read. So you can use while(cin >> bar) and it will read input until it finds an EOF character (either from you pressing CTRL+D in the console or … Read more
Statement a[637][i]++ increases the value of the cell 637/i of two-dimensional array a. assert(0) simply aborts program execution at this point (since condition 0 means false, defining that the assertion is never met). Confer this SO answer for a more detailed explanation. solved In c++, what does an increment to a 2D array? Whats the … Read more
Looking at your code the class Result is not a list which I think you are trying to call with List()? not sure.. To create a List<> you can do it multiple ways see below on a few options. Try these in your Main method. List<string> stringList = new List<string>(); //creates a List<string> called stringList … Read more
The part in is_attaced in which you check the diagonals is not working. It has two basic problems, first you dont check all diagonals all rows – you just check the previous row. And, second, you are doing out of bound accesses, when you are in the first or last column. An implementation that is … Read more
If those strings are the same for all the cases – make them global: string a = “test”; string b = “try”; string c = “compare”; //case 1 for (int i = 1; i <= 2; i++) { if (a == b) { do something }; if (a == c) { do something}; } //case … Read more
It is impossible for this loop to ever end. The only possible way this look could end would be for type1 to be nor, fir, wat, … at the same time. You probably want to replace all ors (||), by ands (&&). In that case your loop would stop at the moment type1 is equal … Read more
what you can is int j = 0,order[] = {0,7,3,4,6,1,5,2}; for(int i = 0;i <256; i +=2) { int x =add(array1[i],array2[order[j%8]]); j++; int y =add(array1[i+1],array2[order[j%8]]); j++; } UPDATE alternate solution can be (if you want without using i+=2) int j = 0,order[] = {0,7,3,4,6,1,5,2}; for(int i = 0;i <256; i ++) { int x =add(array1[i],array2[order[j%8]]); … Read more
This how I finally did it foreach (var item in PisterosLista) { var DatosRegistro = new T_Pisteros { PisteroID = item.PisteroID, PisteroN = item.PisteroN }; var num = db.Insert(DatosRegistro); //insert into new DB } solved How to read IEnumerable List values one by one in Xamarin forms?
Yes, you can and you will run into problems. Since C does not do any boundary checking for you, you are only getting an illusion of it working correctly. You are just getting lucky that no one else is using the piece of memory you are accessing in your *next pointer. So, it may not … Read more
If you split first, you can use the first column as an Id column and search for exact match: string line; using (StreamReader file = new StreamReader(@”db.txt”)) { while ((line = file.ReadLine()) != null) { var values = line.Split(“,”); if (values[0] == “1713”) { label1.Text = values[2]; } } } 0 solved C# Split with … Read more
You should be able to add a collider to the child/head, BoxCollider2D for example. Then you can handle the Collision event and call a method on the parent object. It would look something like: // In the child/head void OnCollisionEnter2D(Collision2D col) { var goomba = transform.Parent.GetComponent<Goomba>(); goomba.Kill(); } This is probably the simplest way, but … Read more
Let’s look at what happens during the first two iterations. We start with: i = 0 d = 4 L = -5 -4 -3 -2 -1 R = 0 The greater element is L[d], so we add that, change the sign of L[i], and decrement d, so now we have: i = 0 d = … Read more
This works. #include <string> #include <iostream> using namespace std; int main() { char title[20] = “testmovie2015.mkv”; int year=0,i=0; for(i=0;title[i]!=’\0′;i++){ if(isdigit(title[i])&&isdigit(title[i+1])&&isdigit(title[i+2])&&isdigit(title[i+3])) { year=1; break; } } if(year) { year=((title[i]-‘0’)*1000)+((title[i+1]-‘0’)*100)+((title[i+2]-‘0’)*10)+(title[i+3]-‘0’); if(year>=2000&&year<=2018) { int k=0; while(k<i) { cout<<title[k]; k++; } } else { cout<<“Year found: “<<year<<“, but out of given range”; } } else { cout<<“No year found … Read more