[Solved] How to count the number of keys in a map [closed]

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Poorly phrased question, but if the values you have as keys (A1, A2, etc…) are String and you want to check by the first letter of the key, you can try something like this: int count = 0; for (String k : myMap.keySet()){ if(k.startsWith(“A”)){ count++; } } 4 solved How to count the number of … Read more

[Solved] Using Java, how can i see if the current time is between 10 and 15 minutes after the hour? [closed]

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The following method should resolve your problem public boolean isInRange() { Calendar calendar = Calendar.getInstance(); int minute = calendar.get(Calendar.MINUTE); return minute >= 10 && minute < 15; } It simple recoveries the minute from the current time and verifies if it is between 10 and 15 0 solved Using Java, how can i see if … Read more

[Solved] Junk values in char* variable

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Strings in C need to be NUL terminated. This means you need to add a zero value byte to the end of the string to indicate the end of the string. Because you have no indication of the end of the string when you view/print the value you are reading on past the end of … Read more

[Solved] Does (p+x)-x always result in p for pointer p and integer x in gcc linux x86-64 C++

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Yes, for gcc5.x and later specifically, that specific expression is optimized very early to just p, even with optimization disabled, regardless of any possible runtime UB. This happens even with a static array and compile-time constant size. gcc -fsanitize=undefined doesn’t insert any instrumentation to look for it either. Also no warnings at -Wall -Wextra -Wpedantic … Read more

[Solved] how to call a method after a delay? [closed]

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you can try: having a bool property called completed. perform selector after delay //(some selector showBusy, some delay) completed = NO; dispatch_queue_t myqueue = dispatch_queue_create(“queue”, NULL); dispatch_async(myqueue, ^{ //your long time operation (must not do any UI changes here) dispatch_async(dispatch_get_main_queue(), ^{ //UI here completed = YES; hideBusy; }); }); -(void)showBusy{ if(!completed)….. } 2 solved how … Read more

[Solved] C# Arrays in List printing to console in formatted way [closed]

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Simplest solution if all arrays have same length: string rowFormat = “{0,15}|{1,3} {2,3}”; Console.WriteLine(rowFormat, “History”, “B”, “W”); Console.WriteLine(new String(‘=’, 25)); for(int i = 0; i < array1.Length; i++) Console.WriteLine(rowFormat, array1[i], array2[i], array3[i]); But I would suggest to use custom type for your data instead of keeping data in three arrays. E.g. (names according to your … Read more

[Solved] In R, how to turn characters (grades) into a number and put in a separate column

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Often you can use a named vector to do these sorts of mapping operations simply: students <- data.frame(name=c(“J. Holly”,”H. Kally”, “P. Spertson”, “A. Hikh”, “R. Wizht”), CRSE_GRADE_OFF=c(“A”,”D”,”E”,”A”,”A”)) scores = c(A=1, B=2, C=3, D=4, E=5, F=6) students$grade <- scores[as.character(students$CRSE_GRADE_OFF)] students # name CRSE_GRADE_OFF grade # 1 J. Holly A 1 # 2 H. Kally D 4 … Read more

[Solved] odds to the first and evens last

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How about this solution #include<stdio.h> int main() { int i; int arr[]={ 2, 1 ,4 ,3 ,6 ,5 ,8 ,7 ,9}; int arr_size = sizeof(arr)/sizeof(arr[0]); int sorted[arr_size]; int sp = 0; for(i=0;i<arr_size;i++){ if(arr[i]&1){ sorted[sp++]=arr[i]; } } for(i=0;i<arr_size;i++){ if(!(arr[i]&1)){ sorted[sp++]=arr[i]; } } for(i=0;i< arr_size ;i++) printf(“%d “, sorted[i]); return 0; } the output is 1 3 … Read more