pow() is a function from the maths library, libm. Your LDFLAGS needs to include “-lm”. See https://man7.org/linux/man-pages/man3/pow.3.html solved unable to compile PuTTY 0.76 on Ubuntu 20.04 [closed]
yes it is possible. You could use a library like Boos.Asio which has an example of chat in its documentation : http://www.boost.org/doc/libs/1_39_0/doc/html/boost_asio/example/chat/posix_chat_client.cpp (client part) 0 solved How to make socket chat in C++?
You need to compute begin and end of your loop, i.e. first and last_exclusive. #include <vector> #include <numeric> #include <iostream> int main(){ std::vector<int> vec(16); std::iota(vec.begin(), vec.end(), 0); size_t first = 3; size_t last_exclusive = 7; //loop using indices for(size_t i = first; i < last_exclusive; i++){ const auto& w = vec[i]; std::cout << w << … Read more
Why instance variables are not initialized before constructor creation. What happens if they are initialized at the class loading itself [closed] solved Why instance variables are not initialized before constructor creation. What happens if they are initialized at the class loading itself [closed]
@AndrewHenle commented: If you want to compare numbers with different numbers of digits and not strings, you need to convert the strings to actual numbers. As it is clear from your comments that you want to compare the numeric value that is contained in the strings, you need to do just that — convert the … Read more
It looks like you are trying to get a count of all users who registered on a certain day. If so, I think you are best off using the group keyword to group together all users by registration date. You can then create a new anonymous type to contain by the date, and the count, … Read more
NO, your function does not behave the same as memset. Your function sets a pointer to NULL and memset sets the values of the data to the value supplied. Different things altogether. 3 solved behavior of memset
The code compiles because it is valid code. It just doesn’t do anything meaningful. sin_start is not a function, it is a variable of type istream_iterator<int>, where istream_iterator is a template class defined in the <iterator> header file. It is being passed std::cin as a parameter to its constructor, so it knows which stream to … Read more
length() probably returns size_t or unsigned long and you are comparing it with signed long. Change long lDelFront = 0, lDelBack = 0; to size_t lDelFront = 0; size_t lDelBack = 0; to avoid signed/unsigned comparison 1 solved warning C4018: ‘
First of all continue is an already occupied keyword, so, better rename your function. The continue statement shall occur only in an iteration-statement and causes control to pass to the loop-continuation portion of the smallest enclosing iteration-statement, that is, to the end of the loop. In our case, you trying to use a non-static member … Read more
char *string1 = strdup(data.c_str()); // do stuff with string1 free(string1); SHA1((unsigned char*)&string1, strlen(string1), (unsigned char*)&digest); There’s your error. You created string1 You used it You freed it You used it again Don’t free what you still need to use. In addition to that: SHA1((unsigned char*)&string1, strlen(string1), (unsigned char*)&digest); What you’re doing here is passing the … Read more
You want to memcpy memcpy(history[index].input, inputBuffer, sizeof(inputBuffer)*sizeof(inputBuffer[0])); 4 solved Copy contents of non null terminated char array into another char array
Your code looks like a weird mixture of C and C++. You are using malloc to allocate dynamic memory, which does not invoke C++ objects’ constructors. You should use new/delete (or even better, smart pointers) if you want to dynamically allocate memory. Any idea why this would be? The reason why your code doesn’t work … Read more
The shorter, more general answer is that you need to bump characters back, and insert a ‘ ‘ in between them. What have you done so far? Does it need to be in place? One (perhaps not optimal, but easy to follow solution) would be making a larger array, copying in alternating letters, something like … Read more
it is exactly like you said: 5*factorial(4)=5*4*factorial(3)=5*4*3*factorial(2)=5*4*3*2*factorial(1) So if it reaches 1 then this will be replaced by 5*factorial(4)=5*4*factorial(3)=5*4*3*factorial(2)=5*4*3*2*1 so the result of the last step goes into the second last step…. where 2*1 will be calculated… After that the 3rd last step gets the value of 2*1 = 2 and multiplies 3 on in … Read more