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Is there a way to iterate over a range of integers?
...ugh it is fully inlined in the setup phase. In the loop phase there is an extra instruction in the loop, but it isn't too bad.
I'd use the simple for loop.
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Replace first occurrence of pattern in a string [duplicate]
Let's say I have the string:
3 Answers
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The static keyword and its various uses in C++
...tic, but can be addressed from the class as well as an instance (like std::string::npos). [Note: you can declare static members in a class, but they should usually still be defined in a translation unit (cpp file), and as such, there's only one per class]
locations as code:
static std::string name...
What is the difference between cout, cerr, clog of iostream header in c++? When to use which one?
... happen.
#include <iostream>
#include <fstream>
#include <string>
void f(std::ostream &os)
{
std::cin.clear(); // clear EOF flags
std::cin.seekg(0, std::cin.beg); // seek to begin
std::string line;
while(std::getline(std::cin, line)) //input from the file ...
using extern template (C++11)
....cpp
template<typename T>
void f(){}
// Explicit instantiation for char.
template void f<char>();
Main.cpp
#include "TemplHeader.h"
// Commented out from OP code, has no effect.
// extern template void f<T>(); //is this correct?
int main() {
f<char>();
return 0...
Why aren't pointers initialized with NULL by default?
... then assign.
So now we have the situation that the compiler has added an extra instruction to the code that initializes the variable to NULL then later the developer code is added to do the correct initialization. Or under other conditions the variable is potentially never used. A lot of C++ devel...
Passing an array by reference
...an be passed by reference OR by degrading to a pointer. For example, using char arr[1]; foo(char arr[])., arr degrades to a pointer; while using char arr[1]; foo(char (&arr)[1]), arr is passed as a reference. It's notable that the former form is often regarded as ill-formed since the dimension...
The split() method in Java does not work on a dot (.) [duplicate]
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java.lang.String.split splits on regular expressions, and . in a regular expression means "any character".
Try temp.split("\\.").
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How to do a LIKE query in Arel and Rails?
...ers[:name].matches("%#{user_name}%"))
PS:
users = User.arel_table
query_string = "%#{params[query]}%"
param_matches_string = ->(param){
users[param].matches(query_string)
}
User.where(param_matches_string.(:name)\
.or(param_matches_string.(:description)))
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With arrays, why is it the case that a[5] == 5[a]?
...left operand wouldn't be a significant burden. (Some languages use "+" for string concatenation; that's certainly not commutative.)
– Keith Thompson
Apr 21 '14 at 18:13
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