,作者说
只有POD类型可以作为省略号“.”的参数。而std::string不是POD类型。
我把它理解为Passing NON-POD type to Variadic function is undefined behavior。是对的吗?
不过,他说的是C/C++标准吗?我试图在n3242 C++规范中找到它。却找不到。
我想知道我的理解是正确的,这是一个标准。
我尝试在Rust中逐行处理文件,并使用Rayon将其并行化。它会抱怨以下几点
rayon::str::Lines<'_>` is not an iterator
= help: the trait `std::iter::Iterator` is not implemented for
= note: required by `std::iter::IntoIterator::into_iter`
到目前为止,代码看起来是这样的
use rayon::prelude::*;
use std::fs;
fn main() {
let file_cont
我写了这个程序:
// splits a sentence into words
#include <iostream>
#include <string>
#include <algorithm>
#include "spacefunc.h"
using std::string;
using std::cout;
using std::endl;
using std::find_if;
int main() {
typedef string::const_iterator iter;
string input =
我在C++中发现了一件有趣的事情。
using namespace std;
struct Person {
string name;
bool check() {
return name == "carol";
}
};
int main()
{
Person p = { "Mark" };
list<Person> l;
l.push_back(p);
if (find_if(l.begin(), l.end(), bind(&Person::check, s
这是来自cppreference.com for std::copy()的报价。
Complexity
1-2) Exactly (last - first) assignments
3-4) Exactly (last - first) applications of the predicate, between 0 and (last - first) assignments (assignment for every element for which predicate is equal to true, dependent on predicate and input data)
考虑以下情况:
typedef struct myStruct
{
int cn;
std::string dn;
} MyStruct;
int main()
{
std::vector<MyStruct> v;
// fill some data
...
...
int c = 1;
std::vector<MyStruct>::iterator it = std::find_if(v.begin(), v.end(),
[c](const MyStruct&
如何在C++11中初始化POD (平原旧数据)的成员变量?
class A {
public:
int theAnswer; // this is not initialized
};
static_assert(std::is_pod<A>::value, "A must be a plain old object");
class B {
public:
int theAnswer { 42 }; // this should initialize to 42
};
static_assert(std::is_pod<B>::
在实现自定义容器时,我需要实现迭代器。当然,我不想为const和non迭代器编写两次代码。我发现详细描述了这样一个可能的实现:
template<class T>
class ContainerIterator {
using pointer = T*;
using reference = T&;
...
};
template<class T>
class Container {
using iterator_type = ContainerIterator<T>;
using const_iterator