这是一些简单的代码.
#include <iostream>
using namespace std;
bool func(char* m)
{
*m = '4';
return true;
}
using namespace std;
int main()
{
char c1 = '3';
cout << "a" << c1 << func(&c1) << c1 << "b" << endl;
return 0;
}
当使用g ++ -O0(4.7.2)编译时,输出为a413b,对于-O2,输出为a414b.对于clang ++ 3.2,输出为a314b.
我在这部分代码中为c ++做了什么未定义的事情吗？
我正在尝试constexpr:
#include <iostream>
constexpr long long fibonacci(const int x)
{
return x <= 1 ? 1 : fibonacci(x - 1) + fibonacci(x - 2);
}
int main()
{
const long long lol = fibonacci(500);
std::cout << lol << std::endl;
}
所以我想lol在编译时计算:
toogy@stewie
» g++ -std=c++14 -g src/test.cc -o test.out
toogy@stewie
» ./test.out
4859788740867454402
它的工作非常好g++.在编译时,它甚至会进行一些记忆,优化这种糟糕的斐波纳契功能,然后fibonacci(500)立即进行计算.
然后我尝试clang++:
toogy@stewie
» clang++ -std=c++1y -g src/test.cc -o test.out
toogy@stewie
» ./test.out
... very long
lol …
考虑以下初始化:
std::string falseString = false;
std::string trueString = true;
有了g++ 5.2.0,编译器抛出警告falseString,而错误trueString.
随着clang++ 3.6 -std=c++11,编译器会引发错误都falseString还有trueString.
Q1)为什么不同的行为gcc即使两个初始化值都是相同的类型(bool)？
Q2)哪个编译器是正确的,为什么？标准说什么？
编辑:
错误:没有可行的从'bool'转换为'std :: string'(又名'basic_string')
警告:将'false'转换为'std :: __ cxx11 :: basic_string <_CharT,_Traits,_Alloc> :: basic_string(const _CharT*,const _Alloc&)的参数1的指针类型[与_CharT = char; _Traits = std :: char_traits; _Alloc = std :: allocator]'[ - Wconversion-null]
我最近将一个类从模板更改为不,并发现在编写using声明从模板化基类继承构造函数时,我不能再省略模板参数.只要我的课没有模板化,我就可以省略这些论点,只要我不能.在下面的可编译片段中bar表示之前的类并buzz表示之后的类.我测试了GCC 5.2和Clang 3.7,它们具有相同的行为.这是编译器错误还是标准？
#include <iostream>
template<class A, class B>
struct foo {
foo(int x) {
std::cout << x << std::endl;
}
};
struct bar : foo<bar, short> {
using foo::foo; // this appears legal
// using foo<bar, short>::foo; // this works too, thought I would need it
};
template<class X>
struct buzz : foo<buzz<X>, short> {
//using foo::foo; // no longer legal for some reason
using foo<buzz<X>, short>::foo; // now do actually need this
};
int …我不明白为什么我没有收到警告(用g ++或clang ++)来返回NULL作为newtstr()下面的对象:
#include<iostream>
using namespace std;
string newstr();
int main()
{
string s = newstr();
cout << s << endl;
return 0;
}
string newstr()
{
return NULL;
}
.
$ g++ -Wall -Wextra -pedantic testptr.cpp
$
但是,我得到一个运行时错误:
$ ./a.out
terminate called after throwing an instance of 'std::logic_error'
what(): basic_string::_S_construct null not valid
Aborted
$
.
$ g++ --version
g++ (GCC) 4.8.0
在科学编程的单元管理环境中,我正在管理以下课程:
template <class UnitName>
class Quantity
{
double value;
public:
Quantity(double val = 0) : value(val) {}
Quantity(const Quantity &) {}
Quantity & operator = (const Quantity &) { return *this; }
double get_value() const noexcept { return value; }
operator double() const noexcept { return value; }
template <class SrcUnit>
Quantity(const Quantity<SrcUnit> &)
{
// here the conversion is done
}
template <class SrcUnit>
Quantity & operator = (const Quantity<SrcUnit> &)
{
// here the conversion is done
return …我试图弄清楚C++ 11规范中是否有任何内容.以下代码的预期行为(此处为GitHub链接):
struct Scalar {
int data;
Scalar(int x) : data(x) {}
int get() {
return data;
}
Scalar &square() {
scale(data);
return *this;
}
void scale(int rhs) {
data *= rhs;
}
};
int main() {
Scalar v(3);
v.square().scale(v.get());
return v.data;
}
这主要是因为发现它在g++和之间做了不同的事情clang++:
$ g++ --version
g++ (GCC) 6.2.1 20160830
Copyright (C) 2016 Free Software Foundation, Inc.
This is free software; see the source for copying conditions. There is NO
warranty; …我正在尝试制作一个范围的线程.
#include <iostream>
#include <thread>
class ScopedThread {
public:
template< class Function, class... Args>
explicit ScopedThread( int id, Function&& f, Args&&... args)
: m_thread( std::ref(f), std::forward<Args>(args)...)
, id(std::move(id)) {
}
int getId() const { return id; }
~ScopedThread() { m_thread.join(); }
private:
std::thread m_thread;
int id;
};
class Worker {
public:
Worker(int id): thd(id, &Worker::work, this) { }
void work() {
for( int i = 0; i < 10; i++)
std::cout << "I am working" << std::endl;
}
private:
ScopedThread …以下代码在VS 2015(更新3)和gcc 6.3(C++ 14)上编译正常,没有任何问题.
#include <string>
#include <locale>
int main()
{
std::u16string ustr = u"Android";
bool var = std::isspace(ustr[0],std::locale());
return 0;
}
但是,在clang/Xcode上它失败并出现以下错误
Error(s):
source_file.cpp:8:10: warning: unused variable 'var' [-Wunused-variable]
bool var = std::isspace(ustr[0],std::locale());
^
In file included from source_file.cpp:2:
In file included from /usr/include/c++/v1/locale:182:
/usr/include/c++/v1/__locale:705:44: error: implicit instantiation of undefined template 'std::__1::ctype<char16_t>'
return use_facet<ctype<_CharT> >(__loc).is(ctype_base::space, __c);
^
source_file.cpp:8:21: note: in instantiation of function template specialization 'std::__1::isspace<char16_t>' requested here
bool var = std::isspace(ustr[0],std::locale());
^
/usr/include/c++/v1/__locale:427:53: note: template is …I was playing around a bit with Concepts offered in C++20 and came up with a simple example, which, to my surprise, does not produce the expected results (please leave any discussion on the usefulness of my example be :-)):
#include <iostream>
#include <type_traits>
#include <vector>
template <typename T>
concept i_am_integral = requires { std::is_integral_v<T> == true; };
template <typename T> requires i_am_integral<T>
using intvector = std::vector<T>;
int main() {
intvector<int> v = { 1, 2, 3 }; // <- …