可变长度数组:如何在 C++ 中创建可变大小的缓冲区
Variable Length Arrays: How to create a buffer with variable size in C++
我目前正在写移动平均线class。
目标是在创建 class Running_Average 的新对象时能够将缓冲区大小指定为构造函数的一部分。
#include <iostream>
#include "Complex.h"
#include <cmath>
#include<stdio.h>
#include<stdlib.h>
#include <windows.h>
using namespace std;
class Running_Average
{
public:
double sum = 0;
double average = 0;
int i;
double Average(void); // Member functions declaration
void AddSample(double);
Running_Average(int);
};
Running_Average::Running_Average(int size) {
int buffersize = size;
double buffer[buffersize] = { 0 };
}
void Running_Average::AddSample(double val) //Add new values to buffer
{
for (i = 9; i>0; i--)
{
buffer[i] = buffer[i-1];
}
buffer[0] = val;
}
double Running_Average::Average(void) //Calculate Average of current values in buffer
{
for (i = 0; i<buffersize; i++)
{
cout << buffer[i] << endl;
sum += buffer[i];
}
average = sum / buffersize;
sum = 0;
return average;
}
int main()
{
double value;
int i;
int f = 0;
Running_Average test;
for (i = (40); i < (50); i++)
{
test.AddSample(i);
}
while (1)
{
i = rand() % 100;
test.AddSample(i);
value = test.Average();
cout << endl;
cout << value << endl;
cout << endl;
Sleep(1000);
}
}
然而,构造函数让我很伤心:
Running_Average::Running_Average(int size) {
int buffersize = size;
double buffer[buffersize] = { 0 };
}
具体来说:
buffer[buffersize]
在 visual studio 中抛出一个错误说:
expression must have a constant size.
我希望用户通过将其值传递给 constructor
.
来指定在创建新对象时要使用的缓冲区大小
如何在不抛出错误的情况下完成这项工作?
感谢您的帮助!
编辑: 已解决!谢谢大家的协助!我设法通过使用 std::vector 来定义可变大小的数组来使函数正常工作。
标准 C++ 没有可变长度数组。 (Why?) The size of an array must be a constant expression. Some compilers have non-standard extensions that allow VLAs, but you shouldn't rely on them. Use std::vector
当您需要一个长度可变且可以调整大小的数组时。
有很多方法可以做到这一点。从好到坏,我想到的是:
1 使用 std::vector
int buffersize = size;
std::vector<double> buffer(buffersize);
2 内置唯一指针或共享指针(取决于用途)
int buffersize = size;
auto buffer = make_unique<double[]>(buffersize) // C++14
int buffersize = size;
auto buffer = make_shared<double[]>(buffersize) // C++14
3 手动分配
int buffersize = size;
double *buffer = new double[buffersize];
// delete [] buffer, must be called later
4 在堆栈上分配(不 建议,并且取决于平台)
int buffersize = size;
double *buffer = alloca(buffersize * sizeof(*buffer));
请注意,在所有这些情况下,您都可以像数组一样索引缓冲区。
expression must have a constant size
double buffer[buffersize] = { 0 };
第一个,buffersize
不是constexpr
。它是一个在运行时改变的变量。
根据 array
的标准定义指定数组大小时:
The constant expression
specifies the bound of (number of elements in)
the array. If the value of the constant expression
is N, the array has
N elements numbered 0 to N-1,
包含 5 个 double
类型元素的数组声明示例应如下所示:
double buffer[5]; // 5 is a literal
constexpr int size = 5;
double buffer[size]; // size is constexpr
其次,buffer
是变长数组(VLA)。
VLA 作为扩展在某些编译器中得到部分支持。
How can I make this work without it throwing an error?
如果你需要长度可变,使用std::vector
并在你的构造函数中初始化它:
class Running_Average {
Running_Average(int size): buffer(size, 0) {}
std::vector<double> buffer;
}
根据您使用 buffer
的方式,我建议使用 std::list<double>
。
将此添加到 Running_Average
的开头:
class Running_Average
{
private:
list<double> buffer;
const size_t MaxBufferSize;
public:
...
构造函数:
Running_Average::Running_Average(size_t size)
: MaxBufferSize(size)
{
}
AddSample()
和 Average()
:
void Running_Average::AddSample(double val)
{
if (buffer.size() == MaxBufferSize)
{
buffer.pop_back();
}
buffer.push_front(val);
}
double Running_Average::Average()
{
double sum = 0;
for (auto a : buffer)
{
cout << a << endl;
sum += a;
}
return sum / buffer.size();
}
我还会删除 sum
、average
和 i
成员变量,而是在使用它们的地方声明它们(如果需要)。
可变长度数组在 C 中有效,但在 C++ 中无效。在 C++ 中,您最好使用 vector
集合,因为它可以让您更好地表示意图,改变数组大小,而不必单独维护当前大小。
以下完整程序为您提供了工作基准,包括测试工具代码:
#include <iostream>
#include <vector>
class RunningValues {
public:
RunningValues(size_t size = 50);
void Add(double val);
double Sum();
double Average();
private:
std::vector<double> dataBuffer;
size_t sizeLimit;
double sum;
};
// Constructor: store limit and zero sum (vector is already empty).
RunningValues::RunningValues(size_t size): sizeLimit(size), sum(0.0) {}
// Add a sample.
void RunningValues::Add(double val) {
// Zero size, disallow adds.
if (sizeLimit == 0) return;
// If we would exceed limit, remove earliest.
if (dataBuffer.size() == sizeLimit) {
sum -= dataBuffer[0];
dataBuffer.erase(dataBuffer.begin());
}
// Add value to end.
sum += val;
dataBuffer.push_back(val);
}
// Get the average (zero if nothing yet added) or sum.
double RunningValues::Average() {
if (dataBuffer.size() == 0) return 0.0;
return sum / dataBuffer.size();
}
double RunningValues::Sum() {
return sum;
}
// Test harness.
int main() {
RunningValues test(10);
std::cout << "Ave = " << test.Average() << ", sum = " << test.Sum() << '\n';
for (int i = 40; i < 50; ++i)
{
test.Add(i);
std:: cout << "Add " << i << ", ave = " << test.Average() << ", sum=" << test.Sum() << '\n';
}
for (int i = 0; i < 20; ++i)
{
int val = rand() % 100;
test.Add(val);
std:: cout << "Add " << val << ", ave = " << test.Average() << ", sum=" << test.Sum() << '\n';
}
}
示例 运行,显示了各个点的平均值和总和,如下所示:
Ave = 0, sum = 0
Add 40, ave = 40, sum=40
Add 41, ave = 40.5, sum=81
Add 42, ave = 41, sum=123
Add 43, ave = 41.5, sum=166
Add 44, ave = 42, sum=210
Add 45, ave = 42.5, sum=255
Add 46, ave = 43, sum=301
Add 47, ave = 43.5, sum=348
Add 48, ave = 44, sum=396
Add 49, ave = 44.5, sum=445
Add 83, ave = 48.8, sum=488
Add 86, ave = 53.3, sum=533
Add 77, ave = 56.8, sum=568
Add 15, ave = 54, sum=540
Add 93, ave = 58.9, sum=589
Add 35, ave = 57.9, sum=579
Add 86, ave = 61.9, sum=619
Add 92, ave = 66.4, sum=664
Add 49, ave = 66.5, sum=665
Add 21, ave = 63.7, sum=637
Add 62, ave = 61.6, sum=616
Add 27, ave = 55.7, sum=557
Add 90, ave = 57, sum=570
Add 59, ave = 61.4, sum=614
Add 63, ave = 58.4, sum=584
Add 26, ave = 57.5, sum=575
Add 40, ave = 52.9, sum=529
Add 26, ave = 46.3, sum=463
Add 72, ave = 48.6, sum=486
Add 36, ave = 50.1, sum=501
如果您希望有一个避免 vector
的解决方案(当向量大小从 0 到 N
然后停留在那里),你可以只使用堆上的裸数组作为循环缓冲区。
该代码略有变化(没有 main
,因为它没有改变):
#include <iostream>
class RunningValues {
public:
RunningValues(size_t size = 50);
~RunningValues();
void Add(double val);
double Sum();
double Average();
private:
size_t count, next, limit;
double sum, *data;
};
RunningValues::RunningValues(size_t size)
: count(0), next(0), limit(size)
, sum(0.0), data(new double[size]) {}
RunningValues::~RunningValues() {
delete[] data;
}
void RunningValues::Add(double val) {
// Zero size, disallow adds.
if (limit == 0) return;
// If we would exceed limit, remove earliest.
if (count == limit) {
sum -= data[next];
--count;
}
// Add value to end.
data[next] = val;
sum += val;
++count;
next = (next + 1) % limit;
}
// Get the average (zero if nothing yet added) or sum.
double RunningValues::Average() {
if (count == 0) return 0.0;
return sum / count;
}
double RunningValues::Sum() {
return sum;
}
基于矢量的解决方案的变化相当小:
- 构造函数不再有向量(很明显)。相反,它有一个固定大小的数组用作循环缓冲区,以及
count
和 next
变量来管理它。
- 您现在需要一个析构函数来清理缓冲区(之前,vector 会自行管理)。
- 项目的添加现在使用
count
和 next
(而不是矢量)来计算如何调整总和并保持相关数据的统计。
- 平均值的计算现在使用
count
而不是矢量大小。
除此之外,它实际上与上面基于向量的代码非常相似。
我目前正在写移动平均线class。
目标是在创建 class Running_Average 的新对象时能够将缓冲区大小指定为构造函数的一部分。
#include <iostream>
#include "Complex.h"
#include <cmath>
#include<stdio.h>
#include<stdlib.h>
#include <windows.h>
using namespace std;
class Running_Average
{
public:
double sum = 0;
double average = 0;
int i;
double Average(void); // Member functions declaration
void AddSample(double);
Running_Average(int);
};
Running_Average::Running_Average(int size) {
int buffersize = size;
double buffer[buffersize] = { 0 };
}
void Running_Average::AddSample(double val) //Add new values to buffer
{
for (i = 9; i>0; i--)
{
buffer[i] = buffer[i-1];
}
buffer[0] = val;
}
double Running_Average::Average(void) //Calculate Average of current values in buffer
{
for (i = 0; i<buffersize; i++)
{
cout << buffer[i] << endl;
sum += buffer[i];
}
average = sum / buffersize;
sum = 0;
return average;
}
int main()
{
double value;
int i;
int f = 0;
Running_Average test;
for (i = (40); i < (50); i++)
{
test.AddSample(i);
}
while (1)
{
i = rand() % 100;
test.AddSample(i);
value = test.Average();
cout << endl;
cout << value << endl;
cout << endl;
Sleep(1000);
}
}
然而,构造函数让我很伤心:
Running_Average::Running_Average(int size) {
int buffersize = size;
double buffer[buffersize] = { 0 };
}
具体来说:
buffer[buffersize]
在 visual studio 中抛出一个错误说:
expression must have a constant size.
我希望用户通过将其值传递给 constructor
.
如何在不抛出错误的情况下完成这项工作?
感谢您的帮助!
编辑: 已解决!谢谢大家的协助!我设法通过使用 std::vector 来定义可变大小的数组来使函数正常工作。
标准 C++ 没有可变长度数组。 (Why?) The size of an array must be a constant expression. Some compilers have non-standard extensions that allow VLAs, but you shouldn't rely on them. Use std::vector
当您需要一个长度可变且可以调整大小的数组时。
有很多方法可以做到这一点。从好到坏,我想到的是:
1 使用 std::vector
int buffersize = size;
std::vector<double> buffer(buffersize);
2 内置唯一指针或共享指针(取决于用途)
int buffersize = size;
auto buffer = make_unique<double[]>(buffersize) // C++14
int buffersize = size;
auto buffer = make_shared<double[]>(buffersize) // C++14
3 手动分配
int buffersize = size;
double *buffer = new double[buffersize];
// delete [] buffer, must be called later
4 在堆栈上分配(不 建议,并且取决于平台)
int buffersize = size;
double *buffer = alloca(buffersize * sizeof(*buffer));
请注意,在所有这些情况下,您都可以像数组一样索引缓冲区。
expression must have a constant size
double buffer[buffersize] = { 0 };
第一个,buffersize
不是constexpr
。它是一个在运行时改变的变量。
根据 array
的标准定义指定数组大小时:
The
constant expression
specifies the bound of (number of elements in) the array. If the value of theconstant expression
is N, the array has N elements numbered 0 to N-1,
包含 5 个 double
类型元素的数组声明示例应如下所示:
double buffer[5]; // 5 is a literal
constexpr int size = 5;
double buffer[size]; // size is constexpr
其次,buffer
是变长数组(VLA)。
VLA 作为扩展在某些编译器中得到部分支持。
How can I make this work without it throwing an error?
如果你需要长度可变,使用std::vector
并在你的构造函数中初始化它:
class Running_Average {
Running_Average(int size): buffer(size, 0) {}
std::vector<double> buffer;
}
根据您使用 buffer
的方式,我建议使用 std::list<double>
。
将此添加到 Running_Average
的开头:
class Running_Average
{
private:
list<double> buffer;
const size_t MaxBufferSize;
public:
...
构造函数:
Running_Average::Running_Average(size_t size)
: MaxBufferSize(size)
{
}
AddSample()
和 Average()
:
void Running_Average::AddSample(double val)
{
if (buffer.size() == MaxBufferSize)
{
buffer.pop_back();
}
buffer.push_front(val);
}
double Running_Average::Average()
{
double sum = 0;
for (auto a : buffer)
{
cout << a << endl;
sum += a;
}
return sum / buffer.size();
}
我还会删除 sum
、average
和 i
成员变量,而是在使用它们的地方声明它们(如果需要)。
可变长度数组在 C 中有效,但在 C++ 中无效。在 C++ 中,您最好使用 vector
集合,因为它可以让您更好地表示意图,改变数组大小,而不必单独维护当前大小。
以下完整程序为您提供了工作基准,包括测试工具代码:
#include <iostream>
#include <vector>
class RunningValues {
public:
RunningValues(size_t size = 50);
void Add(double val);
double Sum();
double Average();
private:
std::vector<double> dataBuffer;
size_t sizeLimit;
double sum;
};
// Constructor: store limit and zero sum (vector is already empty).
RunningValues::RunningValues(size_t size): sizeLimit(size), sum(0.0) {}
// Add a sample.
void RunningValues::Add(double val) {
// Zero size, disallow adds.
if (sizeLimit == 0) return;
// If we would exceed limit, remove earliest.
if (dataBuffer.size() == sizeLimit) {
sum -= dataBuffer[0];
dataBuffer.erase(dataBuffer.begin());
}
// Add value to end.
sum += val;
dataBuffer.push_back(val);
}
// Get the average (zero if nothing yet added) or sum.
double RunningValues::Average() {
if (dataBuffer.size() == 0) return 0.0;
return sum / dataBuffer.size();
}
double RunningValues::Sum() {
return sum;
}
// Test harness.
int main() {
RunningValues test(10);
std::cout << "Ave = " << test.Average() << ", sum = " << test.Sum() << '\n';
for (int i = 40; i < 50; ++i)
{
test.Add(i);
std:: cout << "Add " << i << ", ave = " << test.Average() << ", sum=" << test.Sum() << '\n';
}
for (int i = 0; i < 20; ++i)
{
int val = rand() % 100;
test.Add(val);
std:: cout << "Add " << val << ", ave = " << test.Average() << ", sum=" << test.Sum() << '\n';
}
}
示例 运行,显示了各个点的平均值和总和,如下所示:
Ave = 0, sum = 0
Add 40, ave = 40, sum=40
Add 41, ave = 40.5, sum=81
Add 42, ave = 41, sum=123
Add 43, ave = 41.5, sum=166
Add 44, ave = 42, sum=210
Add 45, ave = 42.5, sum=255
Add 46, ave = 43, sum=301
Add 47, ave = 43.5, sum=348
Add 48, ave = 44, sum=396
Add 49, ave = 44.5, sum=445
Add 83, ave = 48.8, sum=488
Add 86, ave = 53.3, sum=533
Add 77, ave = 56.8, sum=568
Add 15, ave = 54, sum=540
Add 93, ave = 58.9, sum=589
Add 35, ave = 57.9, sum=579
Add 86, ave = 61.9, sum=619
Add 92, ave = 66.4, sum=664
Add 49, ave = 66.5, sum=665
Add 21, ave = 63.7, sum=637
Add 62, ave = 61.6, sum=616
Add 27, ave = 55.7, sum=557
Add 90, ave = 57, sum=570
Add 59, ave = 61.4, sum=614
Add 63, ave = 58.4, sum=584
Add 26, ave = 57.5, sum=575
Add 40, ave = 52.9, sum=529
Add 26, ave = 46.3, sum=463
Add 72, ave = 48.6, sum=486
Add 36, ave = 50.1, sum=501
如果您希望有一个避免 vector
的解决方案(当向量大小从 0 到 N
然后停留在那里),你可以只使用堆上的裸数组作为循环缓冲区。
该代码略有变化(没有 main
,因为它没有改变):
#include <iostream>
class RunningValues {
public:
RunningValues(size_t size = 50);
~RunningValues();
void Add(double val);
double Sum();
double Average();
private:
size_t count, next, limit;
double sum, *data;
};
RunningValues::RunningValues(size_t size)
: count(0), next(0), limit(size)
, sum(0.0), data(new double[size]) {}
RunningValues::~RunningValues() {
delete[] data;
}
void RunningValues::Add(double val) {
// Zero size, disallow adds.
if (limit == 0) return;
// If we would exceed limit, remove earliest.
if (count == limit) {
sum -= data[next];
--count;
}
// Add value to end.
data[next] = val;
sum += val;
++count;
next = (next + 1) % limit;
}
// Get the average (zero if nothing yet added) or sum.
double RunningValues::Average() {
if (count == 0) return 0.0;
return sum / count;
}
double RunningValues::Sum() {
return sum;
}
基于矢量的解决方案的变化相当小:
- 构造函数不再有向量(很明显)。相反,它有一个固定大小的数组用作循环缓冲区,以及
count
和next
变量来管理它。 - 您现在需要一个析构函数来清理缓冲区(之前,vector 会自行管理)。
- 项目的添加现在使用
count
和next
(而不是矢量)来计算如何调整总和并保持相关数据的统计。 - 平均值的计算现在使用
count
而不是矢量大小。
除此之外,它实际上与上面基于向量的代码非常相似。