当我似乎有足够的（虚拟？）可用内存时，我试图理解为什么会出现 std::bad_alloc 异常。本质上，我有一个素数生成器（Eratosthenes 筛子（尚未分段）），我在其中为指标数组更新布尔值，然后为我在命令行上指定的界限下找到的素数更新整数。
我有 1GB RAM（其中一些将被我的操作系统占用（ubuntu 10.04），并且可能其中一些不能用作堆内存（我错了吗？））和 2.8 GB 的交换空间（我相信这是自动的在安装 Ubuntu 时为我设置）
如果我将上限设置为 600000000，那么我要求为我的指标数组提供 0.6 GB 内存和大约 30000000*4 字节（略微高估，因为有 26355867 个小于 500000000 的素数）用于我的素数数组和一些变量这里和那里; 这意味着我要求大约 0.72（+ 可忽略不计）GB 的内存，我认为它应该被我可用的交换空间覆盖（我知道触摸这些东西会荒谬地减慢我的程序速度）。但是我得到了 std::bad_allocs。
谁能指出我在这里缺少什么？（在粘贴我的最后一个错误之前将 long long ints 更改为 ints 的最后一件事是段错误（我的数字远低于 2^31，所以我看不到我溢出的地方） - 仍然试图弄清楚那个)
我的代码如下（并且不会剥夺我自己对更快算法等进行调查的好处。我会很感激这里的任何代码改进！（即，如果我犯了主要的禁忌））
主文件
#include <iostream>
#include <cmath>
#include "Prime.hpp"
#include <ctime>
#include <stdio.h>
#include <cstring>
//USAGE: execute program with the nth prime you want and an upper bound for finding primes --too high may cause bad alloc
int main(int argc, const char *argv[])
{
int a = strlen(argv[1]);
clock_t start = clock();
if(argc != 2)
{
std::cout << "USAGE: Enter a positive inputNumber <= 500000000.\n"
<< "This inputNumber is an upper bound for the primes that can be found\n";
return -1;
}
const char* primeBound = argv[1];
int inputNum = 0;
for(int i = 0; i < strlen(argv[1]); i++)
{
if(primeBound[i] < 48 || primeBound[i] > 57 || primeBound[0] == 48)
{
std::cout << "USAGE: Enter a positive inputNumber <= 500000000.\n"
<< "This inputNumber is an upper bound for the primes that can be found\n";
return -1;
}
inputNum = (int)(primeBound[i]-48) + (10 * inputNum);
}
if(inputNum > 600000000)//getting close to the memory limit for this machine (1GB - memory used by the OS):
//(each bool takes 1 byte and I'd be asking for more than 500 million of these
//and I'd also asking for over 100000000 bytes to store the primes > 0.6 GB)
{
std::cout << "USAGE: Enter a positive inputNumber <= 500000000.\n"
<< "This inputNumber is an upper bound for the primes that can be found\n";
return -1;
}
Prime p(inputNum);
std::cout << "the largest prime less than " << inputNum << " is: " << p.getPrime(p.getNoOfPrimes()) << "\n";
std::cout << "Number of primes: " << p.getNoOfPrimes() << "\n";
std::cout << ((double)clock() - start) / CLOCKS_PER_SEC << "\n";
return 0;
}
Prime.hpp
#ifndef PRIME_HPP
#define PRIME_HPP
#include <iostream>
#include <cmath>
class Prime
{
int lastStorageSize;
bool* primeIndicators;
int* primes;
int noOfPrimes;
void allocateIndicatorArray(int num);
void allocatePrimesArray();
void generateIndicators();
void generatePrimeList();
Prime(){}; //forcing constructor with param = size
public:
Prime(int num);
int getNoOfPrimes();
int getPrime(int nthPrime);
~Prime(){delete [] primeIndicators; delete [] primes;}
};
#endif
Prime.cpp
#include "Prime.hpp"
#include <iostream>
//don't know how much memory I will need so allocate on the heap
void Prime::allocateIndicatorArray(int num)
{
try
{
primeIndicators = new bool[num];
}
catch(std::bad_alloc ba)
{
std::cout << "not enough memory :[";
//if I'm looking for a particular prime I might have over-allocated here anyway...might be worth
//decreasing num and trying again - if this is possible!
}
lastStorageSize = num;
}
void Prime::allocatePrimesArray()
{
//could probably speed up generateIndicators() if, using some prime number theory, I slightly over allocate here
//since that would cut down the operations dramatically (a small procedure done many times made smaller)
try
{
primes = new int[lastStorageSize];
}
catch(std::bad_alloc ba)
{
std::cout << "not enough memory :[";
//if I'm looking for a particular prime I might have over-allocated here anyway...might be worth
//decreasing num and trying again - if this is possible!
}
}
void Prime::generateIndicators()
{
//first identify the primes -- if we see a 0 then start flipping all elements that are multiples of i starting from i*i (these will not be prime)
int numPrimes = lastStorageSize - 2; //we'll be starting at i = 2 (so numPrimes is at least 2 less than lastStorageSize)
for(int i=4; i < lastStorageSize; i+=2)
{
primeIndicators[i]++; //dispense with all the even numbers (barring 2 - that one's prime)
numPrimes--;
}
//TODO here I'm multiple counting the same things...not cool >;[
//may cost too much to avoid this wastage unfortunately
for(int i=3; i < sqrt(double(lastStorageSize)); i+=2) //we start j at i*i hence the square root
{
if(primeIndicators[i] == 0)
{
for(int j = i*i; j < lastStorageSize; j = j+(2*i)) //note: i is prime, and we'll have already sieved any j < i*i
{
if(primeIndicators[j] == 0)
{
numPrimes--;//we are not checking each element uniquely yet :/
primeIndicators[j]=1;
}
}
}
}
noOfPrimes = numPrimes;
}
void Prime::generatePrimeList()
{
//now we go and get the primes, i.e. wherever we see zero in primeIndicators[] then populate primes with the value of i
int primesCount = 0;
for(int i=2;i<lastStorageSize; i++)
{
if(primeIndicators[i] == 0)
{
if(i%1000000 = 0)
std::cout << i << " ";
primes[primesCount]=i;
primesCount++;
}
}
}
Prime::Prime(int num)
{
noOfPrimes = 0;
allocateIndicatorArray(num);
generateIndicators();
allocatePrimesArray();
generatePrimeList();
}
int Prime::getPrime(int nthPrime)
{
if(nthPrime < lastStorageSize)
{
return primes[nthPrime-1];
}
else
{
std::cout << "insufficient primes found\n";
return -1;
}
}
int Prime::getNoOfPrimes()
{
return noOfPrimes;
}
在我阅读的同时，有人对此有任何见解吗？
编辑由于某种原因，我决定开始使用 lastStorageSize ints 而不是 noOfPrime 来更新我的素数列表！感谢 David Fischer 发现了那个！
我现在可以超过 600000000 作为上限