C++總和大數，程序停止？

Question

上午運行這兩個函數做相同的計算「總結前N個整數」，然後比較每個運行時間。該程序工作正常與小輸入，但問題是，當我輸入像1000000大數字，它計算第一個方法「iterativeSum（）」，然後一旦它到達recursiveSum（）它停止工作。

我不確定，但你認爲這可能是因爲cout？

#include <stdio.h> 
#include <iostream> 
#include <ctime> 
#include <cstdlib> 

using namespace std; 

void iterativeSum(int); 
int RecursiveSum(int); 


int main() 
{ 
long long posInt; 
std::cout << "Enter a positive integer: "; 
std::cin >> posInt; 

int start_s=clock(); 
iterativeSum(posInt); 
int stop_s=clock(); 

int start_s1=clock(); 
cout << "\nThe recursive algorithm to sum the first N integers of "<< posInt << " is: "<< RecursiveSum(posInt) << endl; 
int stop_s1=clock(); 

cout << "time: " << (stop_s-start_s)/double(CLOCKS_PER_SEC)/1000 << endl; 

cout << "time: " << (stop_s1-start_s1)/double(CLOCKS_PER_SEC)/1000 << endl; 


return 0; 
} 

void iterativeSum(int posInt) 
{ 
//positive Integer >=0 
int sum = 0; 


//loop through and get only postive integers and sum them up. 
// postcondion met at i = 0 

for(int i = 0; i <= posInt;i++) 
    { 
     sum +=i; 
    } 
    //output the positive integers to the screen 
    std::cout <<"\nThe iterative algorithm to sum the first N integers of " <<posInt <<" is: " << sum << "\n"; 
} 


int RecursiveSum(int n) 
{ 
if(n == 1) // base case 
{ 
    return 1; 
} 
else 
    { 
    return n + RecursiveSum(n - 1); //This is n + (n - 1) + (n - 2) .... 
    } 
} 

Answer 1

您可能需要一個arbitrary precision arithmetic庫像GMPlib，避免arithmetic overflows。你應該害怕stack overflow。

call stack通常是有限的（例如一兆字節）。請參閱this