获取退出值：-1,073,741,571 用简单的代码计算第 n 时刻

Posted 2023-02-22

【中文标题】获取退出值：-1,073,741,571 用简单的代码计算第 n 时刻

【英文标题】：Getting exit value: -1,073,741,571 with simple code to calculate nth moment

【发布时间】：2020-05-03 20:46:30

【问题描述】：

我在计算随机数数组的第 n 时刻（如质心，将是第 1 时刻）时遇到问题。我在eclipse中用C编码，当我尝试用gcc编译时也会出现这个错误。当我运行 N

#include <stdio.h>
#include <math.h>
#include <stdlib.h>
//here we import the three libraries
//we start our main function
int main ()

    setbuf(stdout, NULL);
    //setbuf disables buffering so print statements print properly
    int i,N;

    unsigned int seed;
    double first,second,third,fourth,fifth,sixth,firsttot,secondtot,thirdtot,fourthtot,fifthtot,sixthtot;
    //here we declare the vars to be used
    printf("\nEnter number of iterations and seed");
    printf("\n");
    scanf("%i %u", &N, &seed);
    srand(seed);
    //asks user for input, scans the input, and takes the seed to set a starting point for the rand() function called in the for loop
    //since my R array depends on the user, i declare the array here, after the user inputs the size of the array
    double R[N];
    for (i=0;i<N;i=i+1)
    
        R[i]=(double)rand()/RAND_MAX;
        //printf("%12.8lf \n",R[i]);
    

    //the for loop sets R equal to a random value using our seed with (double)rand()
    printf("\n");
    //here, we have for loops to add up the individual nth moments for each point of the array
    firsttot = 0.0;
    for (i=0;i<N;i=i+1)
    
        firsttot = firsttot + pow(R[i],1);
    
    secondtot = 0.0;
    for (i=0;i<N;i=i+1)
    
        secondtot = secondtot + pow(R[i],2);
    
    thirdtot = 0.0;
    for (i=0;i<N;i=i+1)
    
        thirdtot= thirdtot + pow(R[i],3);
    
    fourthtot = 0.0;
    for (i=0;i<N;i=i+1)
    
        fourthtot = fourthtot + pow(R[i],4);
    

    fifthtot = 0.0;
    for (i=0;i<N;i=i+1)
    
        fifthtot = fifthtot + pow(R[i],5);
    
    sixthtot = 0.0;
    for (i=0;i<N;i=i+1)
    
        sixthtot = sixthtot + pow(R[i],6);
    

    //now, we take the actual nth moment by dividing each total by N;
    first = firsttot/N;
    second = secondtot/N;
    third = thirdtot/N;
    fourth = fourthtot/N;
    fifth = fifthtot/N;
    sixth = sixthtot/N;
    printf("\nThe first moment is:   %lf",first);
    printf("\nThe second moment is:   %lf",second);
    printf("\nThe third moment is:   %lf",third);
    printf("\nThe fourth moment is:   %lf",fourth);
    printf("\nThe fifth moment is:   %lf",fifth);
    printf("\nThe sixth moment is:   %lf",sixth);
    return 0;

【问题讨论】：

整数溢出导致负值

Rustin 不是说 1000000（100 万）吗？您发布的内容是 10000000（1000 万）。无论哪种方式，一个 32 位数字不会溢出 1000 万（32 位整数可以存储 40 亿+）

虽然我怀疑这可能与您建议的溢出有关。

其实我只是注意到R数组是一个基于N值的变长数组。我可以看到，为了足够高的 N 值而炸毁堆栈

有没有办法让它与一个数组一起工作以获取较大的 N 值？我需要能够用 N = 1,000,000 进行计算，它只适用于 N 的值在 10,000 左右

【参考方案1】：

在您的代码中，您在大小为 N:int R[N] 的堆栈上构造一个数组

我怀疑这会导致足够大的 N 值的堆栈溢出。 如果您将行 int R[N] 替换为 int* R = malloc(sizeof(*R) * N);，您能否查看该行为是否重现

使用 malloc 将堆分配您的 R 数组，而不是使用堆栈分配，这将避免可能的堆栈溢出

【讨论】：

成功了！感谢您的帮助，我不知道 malloc，但这绝对解决了问题。