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c - 获取退出值：-1,073,741,571 用简单的代码计算第 n 时刻

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<1000000 的代码时，代码运行良好。但是，当我尝试输入更高的 N 值时，例如 1000000 或 100 万，代码给了我退出值 -1,073,741,571，并且不会打印出应有的时刻。我认为这与记忆有关。 

#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;
}

标签： ceclipsememoryphysicsexit-code

解决方案

在您的代码中，您在 size 堆栈上构造一个数组N：int R[N]

我怀疑这会导致足够大的 N 值导致堆栈溢出。如果将行替换int R[N]为int* R = malloc(sizeof(*R) * N);

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

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