Question

我现在面临一个问题，似乎无法解决它。每当我启动代码时，我都会遇到分段错误。我的目标是掷两个骰子。该值由随机数生成。我想滚动10000次并将值保存在一个数组中，所以我可以在最后创建一个小图表，显示值。我感谢任何有关如何解决问题的帮助和提示。这是我的代码：

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

#define DICE      2
#define DICEEYES  (DICE * 6) -DICE +1
#define COUNT     10000

int diceRoll(int dice);

int main(void)
{
  srand(time(0));
  int valuesOfRoll[DICEEYES] = { 0 };
  for(int i = 0; i < COUNT; i++)
  {
    int index = diceRoll(DICE) - DICE;
    valuesOfRoll[index]++;
  }
  for(int i = 0; i < DICEEYES; i++)
  {
    if(valuesOfRoll[i] < 1) continue;
    printf("The number %2d was rolled %4d times\r\n", i + DICE, valuesOfRoll[i]);
  }
  return 0;
}

int diceRoll(int dice)
{
  int sum;
  for(int i = 0; i < dice; i++)
  {
    sum += rand() % 6 + 1;
  }
  return sum;
}

Answer 1

您应该在函数diceRoll（），

中初始化sum

/* always return value between [dice..dice*6]*/
int diceRoll(int dice)
{
  int sum=0;
  for(int i = 0; i < dice; i++)
  {
    sum += rand() % 6 + 1;
  }
  return sum;
}

由于sum可以具有任何值（在堆栈上），因此您使用的索引值可能超出ValuesOfRoll []数组。要避免的一种方法是使用模数运算符将返回的索引限制为有效范围。

int main(void)
{
  srand(time(0));
  int valuesOfRoll[DICEEYES] = { 0 };
  for(int i = 0; i < COUNT; i++)
  {
    int index = diceRoll(DICE) - DICE;
    /* limit index to [0..DICEEYES] */
    index %= DICEEYES;
    valuesOfRoll[index]++;
  }
  for(int i = 0; i < DICEEYES; i++)
  {
    if(valuesOfRoll[i] < 1) continue;
    printf("The number %d was rolled %4d times", i + DICE, valuesOfRoll[i]);
  }
  return 0;
}

Answer 2

初始化函数中sum的值（最好为0），否则它将保留一些垃圾值。

int diceRoll(int dice)
{
 int sum;
 for(int i = 0; i < dice; i++)
  {
   sum += rand() % 6 + 1;
 }
 return sum;
}

垃圾值可能是一个非常大的数字，因此函数的返回值也是如此。因此，当在此行中使用返回值时：

int index = diceRoll(DICE) - DICE;

然后数组将有一个大数字的索引valuesOfRoll[some_large_integer]++;。这会导致数组外的情况，并导致seg-fault。

Answer 3

稍微迟了一点，故障没有初始化sum和DICEEYES。

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

#define DICE      2
#define DICEEYES  (DICE * 6)
#define COUNT     10000

int diceRoll(int dice)
{
    int i, sum = 0;                         // need to initialise!
    for(i = 0; i < dice; i++)
        sum += rand() % 6 + 1;
    return sum;
}

int main(void)
{
    int i;
    int valuesOfRoll[DICEEYS+1] = { 0 };     // size of array
    srand((unsigned)time(0));

    for(i = 0; i < COUNT; i++)
        valuesOfRoll[diceRoll(DICE)]++;

    for(i = DICE; i <= DICEEYES; i++)
        printf("The number %2d was rolled %4d times\n", i, valuesOfRoll[i]);
    return 0;
}

节目输出：

The number  2 was rolled  264 times
The number  3 was rolled  573 times
The number  4 was rolled  798 times
The number  5 was rolled 1145 times
The number  6 was rolled 1373 times
The number  7 was rolled 1697 times
The number  8 was rolled 1385 times
The number  9 was rolled 1143 times
The number 10 was rolled  835 times
The number 11 was rolled  528 times
The number 12 was rolled  259 times

Answer 4

以下代码

Get-ADComputer -Filter * | ? { $members -notcontains $_.Name }

创建一个骰子图表

4 个答案: