C 分段错误（核心转储）函数

Question

我相信这与我使用roll_die 有关。如果我只是在正文中使用 rand()%6+1 而不是 roll_die 它可以工作，但我想使用该功能。另外，有什么方法可以使用更少的行来打印结果以使我的代码更短？

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

int roll_die() {
    return 1 + rand() % 6;
}

int main()
{

srand((unsigned) time(0));

int roll_die;
int first_die_roll, second_die_roll, total;
int roll_result[13] = {0};
int count = 36000; 
while(count--) {

     first_die_roll = roll_die;
     second_die_roll = roll_die;
     total = first_die_roll + second_die_roll;

    roll_result[total]++;

   }

   printf("Computed die rolling frequencies for 36,000 rolls:\n");

   printf("Sum = 2; Frequency = %d ; Percentage = %f%c\n", roll_result[2], (float)roll_result[2]/36000 * 100, '%');

   printf("Sum = 3; Frequency = %d ; Percentage = %f%c\n", roll_result[3], (float)roll_result[3]/36000 * 100, '%');

   printf("Sum = 4; Frequency = %d ; Percentage = %f%c\n", roll_result[4], (float)roll_result[4]/36000 * 100, '%');

   printf("Sum = 5; Frequency = %d ; Percentage = %f%c\n", roll_result[5], (float)roll_result[5]/36000 * 100, '%');

   printf("Sum = 6; Frequency = %d ; Percentage = %f%c\n", roll_result[6], (float)roll_result[6]/36000 * 100, '%');

   printf("Sum = 7; Frequency = %d ; Percentage = %f%c\n", roll_result[7], (float)roll_result[7]/36000 * 100, '%');

   printf("Sum = 8; Frequency = %d ; Percentage = %f%c\n", roll_result[8], (float)roll_result[8]/36000 * 100, '%');

   printf("Sum = 9; Frequency = %d ; Percentage = %f%c\n", roll_result[9], (float)roll_result[9]/36000 * 100, '%');

   printf("Sum = 10; Frequency = %d ; Percentage = %f%c\n", roll_result[10], (float)roll_result[10]/36000 * 100, '%');

   printf("Sum = 11; Frequency = %d ; Percentage = %f%c\n", roll_result[11], (float)roll_result[11]/36000 * 100, '%');

   printf("Sum = 12; Frequency = %d ; Percentage = %f%c\n", roll_result[12], (float)roll_result[12]/36000 * 100, '%');

   return 0;

}

Answer 1

您在main 中的声明int roll_die; 将roll_die 声明为本地 int 变量，并隐藏（或shadows）同名函数。因此，当您以后有这样的一行时：

    first_die_roll = roll_die; // Copying a local variable

您将 unitialized 局部变量的值分配给 first_die_roll。因此，total 的计算值未定义，对 roll_result[total] 数组元素的引用很可能“超出范围”，从而导致内存访问错误（=“分段错误”）。

修复：一是去掉局部变量的声明；然后，要调用 roll_die 函数，您需要在名称中添加括号，无论您需要它：

//  int roll_die; // Remove this line - it HIDES your "roll_die" function!
//...

    first_die_roll = roll_die(); // Brackets make this a function call.
    second_die_roll = roll_die();

请随时要求进一步澄清和/或解释。

Answer 2

您已经声明了一个局部变量：int roll_die，它掩盖了 函数 roll_die 的定义。

此外，您可以不仅通过提供函数名称来调用函数[或任何函数]。当你只是指定函数时，这是函数本身的地址，不是函数的调用/调用。

因此，而不是：

first_die_roll = roll_die;

你想要：

first_die_roll = roll_die();

而且，您可以使用for 循环进行打印。此外，您可以使用 %% 的格式添加文字 %。

无论如何，这是您的代码已修复并带有错误注释：

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

#define ROLLMAX     13

int
roll_die(void)
{
    return 1 + rand() % 6;
}

int
main()
{

    srand((unsigned) time(0));

// NOTE/BUG: this shadows/masks the actual function
#if 0
    int roll_die;
#endif

    int first_die_roll, second_die_roll, total;
    int roll_result[ROLLMAX] = { 0 };
    int count = 36000;

    while (count--) {
// NOTE/BUG: this does _not_ call roll_die -- it merely sets the values to the
// address of the roll_die function
#if 0
        first_die_roll = roll_die;
        second_die_roll = roll_die;
#else
        first_die_roll = roll_die();
        second_die_roll = roll_die();
#endif
        total = first_die_roll + second_die_roll;

        roll_result[total]++;
    }

    printf("Computed die rolling frequencies for 36,000 rolls:\n");

    for (total = 2;  total < ROLLMAX;  ++total) {
        int result = roll_result[total];
        printf("Sum = %d; Frequency = %d ; Percentage = %f%%\n",
            total, result, (float) result / 36000 * 100);
    }

    return 0;
}