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I completed a given code-example as a homework and it worked fine on my 64 Bit OS X device (compiled with gcc). But it won't execute on my Windows 8 64 Bit machine(compiled with MinGW - gcc). I used gcc sort.c -Wall -Wextra on both machines. Not execute means, that the programm stucks at line 64 in an infinity loop. Further I recognized that this happend after loop reaches 11 in line 64.

It also runs on Codepad (http://codepad.org/BoLhqtzv).

I tryed to use different ways of pointer arithmetics to access the arrays but none of them worked.

The programm sorts an array x of lengh n. In array x is like a bag. So there can appear one number multiple times. The function sort gets the lengh of the Array x, a pointer on the array x and the count of different numbers (10: from 0 to 9). The trick is that the Array muh knows how often which number appears in Array x. This works because the numbers in x are continous in N(atural Numbers).

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

#define M 10

/* This function generates an array of random integers in the range [0,M-1] of length n. */
int* random_array(const int n) {
  int *x;
  int i;

  x = (int*) malloc(n * sizeof(int));

  srand(time(NULL ));

  for (i = 0; i < n; i++) {
    x[i] = rand() % M;
  }

  return x;
}

/* print an array. */
void print_array(const int n, const int *x) {
  int i;

  printf("array: ");
  for (i = 0; i < n && i < 32; i++) {
    printf("%d ", x[i]);
  }
  if (n > 32) {
    printf("...");
  }
  printf("\n");
}

/* check if a given array is sorted in ascending order. */
void is_sorted(const int n, const int *x) {
  int i;

  for (i = 1; i < n; i++) {
    if (x[i - 1] > x[i]) {
      fprintf(stderr, "ERROR: Array is not sorted!\n");
      return;
    }
  }
  printf("Array is sorted!\n");
}

/* n is the length of the array x and m is the same m as on your exercise sheet.
 * In this case m is set to 10. */
void sort(const int n, int *x, int m) {
  /* allocates memory for an zero initialized array */
    int *muh = (int*) calloc(m, sizeof(int));
    int loop; //count variable

    /*counts each number in the array*/
    for(loop = 0; loop < n; loop++){
        muh[x[loop]]++;
    }

    /*Overrides x, Each number appears muh[loop] times at the beginning of line 65*/
    for(loop = 0; loop < n; loop++){
        for(; muh[loop] > 0; muh[loop]--) {
            *(x++) = loop;
       }
    }
}

int main() {
  int *x;
  int n;

  /* set length of the arrays */
  n = 1 << 5;

  /* get a random integer array of length n */
  x = random_array(n);

  /* print the unsorted array */
  print_array(n, x);

  printf("\n");
  printf("sorted array:\n");

  /* sort x by using sort, check if it is sorted and print it out */
  sort(n, x, M);
  is_sorted(n, x);
  print_array(n, x);

  return 0;
}
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2 回答 2

6

你muh有m元素的空间，但是

for(loop = 0; loop < n; loop++){
    for(; muh[loop] > 0; muh[loop]--) {
        *(x++) = loop;
   }
}

你循环通过n. 如果n > m，您有未定义的行为。

 * In this case m is set to 10. */

n = 1 << 5;

表示它咬你。

它似乎在某些系统上有效，但在其他系统上无效，嗯，未定义的行为可以为您做到这一点。

于 2013-04-26T13:59:46.600 回答
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sort(n, x, M);->sort(n, x, n);

loop < n和 muh[0]..muh[9] ( int *muh = (int*) calloc(m, sizeof(int)))

muh[loop]当循环 >= m 时超出范围

于 2013-04-26T14:11:16.800 回答