Question

我不确定为什么我的程序无法在vi上编译。它只打印第一次出现的函数show（var），然后退出并列出一个分段错误和核心转储，但是，它在emacs上编译时没有任何错误，并在快速排序后显示所有字符串。

该程序应该读取我存储在同一目录中的文本文件中的数据，并使用2个比较函数中的一个快速排序（不必有意义，他们只需要功能正常），然后将其打印到屏幕上。

提前致谢。

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


void show(void *array[]){

    int i = 0;
    while(array[i]!=NULL){
       printf("String %d : %s\n",i, array[i]);

      i++;
    }
    printf("\n");
}

  void *readData(void * lineArray[]){

        static const char filename[] = "sampledata.txt";
        FILE *file = fopen ( filename, "r" );
        if ( file != NULL )
        {
            int i ;
            char line [ 128 ]; /* or other suitable maximum line size */

            void *lineadrs ;


            i = 0;

            lineadrs = malloc(sizeof(void) * 1024);
            while ( fgets ( lineadrs, sizeof line, file ) != NULL ) /* read a line */
            {
                lineArray[i] = lineadrs;

                lineadrs = malloc(sizeof(void) * 1024);

                i++;
            }
            fclose ( file );
        }


        else {
            perror ( filename );

            return 0;

        }
        return lineArray ;
    }



void swap(void *v[], int i, int j)
{
    void *temp;
    temp = v[i];
    v[i] = v[j];
    v[j]=temp;
}

//normal compare
int cmp1 (void *first_arg,  void *second_arg)
{

    if ( *(char*)first_arg <  *(char*)second_arg )
    {
        return -1;
    }
    if ( *(char*)first_arg == *(char*)second_arg )
    {
        return 0;
    }
    else    {
        return 1;
    }

}

//reverse the compare
int cmp2  (void * a, void * b)
{
    char *ia = (char *)a; // casting pointer types
    char *ib = (char *)b;
    return *ib  - *ia;
    //return ( *(int *)b + *(int *)a );
}


void QSort(void *v[],int left, int right, int (*compare)(void *first, void *second))
{
    int i, last;
    void swap (void *v[],int ,int);


    if(left >= right){
        return;
    }

    swap(v,left,(left+right)/2);
    last=left;
    for(i=left+1;i<=right; i++){
        if((*compare)(v[i],v[left])<0){
            swap(v,++last,i);
        }
    }

    swap(v,left,last);
    QSort(v,left,last-1,compare);
    QSort(v,last+1,right,compare);
}




int main(){
    void * var[6];
    readData(var);
    printf("Original String:\n");
    show(var);

    QSort(var,0,4,cmp1);
    printf("After cmp 1 which compares alphabetically.\n");
    show(var);

    QSort(var,0,4,cmp2);
    printf("After cmp 2 which compares reverse alphabetically.\n");
    show(var);


    return 0;



}

Answer 1

此代码中的错误列表几乎无法提及

线阵列是固定的。它应该是动态的。读取超过4行文本将通过超出输入数组长度来调用未定义的行为
字符串内容的比较符错误。
内存泄漏很多。

以下代码，我相信，您正在尝试做的事情。我真诚地希望你花时间从中吸取教训。还有几件事情应该做，但差别在于日夜都有。而且我应该警告你，我在网上写了这个并没有给它测试时间，但它应该是正确的。由于我没有您的示例数据，这是我可以做的范围。祝你好运。

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

// read data from a named file one line at a time, storing each
//  in a ever-expanding line array. The return result is the
//  number of lines allocated. The resulting line array is passed
//  as an output parameter
int readData(const char filename[], void ***results)
{
    // default answer: no lines, zero-length
    void **lines = NULL;
    int i=0;


    FILE *file = fopen ( filename, "r" );
    if ( file != NULL )
    {
        char line [ 128 ];
        while ( fgets ( line, sizeof line, file ) != NULL )
        {
            // trim the newline from line buffer
            size_t slen = strlen(line);
            if (slen > 0 && line[slen-1] == '\n')
                line[--slen] = 0;

            // resize lines array
            void **new_lines = realloc(lines, (i+1)*sizeof(*new_lines));
            if (new_lines == NULL)
            {
                perror("Failed to realloc lines array.");
                exit(EXIT_FAILURE);
            }

            // save new line entry, terminate with NULL;
            lines = new_lines;
            lines[i++] = strdup(line);
        }

        fclose ( file );
    }
    else
    {
        perror(filename);
        exit(EXIT_FAILURE);
    }

    // setup output result and return value
    *results = lines;
    return i;
}


// display an array of a specified length
void show(void *array[], int len)
{
    int i=0;
    for (; i<len; ++i)
        printf("String %d : %s\n", i, array[i]);
    printf("\n");
}

//normal compare
int cmp1 (void *first_arg,  void *second_arg)
{
    return strcmp((const char*)first_arg, (const char*)second_arg);
}

//reverse the compare
int cmp2 (void *first_arg, void *second_arg)
{
    return strcmp((const char*)second_arg, (const char*)first_arg);
}

// swap to void* by address
void swap(void **lhs, void **rhs)
{
    void *tmp = *lhs;
    *lhs = *rhs;
    *rhs = tmp;
}

// the simplest quicksort I can fathom
void QSort(void *v[], int len, int (*compare)(void*, void*))
{
    if (len < 2)
        return;

    // swap random element to last slot
    swap(v+(rand() % len), v+(len-1));

    // partition around the pivot value
    int pvt=0,i;
    for (i=0; i<len; ++i)
    {
        if (compare(v[i], v[len-1]) < 0)
            swap(v+i, v+pvt++);
    }

    // swap pivot into place
    swap(v+pvt, v+(len-1));

    // recurse. note the pivot slot is skipped.
    QSort(v, pvt++, compare);
    QSort(v+pvt, len-pvt, compare);
}


int main()
{
    static const char filename[] = "sampledata.txt";

    srand((unsigned)time(NULL));

    void **var = NULL;
    int len = readData(filename, &var);

    if (len > 0)
    {
        printf("Original String:\n");
        show(var, len);

        QSort(var, len, cmp1);
        printf("After cmp 1 which compares alphabetically.\n");
        show(var, len);

        QSort(var, len, cmp2);
        printf("After cmp 2 which compares reverse alphabetically.\n");
        show(var, len);

        // release lines when finished
        while (len-- != 0)
            free(var[len]);
        free(var);
    }

    return 0;
}

程序在VI中导致分段错误，但在emacs中工作正常

1 个答案: