c - 使用 malloc 时对字符串进行排序

Question

如何在不破坏程序的情况下对字符串进行排序？字符串输入必须在不使用 gets(fgets , getsc 等)的情况下完成

int main ()
{

    char **pointsarr;           
    char temp;

    printf("insert strings: \n");      
    scanf("%d",&a);
    printf("declare the size of the strings: \n");          

    pointsarr = malloc(a * sizeof(char*));         



    for (i=0; i<a; i++)
    {

        pointsarr[i]= malloc(b* sizeof(char));      
    }

    printf("input strings: \n");

    for (i=0; i<a; i++)
    {
        scanf("%s", pointsarr[i]);    
    }

    printf("the strings on the 2d board: \n");

    for (int i=0; i<a; i++)
    printf("%s\n",pointsarr[i]);

   /* int n= strlen(pointsarr);

    for (i=0; i<n-1; i++)
    {
        for(j=i+1; j<n; j++)
        {
            if (pointsarr[i]>pointsarr[j])
            {
                temp=pointsarr[i];
                temp=pointsarr[i];                      // whats going wrong on this point?
                pointsarr[i]=pointsarr[j];
                pointsarr[j]=temp;
            }
        }
    }
*/

    free(pointsarr);



}

如果出了什么问题并且它不返回排序后的字符串？

Answer 1

对数组进行排序的最简单方法是使用库函数 qsort()。

示例代码:

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

int comparefunc (const void* p1, const void* p2)
{
    return strcmp (* (const char **) p1, * (const char **) p2);
}

int main ()
{
    int i, a = 4, b = 20;
    char **pointsarr = (char **) malloc(a * sizeof(char*));

    for (i=0; i<a; i++)
        pointsarr[i]= (char *) malloc(b* sizeof(char));      

    strcpy (pointsarr [0], "qwerty");
    strcpy (pointsarr [1], "uiop");
    strcpy (pointsarr [2], "asdf");
    strcpy (pointsarr [3], "ghjkl");

    for (i=0; i<a; i++)
        printf("%s\n",pointsarr[i]);
    printf("-----\n");

    qsort (pointsarr, a, sizeof (char **), comparefunc);

    for (i=0; i<a; i++)
        printf("%s\n",pointsarr[i]);
}

输出:

qwerty
uiop
asdf
ghjkl
-----
asdf
ghjkl
qwerty
uiop

Live demo