我尝试按char对名为results的struct run数组进行排序,但当我打印该数组时,没有排序。看看这个:
struct run {
char name[20], weekday[4], month[10];
(And some more...)
};
typedef struct run run;
int name_compare(const void *a, const void *b)
{
run *run1 = *(run **)a;
run *run2 = *(run **)b;
return strcmp(run1->name, run2->name);
}
int count_number_of_different_persons(run results[])
{
int i = 0;
qsort(results, sizeof(results) / sizeof(run), sizeof(run), name_compare);
for(i = 0; i <= 999; i++)
{
printf("%s\n", results[i].name);
}
// not done with this function yet, just return 0
return 0;
}上面的输出只是按最初放置顺序排列的名称列表
发布于 2011-12-02 17:57:17
int count_number_of_different_persons(run results[])这并不能真正让你在数组上使用sizeof,因为数组会衰减成指针。
这
run *run1 = *(run **)a;看起来也很奇怪,不是吗?
run *run1 = (run*)a;发布于 2011-12-02 17:57:52
一个问题是在name_compare中。试着这样做:
int name_compare(const void *a, const void *b)
{
run *run1 = (run *)a;
run *run2 = (run *)b;
return strcmp(run1->name, run2->name);
}发布于 2011-12-02 19:58:13
检查以下代码:
正如@michel提到的,sizeof( array )提供指针的大小,而不是数组本身的大小,因为当传递array时,它被视为指针。因此,要么将元素数量发送到函数count_number_of_different_persons,要么定义一个元素数量的宏。希望这能有所帮助。:)。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define NOE 3
struct run
{
char name[20];
};
typedef struct run run;
int name_compare (const void *a, const void *b )
{
return strcmp (((run *)a)->name, ((run *)b)->name);
}
int count_number_of_different_persons(run results[], int noOfElements)
{
int i=0;
qsort(results, noOfElements, sizeof (run), name_compare);
for (i=0; i<noOfElements; i++)
printf ("%s\n",results[i].name);
}
int main ( int argc, char * argv[])
{
run a, b, c;
run arg[NOE];
strcpy (a.name, "love");
strcpy (b.name, "you");
strcpy (c.name, "i");
arg[0] = a;
arg[1] = b;
arg[2] = c;
count_number_of_different_persons(arg, sizeof(arg)/sizeof(run));
};https://stackoverflow.com/questions/8354502
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