有什么办法可以用不规则的数组来做到这一点吗?
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
/*function for use in built-in quick sort*/
static int compare(const void *x,const void *y){
return strcmp(*(const char**)x, *(const char**)y);
}
int main(){
FILE *p = fopen("file.txt","w");
char ch = '\0',**c = (char**)calloc(6,sizeof(char*));
int n[6]={0},i=0,j=0;
fprintf(p,"jack\ndanny\njohn\nrachael\nrobin\ntom");
fclose(p);
p = fopen("file.txt","r");
while(1){/*count number of char to create ragged array*/
while((ch=getc(p))!= '\n'){
if(ch == EOF) break;
putchar(ch);
n[i]++;
}
printf(" %d\n",n[i]);
if(ch == EOF) break;
ch = '\0';
i++;
}
ch = '\0';
for(i=0;i<6;i++)/*allocating memory*/
c[i] = (char*) calloc(n[i],sizeof(char));
fclose(p);
i=0;
p = fopen("file.txt","r");
while(1){/*read from file to ragged array*/
while((ch=getc(p))!= '\n'){
if(ch == EOF) break;
*(c[i]+j) = ch;
j++;
}
i++;
j = 0;
if(ch == EOF) break;
ch = '\0';
}
/*using built-in quick sort*/
qsort(*c,6,sizeof(char*),compare);/*why won't this work?*/
for(i=0;i<6;i++)
printf("%s\n",*c[i]);
return 0;
}
最佳答案
for(i=0;i<6;i++)/*allocating memory*/
c[i] = (char*) calloc(n[i],sizeof(char));
您还需要考虑每个字符串后面的 nul 终止符,这应该是
c[i] = (char*) calloc(n[i] + 1,sizeof(char));
请记住,当您读回字符串时,您需要确保它们为空 也终止了。现在不需要,因为 calloc() 将确保您的最后一个字节 string 的值为 0,但通常需要注意。
/*using built-in quick sort*/
qsort(*c,6,sizeof(char*),compare);/*why won't this work?*/
取消引用 c 并将其传递给 qsort 将是错误的,它应该是正确的
qsort(c,6,sizeof(char*),compare);
与 printf 相同, *c[i] 不是 printf %s 格式化程序所期望的 char * 。应该是
printf("%s\n",c[i]);
关于c - 尝试将不规则数组传递给库 qsort 函数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/18238190/