比较函数的原型在原型中指定qsort
功能:
void qsort(void *base, size_t nmemb, size_t size,
int (*compar)(const void *, const void *));
您的比较函数必须兼容,因此您可以这样定义它:
int cmp(const void *a, const void *b) {
/* a and b are pointers to the array of pointers. */
int *aa = *(int * const *)a;
int *bb = *(int * const *)b;
return (aa[1] > bb[1]) - (aa[1] < bb[1]);
}
或者没有强制转换:
int cmp(const void *a, const void *b) {
/* a and b are pointers to the array of pointers. */
int * const *aa = a;
int * const *bb = b;
int ia = (*aa)[1];
int ib = (*bb)[1];
return (ia > ib) - (ia < ib);
}
请注意,您不能使用简单的比较aa[1] - bb[1]
因为它可能会溢出较大的值,并且最多会产生不正确的输出。
此外,您输入的循环不正确:您应该以相反的顺序嵌套循环:
for (int k = 0; k < n; k++) {
for (int j = 0; j < 2; j++) {
scanf("%d", &arr[k][j]);
}
}
这是修改后的版本:
#include <stdio.h>
#include <stdlib.h>
int cmp(const void *a, const void *b) {
/* a and b are pointers to the array of pointers. */
int * const *aa = a;
int * const *bb = b;
int ia = (*aa)[1];
int ib = (*bb)[1];
return (ia > ib) - (ia < ib);
}
int main() {
int t; // test cases
scanf("%d", &t);
for (int i = 0; i < t; i++) {
int n;
scanf("%d", &n); // size of array
int **arr = malloc(sizeof(*arr) * n);
for (int j = 0; j < n; j++) {
arr[j] = malloc(sizeof(*arr[j]) * 2);
}
for (int k = 0; k < n; k++) {
for (int j = 0; j < 2; j++) {
scanf("%d", &arr[k][j]);
}
}
qsort(arr, n, sizeof(arr[0]), cmp);
for (int k = 0; k < n; k++) {
for(int j = 0; j < 2; j++) {
printf("%d\t", arr[k][j]);
}
printf("\n");
}
for (int j = 0; j < n; j++) {
free(arr[j]);
}
free(arr);
}
return 0;
}
您还可以使用实际的二维数组来简化代码:
#include <stdio.h>
#include <stdlib.h>
int cmp(const void *a, const void *b) {
const int *aa = a;
const int *bb = b;
return (aa[1] > bb[1]) - (aa[1] < bb[1]);
}
int main() {
int t; // test cases
scanf("%d", &t);
for (int i = 0; i < t; i++) {
int n;
scanf("%d", &n); // size of array
int (*arr)[2] = malloc(sizeof(*arr) * n);
for (int k = 0; k < n; k++) {
scanf("%d%d", &arr[k][0], &arr[k][1]);
}
qsort(arr, n, sizeof(arr[0]), cmp);
for (int k = 0; k < n; k++) {
printf("%d\t%d\n", arr[k][0], arr[k][1]);
}
free(arr);
}
return 0;
}