1.回調(diào)函數(shù)是什么？

回調(diào)函數(shù)就是?個通過函數(shù)指針調(diào)?的函數(shù)。
如果你把函數(shù)的指針（地址）作為參數(shù)傳遞給另?個函數(shù)，當(dāng)這個指針被?來調(diào)?其所指向的函數(shù)
時，被調(diào)?的函數(shù)就是回調(diào)函數(shù)?；卣{(diào)函數(shù)不是由該函數(shù)的實現(xiàn)?直接調(diào)?，?是在特定的事件或條
件發(fā)?時由另外的??調(diào)?的，?于對該事件或條件進?響應(yīng)

#include <stdio.h>
int Add(int x, int y)
{return x + y;
}
int Sub(int x, int y)
{return x - y;
}
int Mul(int x, int y)
{return x * y;
}
int Div(int x, int y)
{return x / y;
}
void cac(int (*pf)(int ,int) )//用所指向函數(shù)形式接收
{   int m, n;scanf("%d %d", &m, &n);int c=pf(m,n);printf("%d\n", c);
}
int main()
{int input;do {printf("****1.加法****\n");printf("****2.減法****\n");printf("****3.乘法****\n");printf("****4.除法****\n");printf("****0.退出****\n");scanf("%d", &input);switch (input){case 1:cac(Add);break;case 2:cac(Sub);break;case 3:cac(Mul);break;case 4:cac(Div);break;case 0:break;default:printf("輸入錯誤，請重新輸入");break;}} while (input);return 0;
}

2.qsort使用舉例

2.1使用qsort函數(shù)排序整形數(shù)據(jù)

#include <stdio.h>
int paixu(const void* e1, const void* e2)
{return *(int*)e1 - *(int*)e2;
}
int main()
{   int arr[] = { 1,2,5,4,5,7,8,9,6,3 };qsort(arr, sizeof(arr) / sizeof(arr[0]), sizeof(arr[0]), paixu);int i = 0;for (i = 0;i < sizeof(arr)/sizeof(arr[0]);i++){printf("%d ", arr[i]);}return 0;
}

2.2使用sqort排序結(jié)構(gòu)數(shù)據(jù)

#include <stdio.h>
#include <stdlib.h>
struct stu
{char name[20];int age;
};
int pum1(const void* e1, const void* e2)
{return strcmp(((struct stu*)e1)->name, ((struct stu*)e2)->name);
}
int main()
{   struct stu arr[3] = { {"zhangsan",15},{"lisi",19},{"wangwu",20} };qsort(arr, sizeof(arr) / sizeof(arr[0]), sizeof(arr[0]), pum1);}

strcpm比較倆字符串大小，對應(yīng)位置字符的ASCLL值。
按照年齡來排序，想從大到小排只需改變return兩邊的位置

#include <stdio.h>
#include <stdlib.h>
struct stu
{char name[20];int age;
};int pum2(const void* e1, const void* e2)
{return (*(struct stu*)e1).age - (*(struct stu*)e2).age;
}
int main()
{   struct stu arr[3] = { {"zhangsan",15},{"lisi",30},{"wangwu",20} };qsort(arr, sizeof(arr) / sizeof(arr[0]), sizeof(arr[0]), pum2);}

3.qsort函數(shù)的模擬實現(xiàn)

#include <stdio.h>
void parr(const void*e1,const void*e2)
{return *(int*)e1 - *(int*)e2;
}
void swap( char* t1, char* t2,size_t width)
{int tmp=0;for (int i = 0;i < width;i++){tmp = *t1;*t1 = *t2;*t2 = tmp;t1++;t2++;}}
void bubble(void* base, size_t sz,size_t width, int (*pa)(const void* e1,const void* e2))
{int i = 0;for (i = 0;i < sz-1;i++){for(int j=0;j<sz-1-i;j++){if ((pa((char*)base + j * width, (char*)base + (j + 1) * width))>0){swap((char*)base +j*width,(char*)base +(j+1)*width,width);}}}
}
int main()
{   int arr[] = { 1,9,8,5,4,6,7,2,3,0 };bubble (arr, sizeof(arr) / sizeof(arr[0]), sizeof(arr[0]), parr);for (int i = 0;i < sizeof(arr) / sizeof(arr[0]);i++){printf("%d ", arr[i]);}return 0;
}