#include <stdio.h>
#include <string.h>
void valueTwoNumber(int *a,int *b){
int temp;
temp = *a;
*a = *b;
*b = temp;
}
void valueTwoString(char **str,char **str1){
char *temp;
temp = *str;
*str = *str1;
*str1 = temp;
}
//函数:对一个数组进行排序
void sortArray(int a[],int n){
int i,j;
int temp = 0;
for(i = 0;i < n;i++){
for(j = 0;j < n-1-i;j++){
if(a[j] < a[j+1]){
temp = a[j];
a[j] = a[j+1];
a[j+1] = temp;
}
}
}
for(i = 0;i < n;i++){
printf("%d\n",a[i]);
}
}
//写函数:实现数组中能同时被2和3整除的数的和
//找出这个同时能被2和3整除的数的个数并返还
int sumArray(int a[],int n,int *count){
int i,sum = 0;
*count = 0;
for(i = 0;i < n;i++){
if(a[i] % 2==0&&a[i]%3==0){
sum +=a[i];
(*count)++;
}
}
return sum;
}
/*
构造函数要点:
1、首先需要了解函数的实现功能
2、判断函数是否有返回值
3、针对参数传递时,遵守尽量简洁方便、灵活,
*/
//构造一个函数:比较两个字符串的大小,并返还大的字符串
char *compareTwoString(char *str,char *str1){
int k;
k = strcmp(str,str1);
if(k == 0){
return str;
}else if(k < 0){
return str1;
}else{
return str;
}
}
//构造一个函数:比较两个字符串的大小,并返还大的字符串
char *compareTwoStr(char *str,char *str1){
int n;
n = strcmp(str,str1);
if(n == 0){
return str;
}else if(n < 0){
return str1;
}else {
return str;
}
}
//构造一个函数:判断一个字符在这个字符串里边?
int judgeCharInString(char *str,char ch){
int n = 0;
char *s1 = NULL;
s1 = strchr(str,ch);
if(s1 !=NULL){
n = 1;
}else{
n = 0;
}
return n;
}
int judgeStringInString(char *str,char *str1){
char *s;
s = strstr(str,str1);
if(s != NULL){
return 1;
}else{
return 0;
}
}
//构造一个指针函数,通过返还一个指针得到两个整数的和
int *backTwoNumberSum(int a,int b){
int *p;
int sum = 0;
sum = a + b;
p = ∑
return p;
}
//函数指针
/*
int (*sum)(int a,int b){
}
*/
//sum sum1;
typedef (*SumTwoNumber)(int a,int b);//SumTwoNumber是函数指针类 型名称
int sumNumber(int a,int b)
{
return a+b;
}
int chengNumber(int a,int b){
return a * b;
}
//构造一个函数,求1+2+3+---+100的和
int sumOneToThoud(int n){
if(n == 1){
return 1;
}else{
return n+sumOneToThoud(n-1);
}
}
//通过使用递归函数,求100以内能被2 3 5 同时整除的数
int valueToThous(int n){
if(n == 1){
return 0;
}
if(n % 12 == 0){
printf("%d\n",n);
}
return valueToThous(n-1);
}
//使用递归函数,求任何数的阶乘 5! = 5 *4*3*2*1=120
int jiecheng(int n){
if(n == 1)return 1;
else return n * jiecheng(n-1);
}
int backArrayBigValue(int a[],int n){
int max = a[0];
int i;
for(i = 0; i < n;i++){
if(max < a[i])
max = a[i];
}
return max;
}
//abcdef fedcba
char *stringBackString(char str[],int n){
char *p = NULL,*q = NULL;
char s;
p = str;
q = str + n-1;
while(p < q){
s = *p;
*p = *q;
*q = s;
p++;
q--;
}
printf("%s\n",str);
return str;
}
int main(int argc, char **argv)
{
int a[] = {
7,8,45,123,458,1655,87895
};
int sum = backArrayBigValue(a,sizeof(a)/sizeof(int));
printf("sum = %d\n",sum);
char str[] = "helloworld";
int n = strlen(str);
char *s;
s = stringBackString(str,n);
printf("%s\n",*s);
int a = 5;
int sum = jiecheng(20);
printf("sum = %d\n",sum);
int backSum = 0;
SumTwoNumber sum;//创建一个函数指针变量
sum = sumNumber;//为函数指针变量赋值,这个值就是地址,就是函数的名称,就是函数的入口
backSum = (*sum)(10,20);
printf("%d\n",backSum);
int sum = 0;
sum = sumOneToThoud(5);
printf("sum = %d\n",sum);*/
int n = valueToThous(100);
//数组和函数之间的关系
//数组作为形参传递
int a[] = {65,99,12,36,45,67};
int n,count,sum = 0;
n = sizeof(a)/sizeof(int);
printf("n = %d\n",n);
sum = sumArray(a,n,&count);
printf("count = %d sum = %d\n",count,sum);
char *str = "helloworld";
char *str1 = "iiiiii";
char *p;
p = compareTwoStr(str,str1);
printf("%s\n",p);
char *str = "adfasdfsadfsadf";
char ch ='d';
int n;
n = judgeCharInString(str,ch);
if(n == 1){
printf("找到了\n");
}else{
printf("没有找到\n");
}
char *str = "mynameisjack";
char *str1 = "jack";
int n;
n = judgeStringInString(str,str1);
if(n == 1){
printf("找到了\n");
}else{
printf("没有找到\n");
}
数据类型 * 函数的名称(形参){
语句块;
return 指针;
}
int a = 20;
int b = 30;
int *sum;
sum = backTwoNumberSum(a,b);
printf("sum = %d\n",*sum);
//递归函数
函数本身调用函数本身,但是总会有一个条件出口,从而结束该函数调用。
return 0;
}
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