函数的模拟实现

「爱情、让人受尽委屈。」 2023-10-14 14:54 42阅读 0赞

## ![3b4b7ed707794a3fbc1317bc2d3dc1bf.png][]题一： ##

### 模拟实现strncpy ###

#include <stdio.h>
    
    
    void my_strncpy(char* arr2, char* arr1, size_t num)
    
    {
    
    	int i = 0;
    
    	for (i = 0; i < num; i++)
    
    	{
    
    		*(arr2 + i) = *(arr1 + i);
    
    	}
    
    }
    
    int main()
    
    {
    
    	char arr1[] = "hello liangzai";
    
    	char arr2[10] = { 0 };
    
    	//strncpy(arr2,arr1,sizeof(arr2));
    
    	my_strncpy(arr2, arr1, sizeof(arr2));
    
    
    	return 0;
    
    }

### 优解一： ###

char * mystrncpy(char * dst, const char * src, size_t n)
    {
    	int i;
      for (i = 0; src[i] && i < n; i++)
      {
        dst[i] = src[i];
      }
       
      if (i < n)
      {
      	dst[i] = 0;
      }
      return dst;
    }

## 题二： ##

### 模拟实现strncat ###

#include <stdio.h>
    
    #include <string.h>
    
    
    char* my_strncat(char* arr1, const char* arr2, size_t num)
    
    {
    
    	int sz = strlen(arr1);
    
    	int i = 0;
    
    	for (i = 0; i < num; i++)
    
    	{
    
    		*(arr1 + sz + i) = *(arr2 + i);
    
    	}
    
    	return arr1;
    
    }
    
    
    int main()
    
    {
    
    	char arr2[] = "shawanyi ";
    
    	char arr1[20] = "hello ";
    
    	//strncat(arr1,arr2,10);
    
    	my_strncat(arr1, arr2, 10);
    
    
    	return 0;
    
    }

### 优解一： ###

char * mystrncat(char * dst, const char * src, size_t n)
    {
    	char * tmp = dst;
       
      while (*dst)
      {
        dst++;//记录到dst最后一个位置
      }
       
      int i;
      for (i = 0; src[i] && i < n; i++)
      {
        dst[i] = src[i];
      }
       
      dst[i] = 0;
      return tmp;
    }

## 题三： ##

### 模拟实现atoi ###

#include <stdio.h>
    
    #include <stdlib.h>
    
    #include <assert.h>
    
    #include <ctype.h>
    
    
    
    int my_atoi(const char* arr1)
    
    {
    
    	assert(arr1);
    
    	int sum = 0;
    
    	if (*(arr1) == '\0')
    
    	{
    
    		return 0;
    
    	}
    
    	while(isspace(*arr1))
    
    	{
    
    		arr1++;
    
    	}
    
    	int flat = 1;
    
    	if (*arr1 == '+')
    
    	{
    
    		arr1++;
    
    	}
    
    	else if (*arr1 == '-')
    
    	{
    
    		flat = -1;
    
    		arr1++;
    
    	}
    
    	int ret = 0;
    
    	while (isdigit(*arr1))
    
    	{
    
    		ret = ret * 10 + (*arr1 - '0');
    
    		if (ret > INT_MAX)
    
    		{
    
    			return INT_MAX * flat;
    
    		}
    
    		else if (ret < INT_MIN)
    
    		{
    
    			return INT_MIN * flat;
    
    		}
    
    		arr1++;
    
    	}
    
    	if (isdigit(*arr1) == 0)
    
    	{
    
    		return ret * flat;
    
    	}
    
    	return ret*flat;
    
    }
    
    
    int main()
    
    {
    
    	char arr1[] = "-123666a44";
    
    	//int n = atoi(arr1);
    
    	int sz = 0;
    
    	sz = my_atoi(arr1);
    
    	return 0;
    
    }

### 优解一 ###

#include <stdio.h>
    #include <stdlib.h>
    #include <ctype.h>
    //VALID 表示结果合法
    //INVALID 表示结果非法
    enum State
    {
     VALID,//0
     INVALID//1
    };
    //默认结果可能非法，当正确转换后再改为合法
    enum State state = INVALID;
    
    int my_atoi(char *str)
    {
     int flag = 1;
     long long ret = 0;
     assert(str);
     state = INVALID;
      //跳过空白字符
     while(isspace(*str))
     {
      str++;
     }
     if(*str == '\0')
     {
      return 0;
     }
      //跳过正负号
     if(*str == '+')
     {
      str++;
     }
     else if(*str == '-')
     {
      flag = -1;
      str++;
     }
      //开始转换数字字符直到非数字字符
     while(isdigit(*str))
     {
      ret = ret * 10 + flag * (*str-'0');
      if((ret > INT_MAX) || (ret < INT_MIN))
      {
       return 0;
      }
      str++;
     }
      //正常停止
     if(*str == '\0')
     {
      state = VALID;
      return (int)ret;
     }
     else
     {
        //遇到非数字字符
      return (int)ret;
     }

[3b4b7ed707794a3fbc1317bc2d3dc1bf.png]: https://img-blog.csdnimg.cn/3b4b7ed707794a3fbc1317bc2d3dc1bf.png