my_atoi:字符串轉整形
實現:
1 ‘0’-‘9’之外的非數字字符的處理
2 int溢出問題
3 注意轉換後int型的正負問題
my_itoa:整形轉字符串
實現:
1 依次取出每個數字
2 轉化成對應字符
3 放入字符數組中
4 將數組逆置
#include <iostream>
#include <stdio.h>
using namespace std;
///////////////////字符串轉整形/////////////////
//1非數字字符
//2正負號
//3溢出問題
int my_atoi(const char *str)
{
if (str == NULL)
{
cout<<"string is null\n"<<endl;
return 0;
}
const char *s = str;//不改變源字符串
int sign = 1;//整形的正負號
long long result = 0;//用long long保存結果,防止溢出
while (*s != '\0')
{
if(*s == ' '|| *s>'9' || *s<'0')//非數字字符跳過
s++;
if (*s == '-')
sign = -1;
if (*s >='0' && *s <= '9')//字符'0'-'9'轉換成對應數字
{
result = result*10 + *s-'0';//計算結果
if (result >= 2147483647 && sign == 1)
return 2147483647;
if (result >= 2147483648 && sign == -1)
return -2147483647-1;
}
s++;
}
return result*sign;
}
//////////////整形轉字符串///////////////////
//反轉字符串
void resver_str(char *str)
{
int len = strlen(str);
char tmp = 0;
for (int i = 0; i<len/2; i++)
{
tmp = str[i];
str[i] = str[len-1-i];
str[len-1-i] = tmp;
}
}
//1 依次取出每個數字
//2 轉化成對應字符
//3 放入字符數組中
//4 將數組逆置
char buf[12]; //int型變量的範圍-2147483648~2147483647最多12個字符
void my_itoa(int num)
{
memset(buf, 0, sizeof(buf));
int FlagF = 0;//判斷‘-’字符
int i = 0;
if (num < 0)//判斷數字的正負性
{
FlagF = 1;
num = 0 - num;
}
while (num)//倒着取字符,最後要反轉字符串
{
buf[i++] = num % 10 + '0';
num = num / 10;
}
if (FlagF == 1)
{
buf[i++] = '-';
}
buf[i] = '\0';
resver_str(buf);
}
void main()
{
//char *str = "a yrgfy duc JS - 912 34";
//int result = my_atoi(str);
//printf("%d\n",result);
int num = -7654321;
my_itoa(num);
cout<<buf<<endl;
}