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itoa 和 atoi和字符串翻转等操作

2018年05月13日 ⁄ 综合 ⁄ 共 1097字 ⁄ 字号 评论关闭
#include <iostream>
using namespace std;
#include <stdio.h>
#include <string.h>

//输入"12345",输出对应的int值
int myAtoi(const char *number)
{
    int retval = 0;
    while ( '\0' != *number )
    {
        retval = retval *10 + *number - '0';
        number ++;
    }
    return retval;
}

//递归求字符串长度
int myStrLen(const char *src)
{
    if ( *src == '\0' )
    {
        return 0;
    }
    return 1+ myStrLen (src+1);
}

//字符串翻转,根据参数判断是原地翻转还是重新拷贝翻转
void myReverse(char *src,char *dst)
{
    int srcLen = myStrLen(src);
    if ( src == dst ) //reverse in place
    {
        char *start = src;
        char *end = src + srcLen-1;
        while ( start <= end )
        {
            char tmp = *start;
            *start = *end;
            *end = tmp;
            start ++;
            end --;
        }
    }
    else
    {
        while ( srcLen )
        {
            char *end = src + srcLen - 1;
            *dst = *end;
            dst ++;
            srcLen -- ;
        }
        *dst = '\0';
    }
}

//输入int,根据对应的进制转换成相应的字符串
void myItoa( int number, char *result,int radix )
{
    if ( number < 0 )
    {
        *result = '-';
        result ++;
    }
    char *start = result;
    while ( number )
    {
        int lower_number = number % radix;
        *start = lower_number + '0';
        start ++;
        number = (number - lower_number)/radix;
    }
    myReverse(result,result);
    return ;
}

int main()
{
    cout << myAtoi ("123456")<<endl;
    char result[32] = {0};
    myItoa (54321,result,10);
    fprintf(stderr,"%s\n",result);
    memset(result,'\0',32);
    myItoa (543210,result,10);
    char buffer[32];
    myReverse("goodboy",buffer);
    fprintf(stderr,"%s\n",buffer);
    return 0;
}

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