itoa 和 atoi和字符串翻转等操作

#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;

}