面试题目-字符串与整型互转

///////////////////////////////////////////////////////////////////////////////
//
//  FileName    :   atoi_itoa.cpp
//  Version     :   0.10
//  Author      :   Ryan Han
//  Date        :   2014/07/06 23:36
//  Comment     :  
//
///////////////////////////////////////////////////////////////////////////////
#include <stdio.h> // printf
#include <ctype.h> // isspace(), isdigit()
 
int atoi(char s[])
{
    int i, n, sign;
    //skip the space
    for(i=0; isspace(s[i]); i++)
        ;
    //store the sign
    sign = (s[i] == '-') ? -1 : 1;
    //skip the sign
    if(s[i] == '+' || s[i] == '-')
        s++;
    //calculate the number
    for(n = 0; isdigit(s[i]); i++)
        n = 10*n + (s[i] - '0');
    return sign * n;
}
 
void itoa(int n, char s[])
{
    int i, j, sign;
    //store the sign, convert to positive num
    if((sign = n ) < 0)
        n = -n;
    //store the number in reverse order
    i = 0;
    do{
        s[i++] = n%10 + '0';
    }while((n/=10) > 0);
    //store the sign
    if(sign < 0)
        s[i++] = '-';
    //put the end sign
    s[i] = '\0';
    //output in reverse order
    for(j = i; j >= 0; j--)
        printf("%c", s[j]);
}
 
int main()
{
    int i = 0;
    char s[] = "1234";
    i = atoi(s);
    printf("The atoi test: \n");
    printf("The atoi of \"1234\" is: %d", i);
 
    printf("\n");
    printf("The itoa test: \n");
    int j = -5678;
    char t[10];
    itoa(j, t);
 
    return 0;
}