lizhongxu2008

#include<stdio.h>
int atoi(const char *str)
{
    int value=0;
    bool b_plus=true; //判斷符號

    switch(*str) //過濾符號

    {
    case '+':
        str++;
        break;
    case '-':
        b_plus=false;
        str++;
        break;
    default:
        break;
    }

    while('\0' != *str)
    {
        value = (value*10)+(*str-'0');
        str++;
    }
    if(!b_plus)
        value=-value;
    return value;
}

void main()
{
    char str[]="-999";
    int value=atoi(str);
    printf("The result is:%d\n",value);
}


#include<stdio.h>
void itoa(int value, char *str)
{
    if (value < 0) //如果是負(fù)數(shù),則str[0]='-',并把value取反(變成正整數(shù))

    {
        str[0] = '-';
        value = 0-value;
    }
    int i,j;
    for(i=1; value > 0; i++,value/=10) //從value[1]開始存放value的數(shù)字字符，不過是逆序，等下再反序過來

        str[i] = value%10+'0'; //將數(shù)字加上0的ASCII值(即'0')就得到該數(shù)字的ASCII值

    for(j=i-1,i=1; j-i>=1; j--,i++) //將數(shù)字字符反序存放

    {
        str[i] = str[i]^str[j];
        str[j] = str[i]^str[j];
        str[i] = str[i]^str[j];
    }
    if(str[0] != '-') //如果不是負(fù)數(shù)，則需要把數(shù)字字符下標(biāo)左移一位，即減1

    {
        for(i=0; str[i+1]!='\0'; i++)
            str[i] = str[i+1];
        str[i] = '\0';
    }
}

void main()
{
    int value = -1212345;
    char str[10] = {'\0'}; //記得把str全填充為'\0'

    itoa(value, str);
    printf("The result is:%s\n", str);
}