atoi和itoa的实现

int atoi(char* in)
{
int sign=1;
if(*in == '-')
sign=-1;

int sum=0;
in++;
while(*in!='\0')
{
sum=sum*10+(*in-'0');
in++;
}
return sum*sign;
}

void itoaa(int value, char* str)
{
if(value<0)
{
str[0]='-';
value=value*-1;
}

int j=1;
for(int i=value; i>0; i/=10, j++)
{
int n=i%10;
str[j]=n+'0';
}

str[j]='\0';
j--;
int n=j/2;
for(int i=1; i<=n; i++, j--)
{
str[i]=str[i]^str[j];
str[j]=str[i]^str[j];
str[i]=str[i]^str[j];
}
}