CareerCup

int implementAtoI(const string channelNum)
{
int b = 0;
for (int i = 0; i < channelNum.size(); i++)
{
if((int)(channelNum[i])>47 && (int)(channelNum[i]< 58))
{
do
{
int a = 0;
a = (int)channelNum[i]-48;
b = b*10 + a;
i++;
}while (i <= channelNum.size() && (channelNum[i] > 47 && channelNum[i] < 58));

return b;
}
else
break;
}

cout << endl << endl << "reutrning zero " << endl << endl;
return 0;
}

Have that!

int myshiteatoi(char *x)
{
int final = 0;

while(true)
{
if(*x >= '0' && *x <= '9')
{
final *= 10;
final += *x - '0';
x++;
}
else
break;
}

return final;
}

Sorry incomplete code :)
long atoi(const char* str){
const char* p = str;
int sign;

// fail if str is null or length is 0

// get sign if the first char is sign char

// while loop handle each following char
// 1. cut string after incorrect char
// 2. handle overflow case

// return num*sign;

Have to consider special cases
1. parameter may be NULL
2. there might be + or - sign in the beginning of the string
3. each char should be between '0' and '9'
4. may need to use linked list for int overflow
5. parameter char may not have NULL character in it

and what else?