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  int* Bob = atoi(BobPtr.c_str());
                    ^ this str has a value i want to covert to a int*
  Any ideas on how to do this, in C, or ASM ?

  P.S. How do you guess do you EPCs?
  

I think you ment
int Bob = atoi(BobPtr.c_str());

well, you need to write a parser for it: it is VERY simple, like this:

int atoi(char *buf){
int n=0;
int i;
for(i=0;;i++){
  int a;
  if( (a=char2int(buf[i])) == -1)break;
  n = n*10+a;
}
return n;
}

int char2int(char a){
switch(a){
case '0':return 0;
....
case '9':return 9;
default: return -1;
}
}

Anton

Doesn't C have the ability to convert char to there ASCII equalent???? Then if looking at it to make the number '1' just minus the places amount of the value '0' hold in ASCII???? Just a guess though....

Benjamin

int atoi(char * z){
int x,i,n=0,h=0;
x=sizeof(z);
for(i=x-2;i>=0;i--){
h+=((int)z[i]-(int)'0') * (int)pow(10,n);
n++;
}
return h;
}
Good example of what I am talking about... Part of my old/new OS titled Nodis...

First, It will parse "123dasdad" incorectly.
Second, (int)z[i]-(int)'0' is a programming trick, i will confuse the reader who asked the question.
Third, It is not a good programming style to asume something about lowlevel architecure. For example, if it used unnicode chars, then you will have a problem.

Anton.

Quick and dirty atoi()-like example

int     matoi(const char *p)
{
        int     i, result = 0, x = 0;

        if(p[i = 0] == '-') { x++; i++; }
        while(p[i])
        {
                if((p[i] - '0') > 9)
                {
                        i++;
                        continue;
                }
                result = (result * 10) + (p[i++] - '0');
        }

        return (x ? -(result) : result);
}

No you guys don't understand my problem, look to my latest post as of 10/17/2002