BleepingComputer.com: String to Int

How do I get this function to work with a single number?

int GetIntVal(string strConvert, int location) { 
  int intReturn; 
 
  intReturn = atoi(strConvert.c_str()); 

  return(intReturn); 
}

I don't want it to read the whole string I just want it to read one part of the string

I am building an equation solver

string equation;
	 cout << "Enter your equation, one variable (x(lowercase)), no exponents: ";
	 cin >> equation;
	 for(int i = 0; i < equation.size(); ++i)
	 {
			 if(equation[i] == 'x')
			 {
						  int coef = GetIntVal(equation, i - 1);
						  cout << coef;
			 }
	 }

ANd I want it to read the coefficient right before the x but when I change the atoi function, it says c_str is not defined (I have no idea what the function does though)

This post has been edited by sausage: 22 March 2009 - 04:39 PM

Accidental Double Post deleted

This post has been edited by sausage: 22 March 2009 - 04:40 PM

Hello

Here's one way to approach it:

Make a string with just the one number:
char temp[2];
copy the single number you want in there:
temp[0] = strConvert[location];
Null terminate:
temp[1] = NULL;
Call atoi:

return atoi(temp);

Billy3

Quote

Which function don't you understand, atoi or c_str?

Billy3

Billy O, on Mar 22 2009, 04:59 PM, said:

both.......

Alright... what is a string typically in C (not C++)?

Billy3

its an array of characters... why?

Because that character array is what std::basic_string returns when you call c_str(). It's a way of getting a C-style string out of a std::basic_string object. (Note that std::string is std::basic_string<char>. std::wstring is std::basic_string<wchar_t> -- since it's a template class it works the same way for both plain characters and wide characters)

atoi turns a character array which is a string of numbers, and converts it to an integer representation. http://www.cplusplus.com/reference/clibrar...tdlib/atoi.html

If you can use the boost conversions library, you can accomplish the same thing without messing with c-style character arrays at all using lexical_cast. For example:

return boost::lexical_cast<unsigned int, std::string>(strConvert.substr(location,1));

Billy3

Ok... the top part doesn't make much sense but I understand atoi now

but how would I right an atoi function using an exact position inside the string?

but how would I right an atoi function using an exact position inside the string?

Instead of pointing atoi at the existing string, which wont do what you want, create a new string containing the number you want.

Billy3

yeah that works, thanks :-)