Overloads cannot differ by return type alone. The only way to achieve this is to use output arguments. Since the implementation is exactly the same regardless of the output type, you can use a function template to generate the overloads.
#include<iostream>
#include<sstream>
template<typename T>
bool convert(std::string& s, T& value)
{
std::stringstream ss;
ss << s;
if (ss >> value)
return true;
return false;
}
int main()
{
int i;
float f;
std::string s {"3.14"};
if (convert (s, i))
std::cout << '"' << s << "" = " << i << std::endl;
if (convert (s, f))
std::cout << '"' << s << "" = " << f << std::endl;
}
Output:
"3.14" = 3
"3.14" = 3.14
[ string toupper $str ] or [ string tolower $str ]
This is not a question.
One way to do this is to convert the number to a String, then use the corresponding String method to find out the length of the String.
Write a loop, in which you divide by each of the digits; if it's NOT divisible, set the variable "result", which you initiall set to "true", equal to "false". The tricky part is extracting the digits; one way to do this is to convert the number to a string, and use string functions. You can also repeatedly divide by 10, and use the remainder. That is, first take the remainder and test it; then divide the remaining number by 10.
std::string input = ""; std::getline (std::cin, input); // get input from stdin std::stringstream ss (input); // place input in a string stream integer num = 0; if (ss >> num) // extract integer from string stream { // Success! } else { // Fail! }
[ string toupper $str ] or [ string tolower $str ]
This is not a question.
One way to do this is to convert the number to a String, then use the corresponding String method to find out the length of the String.
You would use the ToUpper and ToLower functions. string name = "XVengeance"; string upperName = name.ToUpper(); string lowerName = name.ToLower(); Console.WriteLine(upperName); Console.WriteLine(lowerName); Console.ReadLine(); I don't think I'm supposed to do your homework for you, but this code should get you started on making a more dynamic program.
Write a loop, in which you divide by each of the digits; if it's NOT divisible, set the variable "result", which you initiall set to "true", equal to "false". The tricky part is extracting the digits; one way to do this is to convert the number to a string, and use string functions. You can also repeatedly divide by 10, and use the remainder. That is, first take the remainder and test it; then divide the remaining number by 10.
Use the tolower() function. Example: char* a = 'X'; a = tolower(a); printf("%c", a);
Copy and reverse the string. If the reversed string is equal to the original string, the string is a palindrome, otherwise it is not. When working with strings that hold natural language phrases (including punctuation, whitespace and so on) we must remove all the non-alphanumerics and convert the remainder to a common case, such as lower-case, prior to copying and reversing the string.
You can write 2X, where X is the length of the string.
std::string input = ""; std::getline (std::cin, input); // get input from stdin std::stringstream ss (input); // place input in a string stream integer num = 0; if (ss >> num) // extract integer from string stream { // Success! } else { // Fail! }
what is string
// Let's assume we're sorting the characters in String toSort // convert the String to an array of characters char[] chars = toSort.toCharArray(); // let Java do the sorting for you Arrays.sort(chars); // recreate the original String with the newly sorted array of characters toSort = new String(chars);
The safest way is to capture the input as a string and then convert the string to an integer. The reason is that all standard input (regardless of where it comes from) is done through character streams and it's safer to capture this input using a string rather than trying to perform conversions on the incoming data directly.