How to fix iteration without a cause (with an input); in c++? - c++

Question

Just on a starting note, Please excuse the lack of c++ conventions and logic shown in this code. I'm still trying to get around formatting syntax for c++..
With this small part of a larger application building project, I am trying to create a 'data validation' type subroutine with integers and strings being the main test cases, the input coming from the intended user.
The problem is that each character of the variable input is being iterated over, even if the 'if statements' are outside of the 'for loop' (or in some other cases while loops).
An example of the error being:
Enter a value:f1o2o3
Your value is a string
Your value is a integer
Your value is a string
Your value is a integer
Your value is a string
Your value is a integer
The extra lines after the first Your value is a string and
Your value is a integer are unwanted.
I know that I could just instead return the boolean variable letexists or intexsits, with them both being iterated without a problem.
However this 'character iterating' problem has been a nuisance in other parts too and I cannot seem to find a definite, understandable fix anywhere.
I am used to using python so this 'iterating over every character while passing parameters' is quite new to me.
I have tried looking into what it could be for multiple days now. Both: searching online and asking others (in person showing them code that has problems) about the 'iteration over each character' problem, however, none seem to know why this happens.
I have solved this problem in the past by breaking a while(true) loop after the intended data value has been received, however I understand this is really bad practice and therefore want to figure out how I can improve data validation.
#include <iostream> // for std::cin, std::cout
#include <string> // for string datatype
#include <algorithm> // for std::find
#include <iterator> // for std::begin, std::end
using namespace std;
// If the 'cout's are changed to returning functions I will change this to a function itself (using int datacheck(...){...})
void datacheck(string &i) {
const char nums[] = { '0','1','2','3','4','5','6','7','8','9','\0' };
const char alph[] = { 'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z','\0' };
for (char& c : i) {
bool intexists = find(begin(nums), end(nums), c) != end(nums);
bool letexists = find(begin(alph), end(alph), c) != end(alph);
if (letexists || (letexists && intexists))
cout << "Your value is a string" << endl;
// This will be changed for a return value
else if (!letexists && intexists)
cout << "Your value is a integer" << endl;
// This will be changed for a return value
else
cout << "Your value has an erogenous input (Special characters or non-letter/ number related stuff.";
}
}
int main(void) {
string checkedvalue;
cout << "Enter a value: ";
cin >> checkedvalue;
cin.clear();
cin.ignore(512, '\n');
datacheck(checkedvalue);
return 0;
}
I have realised that it may be because each character is being sent through individually into the subroutine, however, if this is the case I am still unsure how to fix this.
Just to end off...
Overall, an explained solution for why the iterating happens and how it can be fixed will be greatly appreciated. If possible for general cases, however, just this case will be extremely helpful nonetheless.
Thank you.

Answer 1

I have to admit, i dont understand all of what you write. What I did understnad is
The problem is that each character of the variable input is being
iterated over, even if the 'if statements' are outside of the 'for
loop'
And that is wrong!
for (char& c : i) {
bool intexists = find(begin(nums), end(nums), c) != end(nums);
bool letexists = find(begin(alph), end(alph), c) != end(alph);
if (letexists || (letexists && intexists))
cout << "Your value is a string" << endl;
// This will be changed for a return value
else if (!letexists && intexists)
cout << "Your value is a integer" << endl;
// This will be changed for a return value
else
cout << "Your value has an erogenous input (Special characters or non-letter/ number related stuff.";
}
All this above is "the for loop". In simple terms the syntax for a (ranged based) for loop is
for (char& c : i) {
// body of the loop
}
Your code will be more clear and readable if you use {} also for the if, even if not needed and proper intendation:
for (char& c : i) {
bool intexists = find(begin(nums), end(nums), c) != end(nums);
bool letexists = find(begin(alph), end(alph), c) != end(alph);
if (letexists || (letexists && intexists)) {
cout << "Your value is a string" << endl;
} else if (!letexists && intexists) {
cout << "Your value is a integer" << endl;
} else {
cout << "Your value has an erogenous input (Special characters or non-letter/ number related stuff.";
}
}

Answer 2

In C++, the standard provides different ways to checks if an string contains alpha/numeric values:
Check if all characters are numeric: std::isalnum
Check if all characters are alpha: std::isalpha
So you can replace part of your code with:
const all_numeric = std::all_of(std::begin(str), std::end(str), [](const auto c) {
return std::isalnum(c);
})
const all_alpha = std::all_of(std::begin(str), std::end(str), [](const auto c) {
return std::isalpha(c);
})
So to solve your problem, you can start by creating a helper function that checks if a string is a number by using standard functions like strtool or recycle what we did later:
bool is_number(const std::string& s) {
return !s.empty() && std::find_if(s.begin(),
s.end(), [](char c) { return !std::isdigit(c); }) == s.end();
}
Now that you know if the string is a number or not, you can check if your string contains any weird character, and discard it.
Updates:
void datacheck(string &i) {
const auto intexists = is_number(i);
if (intexists) {
// do whatever
return;
}
const all_alpha = std::all_of(std::begin(str), std::end(str), [](const auto c) {
return std::isalpha(c);
});
if (all_alpha) {
// do whatever
return;
}
}