trouble compiling a parse integer program when calling the parse int function - c++

Question

Currently, I'm having some trouble implementing a parse integer program I've made. I get two yellow lines appearing under the function when it is called in the program
using namespace std;
char parseInt(char userInput[40], char& integers);
int main()
{
char userInput[40];
char integers[40];
cout << "After you enter string input, the program will parse all the integers "
<< "inside the string and display them. \n Enter a string: \n";
cin.ignore();
cin.getline(userInput, 40);
parseInt(userInput, integers); //Here lies the problem
cout << "The integers inside the string you entered were: \n" << integers;
return 0;
}
char parseInt(char userInput[40], char& integers)
{
int j = 0;
for(int i = 0; i<sizeof(userInput); i++)
{
if( userInput[i] == 1 || userInput[i] == 2 || userInput[i] == 3 ||
userInput[i] == 4 || userInput[i] == 5 || userInput[i] == 6 ||
userInput[i] == 7 || userInput[i] == 8 || userInput[i] == 9 ||
userInput[i] == 0 )
{
integers = userInput[i];
j++;
}
}
}
When the parse Int function is called, the two error messages I'm getting are :
-cannot bind rvalue (char) (char*) (&integers) to '(char&)'
-invalid conversion from 'char*' to char [-fpermissive]
I'm just having a hard time understanding exactly what the error codes are trying to say, I'm trying to understand them more.

Answer 1

Many bugs here.
sizeof(userInput) doesn't work like you expect (user4581301 mentioned this in a comment). userInput is a parameter, and due to parameter type adjustment.
The type of a function is determined using the following rules. The type of each parameter (including function parameter packs) is determined from its own decl-specifier-seq and declarator. After determining the type of each parameter, any parameter of type "array of T" or of function type T is adjusted to be "pointer to T". After producing the list of parameter types, any top-level cv-qualifiers modifying a parameter type are deleted when forming the function type.
As a result, your loop only goes up to sizeof (char*).
You are reading in a string but checking the numeric value of each character. Don't expect that '1' == 1, it's not true on any character encoding in common use. (EloyBG mentioned this one)
You declared integers as a reference to just one integer, but tried to pass a whole array. (Ilan mentioned this)
You're trying to stuff many values into a reference to just one integer. (EloyBG mentioned this too, but his fix is misleading for the same reason as #1)
Reaching the end of a non-void function is illegal, you have to return a value or leave abnormally (user4581301 mentioned this too)
The C way to fix it would be to pass the length of the array as a separate parameter. In C++, you can use a reference to array to either enforce a fixed length or even infer the actual length:
void parseInt(char (&userInput)[40], char (&integers)[40])
Parameter adjustment only changes arrays, not references to arrays, and therefore your sizeof attempt will again work.

Answer 2

In parseInt, integers is declared like single character but you are passing an array. Furthermore, you compare userInput as integer instead of as ASCII.
void parseInt(char userInput[40], char integers[40])
{
int j = 0;
for(int i = 0; i<sizeof(userInput); i++)
{
if( userInput[i] == '1' || userInput[i] == '2' || userInput[i] == '3' ||
userInput[i] == '4' || userInput[i] == '5' || userInput[i] == '6' ||
userInput[i] == '7' || userInput[i] == '8' || userInput[i] == '9' ||
userInput[i] == '0' )
{
integers[j] = userInput[i];
j++;
}
}
}
If you use your code,
It return a single character by reference, but not return a chain of characters.
char parseInt(char userInput[40], char& integers)
{
int j = 0;
for(int i = 0; i<sizeof(userInput); i++)
{
Here, you are comparing the userInput with the first elements of the ASCII table.
if( userInput[i] == 1 || userInput[i] == 2 || userInput[i] == 3 ||
userInput[i] == 4 || userInput[i] == 5 || userInput[i] == 6 ||
userInput[i] == 7 || userInput[i] == 8 || userInput[i] == 9 ||
userInput[i] == 0 )
{
In the next line, you change the value of the single character but you do not adding a new character, because it is not a array (I supuse you got the idea because you declare j)
integers = userInput[i];
j++;
}
}
}
In the main,
cin.ignore()
is unnecesary,it makes that you lose the first character of the chain.
Declare integers like an array of characters
Compare userInput as ASCII not as integer (you can compare it as integer but you have to use the integer that corresponds with the character on the ASCII table)
Delete the cin.ignore() line

Answer 3

You are passing in char integers[40] into a function that it's second parameter requires a char to be passed by reference.
change char& to char[40]
char parseInt(char userInput[40], char integers[40])