C++ Passing a local variable [closed] - c++

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I can not figure out where the pointer is wrong in this code. However, I receive the error that the code does not have a pointer-to function.
#include <iostream>
using namespace std;
char uppercase (char ch) {
if ((ch >= 'a') && (ch <= 'z')) {
return ch - 'a' + 'A' ;
cout << "Your capital letter is " << ch << endl;
} else {
return ch;
cout << "Your original letter is: " << ch << endl;
}
}
int main(int& ch){
cout << "Please enter a lowercase letter between a to z: ";
cin >> ch;
char uppercase;
char outChar;
char inChar;
outChar = uppercase(inChar);
system("pause");
}

Answer 1

int main(char&) is not strictly-conforming. It may be provided by the implementation but I don't know of any platform doing this. On a hosted implementation, use int main() or int main(int argc, char** argv) instead.
Building up on the 1st note, declare ch in the function as a local variable and use char not int:
char ch;
Or remove it completely, as described in the 4th point.
You call uppercase on an uninitialized variable (inChar), resulting in undefined behavior because uppercase reads from it. Remove the ch variable and use cin on inChar instead.
You should probably exchange the return ch; with the cout-statement in the uppercase function. The cout-statement is dead code, meaning it will never be executed because the function returns beforehand.

Answer 2

you have local variable with same name as function name: uppercase
get rid of local variable
also fix main function to following signature.
int main(int argc, char *argv[])
{
....
}
ex:
#include <iostream>
using namespace std;
char uppercase (char ch) {
if ((ch >= 'a') && (ch <= 'z')) {
return ch - 'a' + 'A' ;
}
else {
return ch;
}
}
int main(int argc, char* argv[]){
cout << "Please enter a lowercase letter between a to z: ";
char inChar;
cin >> inChar;
char outChar = uppercase(inChar);
return 0;
}

Answer 3

I can see quite a few problems in your code:
Read Why System is evil. In short words, using system("pause") tends to be problematic. There are many other ways to make your code stop (e.g. another cin).
As one of the above comments stated, the general way of opening the main program is by using int main() or int main(int argc, char** argv).
Let's say that your code works... Then you're asking for an executable input "ch" which after that is changed again by asking the same input again during the program's run-time! I proppose that you use point 2. and what #cad said.
The true answer to your problem! You are using as an input to uppercase(char) a variable that has not been filled "InChar", but the character you are filling by user input is "ch".
Hope this helps.