Return the complete uppercase/lowercase converted string - c++

Question

I am writing a program to convert all lowercase letters in a string to uppercase letters, vice versa.
However, I found that my program cannot return the whole converted string.
Here are outcomes and my program code.
Input: "a Survey", Output: "A".
Input: "Hello", Output: "hELLO".
#include <iostream>
#include <cstring>
using namespace std;
int main() {
char inWord[50], outWord[50];
int j = 0;
cin.getline(inWord, 50);
for (j = 0; j < strlen(inWord); j++)
{
//upper to lower
if (inWord[j]>='A'&&inWord[j]<='Z')
outWord[j] = inWord[j] + 'a' - 'A';
//lower to upper
else
outWord[j] = inWord[j] + 'A' - 'a';
}
cout << outWord;
return 0;
}

Answer 1

You need to take into account characters that are not letters. Now you are converting the space to 0, which is the way cout knows he needs to stop reading from the string. This is a simple solution:
#include <iostream>
#include <cstring>
using namespace std;
int main() {
char inWord[50], outWord[50];
cin.getline(inWord, 50);
for (int j = 0; j < strlen(inWord); j++)
{
char c = inWord[j];
//upper to lower
if (c>='A' && c<='Z')
outWord[j] = c + 'a' - 'A';
//lower to upper
else if (c>='a' && c<='z')
outWord[j] = c + 'A' - 'a';
else
outWord[j] = c;
}
cout << outWord;
return 0;
}
And if you are programming in C++, you should probably replace those char[50] with std::string, and avoid that using namespace std:
#include <iostream>
#include <string>
int main() {
std::string inWord, outWord;
std::getline(std::cin, inWord);
outWord.reserve(inWord.size());
for (int i = 0; i < inWord.size(); i++)
{
char c = inWord[i];
//upper to lower
if (c>='A' && c<='Z') {
outWord.push_back(c + 'a' - 'A');
}
//lower to upper
else if (c>='a' && c<='z') {
outWord.push_back(c + 'A' - 'a');
}
else {
outWord.push_back(c);
}
}
std::cout << outWord;
return 0;
}

Answer 2

In ascii,
' ' is not in 'A'-'Z' range, so second branch is taken.
By hazard, its value is 'a' - 'A'. resulting in '\0' (nul terminator of C-string).
To fix that issue, you have to handle a third case:
const std::size_t size = strlen(inWord)
for (std::size_t j = 0; j < size; j++)
{
//upper to lower
if (inWord[j]>='A'&&inWord[j]<='Z')
outWord[j] = inWord[j] + 'a' - 'A';
//lower to upper
else if (inWord[j]>='a'&&inWord[j]<='z')
outWord[j] = inWord[j] + 'A' - 'a';
else
outWord[j] = inWord[j];
}
'a'-'z', 'A'-'Z' are not guaranteed to be contiguous ranges (it is for Ascii, not for EBCDIC), so you might use standard methods instead (even if their interfaces is error prone :( ):
const std::size_t size = strlen(inWord)
for (std::size_t j = 0; j < size; j++)
{
unsigned char c = static_cast<unsigned char>(inWord[j]);
if (std::isupper(c))
outWord[j] = static_cast<char>(std::tolower(c));
//lower to upper
else if (std::islower(c))
outWord[j] = static_cast<char>(std::toupper(c));
else
outWord[j] = inWord[j];
}

Answer 3

If you want to convert all letters to for instance to the opposite case you would need to check which case the string is after input. (for instance by counting the upper case characters).
Right now you are inverting the case of each letter, as you do an if else comparison of each letter in the loop. If you execute only one of the 2 branching conditions it will do an conversion to the same case.
A few words on how you might want to improve your style:
You can define j inside the for loop if you are not using the int outside of the loop
for (int j = 0; j < strlen(inWord); j++){}
'a' - 'A' and 'A' - 'a' do have the same absolute value and do not change during the loop, so you could define a variable before the loop and then add or subtract inside of the loop respectively.
It is also useful to know that C++ treats characters as 8bit numbers in the case of ascii, this is the reason why the offset calculation works:
Link to an ASCII table