Why is it not printing in the right order?

Question

#include <iostream>

using namespace std;

int output_1(string s, char c);
int output_2(string s, char c);

int main() {

    string str;
    char c;

    cout << "Enter a string: ";
    getline(cin, str);

    cout << "Enter a character: ";
    cin >> c;

    cout << "[1] the number of times the character appears: " << output_1(str, c) << endl;
    
    cout << "[2] That character is found at Index/Indices: " << output_2(str, c) << endl;

    return 0;
}

int output_1(string s, char c) {

    int count = 0;

    for (int x = 0; x < s.length(); x++)

        if (s[x] == c)
            count++;

    return count;
}

int output_2(string s, char c) {
    
    for (int x = 0; x < s.length(); x++) {
        if (c == s[x]){
            cout << x << " ";
        }
    }
}

Why is the second output not in order and have an extra 0 at the end?

Enter a string: test
Enter a character: t
[1] the number of times the character appears: 2
0 3 [2] That character is found at Index/Indices: 0

Answer 1

The problem is this line:

cout << "[2] That character is found at Index/Indices: " << output_2(str, c) << endl;

The first thing to note is that << is just syntactic sugar for a function call. So we can re-write it to look like this:

cout.op<<("[2] That character is found at Index/Indices: ")
    .op<<(output_2(str, c))
    .op<<(endl);

Because they are chained statements you can think of this as:

std::ostream& t1 = cout.op<<("[2] That character is found at Index/Indices: ");
std::ostream& t2 = t1.op<<(output_2(str, c));
std::ostream& t3 = t2.op<<(endl);

So what we have are three method calls to op<<() each passing one parameter. Before calling these methods that the compiler must get the operands to the calls. The first one is easy as it is a literal string, the second one requires a function call. So the compiler must call the function output_2() before it can call op<<() and the third one is an function (easy to pass the address).

But the compiler can sequence that call to output_2() at any point (as long as it happens before the call to op<<(). So it looks the compiler is doing this:

<Seq Start>    
void&         x  = output_2(str, c); // yes I know thats not legal C++ just for explaining.
std::ostream& t1 = cout.op<<("[2] That character is found at Index/Indices: ");
std::ostream& t2 = t1.op<<(x);
std::ostream& t3 = t2.op<<(endl);
<Seq End>

Note: The compiler could place the call anywhere between <Seq Start> and <Seq End> (before it is needed) it's just an implementation detail. Now the call to output_2() also actually outputs to the std::cout followed by a int (which is not returned (undefined behavior by itself)).

Answer 2

Instead of printing value in main, Try this :

//call this from main 

    output_1(str, c);
    output_2(str, c);

Then:

void output_1(string s, char c) {

    int count = 0;

    for (int x = 0; x < s.length(); x++)

        if (s[x] == c)
            count++;

    cout<< count<<endl;
}

void output_2(string s, char c) {
    
    for (int x = 0; x < s.length(); x++) {
        if (c == s[x]){
            cout << x << " ";
        }
    }
}

Answer 3

This is UB. output_2 should return an int, but doesn't return anything. Now you get a 0, but it could be anything.

As you're building op an ostream in output_2, you could use a stringstream internally and convert that to a string for the output.

#include<string>
#include<iostream>
#include<algorithm>
#include<sstream>

int output_1(std::string const& s, char c) {
    return std::count(cbegin(s), cend(s), c);
}

std::string output_2(std::string const& s, char c) {
    std::stringstream os;
    for (int x = 0; x < s.length(); ++x) {
        if (c == s[x]){
            os << x << " ";
        }
    }
    return os.str();
}

int main() {
    std::string str = "test";
    char c = 't';

    std::cout << "[1] the number of times the character appears: " << output_1(str, c) << '\n';
    
    std::cout << "[2] That character is found at Index/Indices: " << output_2(str, c) << '\n';
}

Answer 4

You issue is that output_2 lies. The return type of int says the function returns a value. There is no return statement, so there is undefined behavior.

My suggested solution:

void output_2(string s, char c)
{
    for (int x = 0; x < s.length(); x++)
    {
        if (c == s[x])
        {
            cout << x << " ";
        }
    }
}

int main()
{
  //...
    cout << "[1] the number of times the character appears: " << output_1(str, c) << endl;
    
    cout << "[2] That character is found at Index/Indices: ";
    output_2(str, c);
    cout << endl;

    return 0;
}

In the above code fragments, the return type of output_2 is set to void since it doesn't return a value.

In the main function, the output_2() is removed from the cout statement (since output_2() doesn't return a value), and set as a separate function call.

Answer 5

Running through output_2():

int output_2(string s, char c) {

for (int x = 0; x < s.length(); x++) {
    if (c == s[x]){
        cout << x << " ";
    }
}

}

First pass (t) prints 0.

Second pass (e) prints nothing.

Third pass (s) prints nothing

Fourth pass (t) prints 3

You wrote output_2() as returning int, but actually you should be returning an array of integers and printing those (or something similar). The return value of a function that is supposed to return an int but has no return line is undefined. The behaviour of the whole program is also undefined because of this missing return statement. Anyway, in your specific case, it looks like output_2() returned 0 and so finally after control was returned to main it printed

cout << "[2] That character is found at Index/Indices: " << output_2(str, c) << endl;

where output_2(str, c) prints a zero (in an undefined manner). I think one thing that you were missing (with respect to order) is that the line

cout << "[2] That character is found at Index/Indices: " << output_2(str, c) << endl;

first allows output_2() to print 0 and 3 and then starts to print this line.