I don't understand how my compiler got this output - c++

Question

Note that I am using a Turbo C++ compiler because we are supposed to learn only Turbo C++ for our school syllabus. Which is why, the cout statement is evaluated from right to left in this case.
Program
#include <iostream.h>
#include <string.h>
void func(char *s, char t[]) {
strcpy(t, "Have fun");
s = "Be\0Cool";
cout << s[0] << ++s << s++ << --s << strupr(s+2) << ++s << s++ << s;
}
int main() {
char x[] = "Hello World!!!", y[] = "Hello World";
func(x, y);
cout << x << y;
return 0;
}
Output
CCOOLeeOOLBeBeHello World!!!Have fun
I feel the output should be:
CCooleeOOLBeBeHello World!!!Have fun
Because in the ++s part of the cout statement (second position), the pointer is at index 3 of the string s, so only 'Cool' should be printed. Instead 'COOL' is being printed. Why does this happen?

Answer 1

Tests with Visual Studio 2019
For comparison purpose, in Visual Studio 2019 (DEBUG), if we make required change to compile the code, then the program crash because we try to modify a constant string ("Be\0Cool").
If we do additional change to avoid the crash (by using a local array), the output is:
CCoOLeCoOLOLCoOLBeCoOLHello World!!!Have fun
If we split cout << s[]…; line to multiple calls to cout (one before each <<), then the output would be:
BeeeCOOLCOOLHello World!!!Have fun
Or if we add a line after each output, we get:
B
e
e
e
COOL
COOL
Hello World!!!
Have fun
Trying to understand the output of Turbo C++
If we then reverse each call to cout to start with the last (i.e. cout << s<< endl;) and end with first (cout << s[0] << endl), then we get:
Be
Be
OOL
e
e
COOL
C
Hello World!!!
Have fun
If we manually write that starting with the third last line and up and then the two last line in order without space, we get:
CCOOLeeOOLBeBeHello World!!!Have fun
Which is exactly what you got as an output.
Thus, it appears that Turbo C++ evaluate every expression from the right to the left.
Notes about required changes to compile (and run)
<iostream.h> is not available so I have to use <iostream> instead.
#define _CRT_SECURE_NO_WARNINGS must be added at top because some functions are not secure (and won't compile by default).
using namespace std; to avoid making more change to the code.
Add cast in s = (char *)"Be\0Cool"; so that line compile.
This lead to a crash because data is constant and we try to modify it.
Remove the cast and instead write char data[] = "Be\0Cool"; s = data;
The program run but the output is CCoOLeCoOLOLCoOLBeCoOLHello World!!!Have fun
In fact, this is undefined behavior. It just happen to be the actual output.
Undefined behavior
Some things are not defined by the standard and thus are not required to work a certain way. Well, as expected if read-only memory is not supported, it works like read-write memory.
For the order of evaluation, common possibilities are:
left to right
right to left
whatever is more optimal
Also since a variable is modified more than once, the value of s is not defined during the evaluation and afterward. The easy to remember rule is to avoid modifying the same variable more than once in a single expression.
About strupr
That function modify the string up to the terminating null character. In your case, it would convert to uppercase every letters from whatever value s has at the moment of the call.