C++ Pass by Reference and Pass by Value Functions side effect?

Question

I understand why this works as it does

#include <iostream>
using namespace std;

int additionFive (int a)
{
    a = a - 5;
    return a;
}

int subtractFive (int &a)
{
    a = a -5;
    return a;
}

int main()
{
    int local_A = 10;

    cout << "Answer: " << additionFive(local_A) << endl;
    cout << "local_A Value "<< local_A << endl;

    cout << "Answer: " << subtractFive(local_A) << endl;
    cout << "local_A = Value "<< local_A << endl;

    return 0;
}

OUTPUT:

Answer: 5
local_A Value 10
Answer: 5
local_A = Value 5

But I dont understand why this change of syntax changes the answer (simply putting the arithmetic and print out on the same line)

#include <iostream>
using namespace std;

int additionFive (int a)
{
    a = a - 5;
    return a;
}

int subtractFive (int &a)
{
    a = a -5;
    return a;
}

int main()
{
    int local_A = 10;

    cout << "Answer: " << additionFive(local_A) << " local_A Value: "<< local_A << endl;
    cout << "Answer: " << subtractFive(local_A) << " local_A = Value: "<< local_A << endl;

    return 0;
}

OUTPUT:

Answer: 5 local_A Value: 10
Answer: 5 local_A = Value: 10

Answer 1

You're running into undefined behavior . The second version modifies the value of a which you're reading in the second cout 2 times, with no sequence points in between the reads.

First version:

cout << "Answer: " << subtractFive(local_A) << endl;
//                              |                  |
//                  reads and modifies local_A     |
//                                           sequence point
cout << "local_A Value ="<< local_A << endl;
//                             |
//                       reads local_A

Second version:

cout << "Answer: " << subtractFive(local_A) << " local_A Value: "<< local_A << endl;
//                             |                                       |
//                  reads and modifies local_A                   reads local_A

Answer 2

Well the behavior of the second code is completely system/compiler depended. On Dev C++ the second code is giving the same output as first one. It depends on how the compiler builds the cout statement in program assembly...