C++ Beginner Assignment Operators - c++

Question

i'm a beginner in C++ and i'm doing this example code of assignment operators. I don't know what i'm doing wrong here.
#include <iostream>
using namespace std;
int main()
{
int x = 200; int y = 100;
x += y; cout << x << endl; //x += y; // x = x + y, x = 200 + 100, output 300.
x -= y; cout << x << endl; //x -= y; // x = x - y, x = 200 - 100, output 100.
x /= y; cout << x << endl; //x /= y; // x = x / y, x = 200 / 100, output 2.
x %= y; cout << x << endl; //x % y; // x = % y, x = 200 % 100, output 0.
return 0;
}
i'm getting the results for
x -= y = 200
x /= y = 2
x %= y = 2
when its suppose to be
x -= y = 100
x /= y = 2
x %= y = 0
its actually adding up the previous results. How do i stop the code from adding the result to the next line? Thanks!

Answer 1

As you mentioned in your comments in your code, the result of x += y is equivalent to x = x + y, which means the value of x will change. So it is no surprise that the new value of x, 300, is used in the next mathematical assignment.
If you want to avoid that, consider saving the result of x + y to another variable.
int x = 200;
int y = 100;
int x_addition = x + y;
std::cout << x_addition << std::endl;
Or, if its only usage is in displaying the result of the addition, do it all in one line.
int x = 200;
int y = 100;
std::cout << x + y << std::endl;

Answer 2

When x and y are variables and U is an operation x U= y is equivalent to:
x = x U y;
which means the original variable is being modified and assigned with the result.
You code is equivalent to:
x = x + y;
cout << x << endl; // x is now 300
x = x - y;
cout << x << endl; // x is now 200
x = x / y;
cout << x << endl; // x is now 2
x = x % y;
cout << x << endl; // x is now 2
If you want not to change x you want can save x in a temporary variable or just print the result:
int x = 200;
int y = 100;
cout << x - y << endl; // x unchanged
cout << x + y << endl; // x unchanged
cout << x / y << endl; // x unchanged
cout << x * y << endl; // x unchanged
Or:
int x = 200;
int x = 100;
int result = x + y;
cout << result << endl;
result = x - y;
cout << x << endl;
result = x / y;
cout << result << endl;
result = x % y;
cout << result << endl;
Note: since the = operator returns the value after the assignment cout << x += y << end can compile and print the assigned value.

Answer 3

Every time you are doing assignment like x += y;, it changing the value of x.
x += y;
is equivalent to
x = x + y;
Which means assignment. Value of x is changing. It's now 300. Same thing happening for latter assignments too.
Note: if you don't want to change the value of x and y, better do the assignments in other variable.
int x = 200, y = 100, t;
t = x + y; cout << t << endl;
t = x - y; cout << t << endl;
t = x * y; cout << t << endl;
t = (int) x / y; cout << t << endl;

Answer 4

You are expecting the original value of x and y to be kept forever, but it isn't. Assignments like x += y; or x = x + y; change the value of the variables (in this case x, that is, the one on the left side of the equal sign). The previous value is overwritten and therefore lost.
If you don't want to change the original values, you have to either assign the result to other variables (for example, one called temp, that you can overwrite every time), or you don't assign it at all, and you simply send to cout the result of the calculation.
As an example with a temp variable:
int x = 200; int y = 100; int temp;
temp = x + y; cout << temp << endl; //temp = x + y, temp = 200 + 100, output 300.
temp = x - y; cout << temp << endl; //temp = x - y, temp = 200 - 100, output 100.
Or, doing away with variables entirely:
cout << x + y << endl; // 200 + 100, output 300
cout << x - y << endl; // 200 - 100, output 100
Another way to see that the variables are overwritten (well, in this case it's only x) is to declare them as const:
const int x = 200; const int y = 100;
The program won't compile, because it tries to change constant variables.

Answer 5

Try resetting x back to 200 after each assignment. Like this:
#include <iostream>
using namespace std;
int main()
{
int x = 200; int y = 100;
x += y; cout << x << endl;
x = 200;
x /= y; cout << x << endl;
x = 200;
x %= y; cout << x << endl;
return 0;
}

Answer 6

You are using the accumulation operators so you're changing the value of 'x'.
x += y; // x = 200 + 100 = 300
x -= y; // x = 300 - 100 = 200
x /= y; // x = 200 / 100 = 2
x %= y; // x = 2 % 100 = 2
Instead you could do:
int x = 200;
int y = 100;
cout << x + y << endl;
cout << x - y << endl;
cout << x / y << endl;
cout << x % y << endl;
Or:
int x = 200;
int y = 100;
int result;
result = x + y;
cout << result << endl;
result = x - y;
cout << result << endl;
result = x / y;
cout << result << endl;
result = x % y;
cout << result << endl;

Answer 7

You are not counting on Addition operations value
x=200,y=100;
x+=y;//x=200+100=300,x=300,y=100
x-=y;//x=300-100=200,x=200,y=100
Here you are forgetting that there is ADDITION OPERATION before subtraction