I am writing a program in C++. I am having a problem with the code when I assign multiple values to a single variable. I am confused about the particular logic as there is no decrement operator in the code but when i assign the multiple values its logic is not completely understood by me:
int main()
{
int j = 1;
int i = (j+2, j+3, j++);
cout<<"value is "<<i;
getch();
return 0;
}
Output is 1. I don't know one is assigned to i.
When you write int i=(j+2,j+3,j++);, it's basically like you wrote int i=j++;.
That means:
1) since the value of j is 1, i will have the same value;
2) after that line j will be incremented.
As others have pointed out, you cannot do what you want to do, and the comma operator is coming into play. To be crystal clear, this line:
int i=(j+2,j+3,j++);
Is performing these steps:
- j+2 is evaluated. This evaluates to 3, which is discarded as it is not assigned to anything.
- j+3 is evaluated, see step 1.
- j++ is evaluated. This increments j, whose value becomes 2, but evaluates to the OLD value of j, which is 1.
- The result of step 3 is assigned to i.
variable j is post increment so the current value of j is put into i and then j is incremented.
You can think of this as being:
int j=1;
int temp = j + 2;
int temp1 = j + 3;
int i=j;
j++;
Where the variables temp and temp1 do not really exist and are thrown away.
Let consider these definitions
int j=1;
int i=(j+2,j+3,j++);
(j+2,j+3,j++) is an expression with the comma operator. Its result is the last subexpression. Subexpressions j + 2 and j + 3 do not influence neither j nor i. The only subexpression that influence j and i is the last. It is j++. The value of postincrement operator is the value of the original variable before the increment. So the value of j++ is 1. And i will get this value. At the same time j will be incressed and get value 2.
Read: Comma Operator: ,
The comma operator
,hasleft-to-right associativity. Two expressions separated by a comma are evaluated left to right. The left operand is always evaluated, and all side effects are completed before the right operand is evaluated.
The expression:
i = (j + 2, j + 3, j++);
is in effects equivalent to (because evaluation of j + 2 and j + 3 has no side-effects):
i = j++;
So first value of j is assigned to i that is 1 then j incremented to 2.
Here j + 2 and j + 3 are evaluated before j++ but it has no effect.
Important to note parenthesis ( ) in expression over write the precedence so first , operator evaluated within parenthesis and then = evaluates at last.
To understand the output: look at precedence table , have lower precedence than =. In you expression you have overwrite the precedence using parenthesis.
Edit: on the basis of comments:
Suppose if expression is:
int i = (j++, j+2, j+3);
In this expression j++ first increase value of j because of ++ operation to 2 then j + 2 evaluates but this sub-expression has no side effects finally j + 3 evaluates = 5 and 5 is assigned to i. so value to i at the end is 5.
So here int i =(j++, j+2, j+3); is NOT equivalent to just i = j + 3 because j++ has side effect that change value of j.
In sort if you have an expression like a = (b, c, d) then first expression b evaluates then expression c evaluates then expression d evaluates due to left-to-write associativity of , operator then final value of expression d is assigned to variable a.
i = j + 3int i = (j++, j+2, j+3); (in first comment) if j = 1 first j++ incremented to 2 then because of i = j + 3, i will be 5