This question already has answers here:
How does the Comma Operator work
(9 answers)
Closed 8 years ago.
Why is a assigned the value 3? Does the compiler simply take the last value from the list?
int a;
a=(1,2,3);
printf("%d",a);
How does compiler parse this statement or how it works internally?
Comma in (1,2,3) is a comma operator. It is evaluated as
a = ( (1,2) ,3 );
Comma operator is left associative. The result/value of the expression (1,2,3) is the value of the right operand of comma operator.
As pointed out in the comments, it's because you're using the comma operator. That means the 1 and 2 are evaluated and discarded. The three is the only thing left to be assigned. Without the parenthesis it would most likely be assigned as 1.
Related
This question already has answers here:
What does the comma operator , do?
(8 answers)
Closed 6 years ago.
int main()
{
switch(1,2)
{
case 1:printf("1");break;
case 2:printf("2");break;
default: printf("error");break;
}
}
Is this valid in c?
I thought it shouldn't be , but when I compiled it , it shows no error and produces output 2.
Yes, this is valid, because in this case, the , is a comma operator.
Quoting C11, chapter §6.5.17, Comma operator, (emphasis mine)
The left operand of a comma operator is evaluated as a void expression; there is a
sequence point between its evaluation and that of the right operand. Then the right
operand is evaluated; the result has its type and value.
This (evaluates and) discards the left operand and uses the value of the right (side) one. So, the above statement is basically the same as
switch(2)
Just to elaborate, it does not use two values, as you may have expected something like, switching on either 1 or 2.
This question already has answers here:
What does the comma operator , do?
(8 answers)
Closed 7 years ago.
Guys when I attended a C quiz online, I came across a statement while(s++,6); where s is initialized to zero. I don't know what the while loop will exactly do when there is a comma operator in between. When I ran it on my gcc compiler it ran with no output. But when I changed the while condition as while(1,s++) it returned s value as 1. Can anyone tell me what is happening at that while.
The comma operator evaluates the left side and then throws away the result. The while condition keeps looping for any value other than zero. The first will be an infinite loop; the second will increment s and then stop.
I suspect in this case the comma was a typo and they meant to type a less-than.
From C11 standard §6.5.17:
The left operand of a comma operator is evaluated as a void expression; there is a sequence point between its evaluation and that of the right operand. Then the right operand is evaluated; the result has its type and value.
This means that 1,s++ evaluates 1 (so nothing happens), then it evaluates s++ and returns the result of that expression only.
So that's expression is equivalent to while (s++). If the left-hand side of a comma expression doesn't have any side effects, like in your situation, then you can remove it.
This question already has answers here:
What does the comma operator , do?
(8 answers)
Closed 7 years ago.
I have this C code line:
int a;
a = (1, 2, 3);
printf("%d", a);
Why the value 3 is printed? (the last one).
The comma operator evaluates all its "members" but returns the value of the last expression.From the C11 standard:
The left operand of a comma operator is evaluated as a void
expression; there is a sequence point between its evaluation and that
of the right operand. Then the right operand is evaluated; the result
has its type and value.
This question already has answers here:
Using comma operator in c
(1 answer)
Not able to understand the reason for output
(3 answers)
Closed 8 years ago.
#include<stdio.h>
int main()
{
int x=10,y=12;
printf("%d",(x,y));
return 0;
}
The output of the program is 12. How?
The expression that you are evaluating is:
x,y
This expression uses the comma operator. The standard (6.5.17 Comma operator) says:
The left operand of a comma operator is evaluated as a void expression; there is a
sequence point between its evaluation and that of the right operand. Then the right
operand is evaluated; the result has its type and value.
So, in your code, x,y evaluates to y, which has a value of 12.
For a more expansive discussion, I refer you to cppreference.com. Although that discusses C++, the discussion for this operator is valid in the context of C. Particularly relevant to your question is this section:
The comma in various comma-separated lists, such as function argument lists (f(a, b, c)), initializer lists int a[] = {1,2,3}, or initialization statements (int i, j;) is not the comma operator. If the comma operator needs to be used in that context, it has to be parenthesized: f(a, (n++, n+b), c).
And that's exactly the situation in your question. If you had written:
printf("%d", x, y);
then there would have been no use of the comma operator, and you would have supplied one more argument to printf than format specifier.
You're by chance using the comma operator.
In the C and C++ programming languages, the comma operator (represented by the token ,) is a binary operator that evaluates its first operand and discards the result, and then evaluates the second operand and returns this value (and type).
That said,
printf("%d",(x,y));
is functionally equivalent to
printf("%d", y);
it is because first (x,y) is evaluated.
inside () the expression is x,y they are evaluated from left to right since the Associativity of Comma operator is left to right, so the last value of evaluating (x,y) is y.
read operator precedence and associativity rule and how expressions are evaluated under operator precedence to understand these type of expressions
This question already has answers here:
What does the comma operator , do?
(8 answers)
C comma operator
(4 answers)
Closed 8 years ago.
Compiled using GCC on a Fedora 20 Desktop, the following code outputs 10.
int x=10;
int y=5;
printf("%d",(y,x));
In C, (y,x) means evaluating x and y and returning only x. EG. From wikipedia:
the comma operator (represented by the token ,) is a binary operator that evaluates its first operand and discards the result, and then evaluates the second operand and returns this value (and type).
That´s the "comma operator" (not to be confused
with the comma when separating function parameters.)
The "result" is only the last part, but if the other parts are functions etc.,
they will be executed too.
In C (exp1,exp2)
First exp1 is evaluated, then exp2 is evaluated, and the value of exp2 is returned for the whole expression.
(exp1, exp2) is like (exp1 && exp2) but both exp1 and exp2 will
always be evaluated, whatever exp1 returns.
(exp1, exp2) is like { exp1; exp2; } but it can be used as an
expression in a function call or assignment.