C Programming - Expressions - Discussion

In Let Us C book it describe that the comma operator has Right_Left associative. Please discuss this

I don't understand this. Can anybody explain!

Which means this operator select the values from right to left ie(here it takes the right most value 5 and ommit the others).
so i+5=7.

The associativity of comma is left to right. So it will take the first value from set i.e. 2 and put it to j i.e. j = 3. Now it will move to next element i.e., 3 and calculates i+3 = 4. But j is an integer which can only take one value and it will discard the present value and updated to j= 4. This process goes on and at last we left with j = 2+5 = 7.

// cpp_comma_operator.cpp.

#include <stdio.h>
int main () {
int i = 10, b = 20, c= 30;
i = b, c;
printf("%i\n", i);

i = (b, c);
printf("%i\n", i);
}

Output:
20
30

@Prasant.

The let us C book has given the associativity wrong. The associativity of ", " is left to right. Hence 5 is taken and 5+2 = 7.

#include<stdio.h>
void main()
{
int x;
x=10,20,30;
printf("%d",x);
return 0;
}

Why x = 10?

i+5 means 2+5=7.

int a=1, b=2, c=3, i=0; commas act as separators in this line, not as an operator.
a=1, b=2, c=3, i=0.

i=(a, b); stores b into i.
a=1, b=2, c=3, i=2.

i=a, b; stores a into i equivalent to (i = a), b;
a=1, b=2, c=3, i=1.

i=(a += 2, a + b); increases a by 2, then stores a+b = 3+2 into i.
a=3, b=2, c=3, i=5.

i = a += 2, a + b; increases a by 2, then stores a to i, and discards unused.
a+b r value equivalent to (i=(a += 2)), a + b;
a=5, b=2, c=3, i=5.

i = a, b, c; stores a into i, discarding the unused b and c r values.
a=5, b=2, c=3, i=5.

i = (a, b, c); stores c into i, discarding the unused a and b r values.
a=5, b=2, c=3, i=3.

return a=4, b=5, c=6; returns 6, not 4, since comma operator sequence points.

Following the keyword 'return' are considered a single.

Expression evaluating to r value of final sub expression c=6.

return 1, 2, 3; returns 3, not 1, for same reason as previous example.

return (1), 2, 3; returns 3, not 1, still for same reason as above.

This example works as it does because return is a keyword, not a function call. Even though most compilers will allow for the construct return (value), the parentheses are syntactic sugar that get stripped out without syntactic analysis.

@Vvv.

Try and change one statement to,
x=(10, 20, 30);

And see result yourself.