Question

The foll basic code on gcc compiler outputs 3.

But when I replace the {} by () then it outputs 4

     int i={3,2,4};printf("%d",i); 

Can someone explain this behaviour? The , operator follows left to right evaluation but it doesn’t in case of {}.

Answer 1

When you use {3,2,4} you are using scalar initialization, but you are providing more elements than what you should. This will trigger a warning by the compiler, but yeah it will both compile and execute, and the first element on the scalar initialization will be attributed to the variable i.

When you use (3,2,4) you are basically using the comma operator (the parentheses are needed because you are initializing the variable). The comma operator evaluates the first operand and discards the result, then evaluates the second operand and returns the result. So here you evaluate 3, discard it, then evaluate 2, return it (this return value is used as left operand of the second comma) and finally evaluate 4, and return it, so the 4 is attributed to the variable i.

Here’s another interesting operation:

int i;
i = 3,4;
printf("%d\n",i);

This will print 3, because first i=3 will be evaluated (as this is left operand of the comma), and its result will be discarded. Then 4 will be evaluated and its result returned, but not attributed to the variable.

And to make things more interesting:

int i,j;
j = (i = 3,4);
printf("%d %d\n",i,j);

This will print “3 4”. In other words, everything that I described above will happen, but when the 4 is evaluated and returned it will now be attributed to j.