Is writing 3 instructions separated by comma `,` undefined behaviour?

Question

I think that I saw somewhere that writing more than 1 instruction separated by a comma , is undefined behavior.

So does the following code generate undefined behavior?

for (i=0, j=3, k=1; i<3 && j<9 && k<5; i++, j++, k++) {
    printf("%d %d %d\n", i, j, k);
}

because there are 3 instructions separated by a comma , :

i++, j++, k++

Answer 1

writing more than 1 instruction separated by comma , is undefined behaviour.

Nope, it's not the general case.

In your case, i++, j++, k++ is perfectly valid.

FWIW, as per 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; [...]

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[Note]: You might have got confused by seeing something along the line of

  printf("%d %d %d", i++, ++i, i);

kind of statement, but do note, there the , is not a comma operator altogether (rather, a separator for supplied arguments) and the sequencing does not happen. So, those kind of statements are UB.

Again, referring to the standard, footnote 3 for the same chapter

^{As indicated by the syntax, the comma operator (as described in this subclause) cannot appear in contexts where a comma is used to separate items in a list (such as arguments to functions or lists of initializers).}