Unexpected output printing a and a++ inline in C

Asked Jun 27 '22 at 19:14

Active Jun 27 '22 at 19:14

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I am running the following type of C code

int a = 0;
printf("%d %d" , a , a++);

this code is returning me the output

1 0

As far as I understand a++ would append the value of a after it is read, and since a gets printed first the output should be 0 0 according to me. Where am I going wrong in understanding this?

asked Jun 27 '22 at 19:14

AshAR

3

This is undefined behavior, the order of evaluation of arguments is unspecified. – Barmar Jun 27 '22 at 19:15
Does that mean that we cannot comment on the output? – AshAR Jun 27 '22 at 19:15
1

It's evaluating `a++` first and then `a`. So `a` returns the incremented value. – Barmar Jun 27 '22 at 19:16
But any result is possible since it's undefined behavior. – Barmar Jun 27 '22 at 19:17
Exactly. Try running this: `int foo(void) {printf("Foo\n"; return 1;} int bar(void) {printf("Bar\n"); return 2;} int main() {printf("%i %i\n", foo(), bar()); return 0;}` – Daniel Walker Jun 27 '22 at 19:19
@AshAR Yes, that's what it means. – Paul Sanders Jun 27 '22 at 20:29
You are confusing what parameter gets *printed* first with what parameter gets *evaluated* first. – David Schwartz Jun 27 '22 at 21:30

Unexpected output printing a and a++ inline in C

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