c - Explain the order of evaluation in printf

Question

This question already has answers here:

Answer 1

The evaluation order of function parameters is unspecified.

From c99 standard:

6.5.2.2 Function calls

10/ The order of evaluation of the function designator, the actual arguments, and subexpressions within the actual arguments is unspecified, but there is a sequence point before the actual call.

This is, however, only a part of the problem. Another thing (which is actually worse, since it involves undefined behavior) is:

6.5 Expressions

2/ Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression. Furthermore, the prior value shall be read only to determine the value to be stored.

In our case all the arguments evaluation is between only 2 sequence points: the previous ; and the point before the function is entered but after all the arguments have been evaluated. You'd better not write a code like this.

C standard is pretty relaxed in some places to leave room for optimizations that compilers might do.