Explain the order of evaluation in printf [duplicate]

Question

main()
{
    int i=5;
    printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}

Output is 45545, but I don't how it is working. Some say that the arguments in a function call are pushed into the stack from left to right.

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.

Answer 2

The order in which the parameters to a function are passed is not defined in the standard, and is determined by the calling convention used by the compiler. I think in your case, cdecl calling convention (which many C compilers use for x86 architecture) is used in which arguments in a function get evaluated from right to left.