Question

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Compilers and argument order of evaluation in C++
cout << order of call to functions it prints?

1. This:

   int k=3;
   printf("%d %d %d",k++,k,++k);
   

   Gives output as 4 4 4 because they are pushed into the stack as:

   %d%d%d
   
   4  -- for k++    
   4  --for k    
   4  --for ++k
   

   Right?

2. This:

   int k = 3;
   cout << k++ << k << ++k;
   

   Is actually repeated function calls, so it’s equivalent to:

   ( ( (cout << k++) << k) << ++k);
   

   So, I suppose first of all k++ then k and then ++k must always be executed in this order, right? I believe a function call is a sequence point, but the outputs vary on different implementations. Why is this so?

Answer 1

This is undefined because there is no sequence point between the , in the printf statement. Without a sequence point the compiler is free to order writes to the memory location k as it wills.

Now you may be wondering ‘what the hell is a seqeunce point’ and why is it relevant? Basically a sequence point is a point in the code where the memory location in question, in this case k has been modified at most once. There is a fuller description here:https://isocpp.org/wiki/faq/misc-technical-issues#double-mod-betw-seq-pt

As you can see from the FAQ, the , in the printf does not introduce a sequence point.

In the case of cout this is different because there are 3 function calls to operator >>. A function call introduces a sequence point therefore the modifications to memory location k have a defined order. However (and this was a point I missed but Cubbi pointed out) because C/C++ doesn’t define the order of evaluation of function arguments those arguments, even if they are functions, can be evaluated in any order the compiler defines. So in the expression:

f(h(), g())

Whether h() or g() is evaluated first is undefined: http://www.stroustrup.com/bs_faq2.html#undefined. So this is the reason why even in the case of cout you are getting different results from different compilers, basically because the cout << k++ << k << ++k translates to cout.operator<<(k++).operator<<(k).operator(++k) which is effectively an expression like this: f(h(g(cout, k++), k), ++k) and each function argument is evaluated in an unspecified order.