Why does the function call get skipped when there are 2 function calls on the same line?

Question

I'm stumped on why the line:

cout << "Your average in this course is " << average() << " and your letter grade is " << convert_to_letterGrade() << "\n";

does not work, whereas the following works (one cout is split into 2):

cout << "Your average in this course is " << average();
cout << << " and your letter grade is " << convert_to_letterGrade() << "\n";

This code is in a void function that is a public member of the class Grades. average() calculates and stores the result in the private member variable averageScore, and also returns averageScore. convert_to_letterGrade() returns a char value based on the value of averageScore and an error message if the average score is not reasonable, and both functions are private members of the class.

In the first case, I get the error message from convert_to_letterGrade() first, THEN followed by what the cout statement was supposed to print. The cout statement prints the correct averageScore value, but when I stepped into convert_to_letterGrade(), the function was still using a garbage value for averageScore, whereas the second one works perfectly fine and convert_to_letterGrade() is also using the correct averageScore value.

Why is this?

If there would be a way to test and mess around with it it would be easier. Genereally, i would prefer using `std::endl;` instead of `"\n"` — Khaldor, May 02 '15 at 00:07
@Khaldor Why? It's not always necessary to flush the output buffers. — Captain Obvlious, May 02 '15 at 02:30
@CaptainObvlious *Generally i prefer* simply me. And i don't know. Picked up the habit to try to get used to c++ namespaces when i started. It stuck. — Khaldor, May 02 '15 at 03:26

Ben Voigt · Answer 1 · 2015-05-02T00:12:29.077

5

convert_to_letterGrade() returns a char value based on the value of averageScore and an error message if the average score is not reasonable

Based on your description of the behavior, this is not actually true.

Your current function returns a char value if the score is reasonable, or prints an error message directly to cout if unreasonable.

Fixing it is as simple as doing what you said, making the error string a return value so that the caller controls when it gets printed.

In addition to that, you have the problem of convert_to_letterGrade() relying on the variable set by the other function. As gnasher729 mentions in his answer, there is no guarantee that average() gets called before convert_to_letterGrade() unless you put a sequence point in between.

edited May 02 '15 at 00:12

answered May 02 '15 at 00:09

Ben Voigt

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@ruisen This is the answer you're looking for. – Khaldor May 02 '15 at 00:11
@Khaldor: Both issues need to be dealt with, ultimately. – Ben Voigt May 02 '15 at 00:12
Which is why this answer, including both problems, is viewed as correct by me. :-) – Khaldor May 02 '15 at 00:14
This answer makes no sense. The function either returns `char` or it doesn't. It can't do both. – Lightness Races in Orbit May 02 '15 at 03:28
@LightnessRacesinOrbit: A `char` can be returned inside a `std::string`. So can an error message. Think before you say "It can't" – Ben Voigt May 02 '15 at 05:37
@BenVoigt: Lol that's pretty weak and you know it. It's like saying `std::vector>> foo();` "returns int" because the vector itself has an int member. – Lightness Races in Orbit May 02 '15 at 14:47

score 5 · Accepted Answer · answered May 02 '15 at 00:10

You should be aware that functions within an expression can be called in any order.

I would assume that your implementation of convert_to_letterGrade depends on average() being called first, and that isn't guaranteed in your first statement.

Writing a function that only gives a correct result if another function has been called before is a dangerous practice. What would happen if I only wanted to print the grade as a letter? Would I have to remember to call average () first? That is an awful practice. Or what if you wanted to print the letter score first, followed by the average?

Ah, I see, Thanks :) I always thought it was evaluated left to right. — ruisen, May 06 '15 at 22:01

Why does the function call get skipped when there are 2 function calls on the same line?

2 Answers2