formatting string with incrementing integer c++

Question

Is there a quick way to do the following without the repetition of i++

int i = 0;
printf("foo-%d,blah-%d,orange-%d", i++, i++, i++);

I need to do this with 8 or more pairs and the code looks horrible..

Maybe you will need to create you own function in order to keep your code as clean as possible. Take a look at vprintf (http://www.manpagez.com/man/3/vprintf/) and variable arguments functions (http://stackoverflow.com/questions/1657883/variable-number-of-arguments-in-c) — Neozaru, Sep 26 '12 at 16:48
One question under both [c++] and [printf] ... wow O_ó ... jsut to sound less obnoxious ... it's plain c, it should not be confused, even though it compiles. — luk32, Sep 26 '12 at 17:09
Why not `printf("foo-1,blah-2,orange-3");` or `int x = ...; printf("foo-%d,blah-%d,orange-%d", x, x + 1, x + 2);`? — PiotrNycz, Sep 26 '12 at 21:39

score 4 · Answer 1 · answered Sep 26 '12 at 16:42

4

The order of evaluation of arguments is unspecified. Also, you can't modify a variable twice without an intervening sequence point. So do this instead:

printf("foo-%d,blah-%d,orange-%d", i, i+1, i+2);
i+=3;

answered Sep 26 '12 at 16:42

David Schwartz

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I didn't know the rule about modifying a variable twice.. thanks. (the order was unimportant). – AlasdairC Sep 26 '12 at 16:43
1

Really? You don't care if the output is `foo-1,blah-2,orange-3` or `foo-3,blah-1,orange-2` or `foo-2,blah-3,orange-1` or ...? – David Schwartz Sep 26 '12 at 16:44
@DavidSchwartz +1 ...The `foo-1,blah-1,orange-1` is also possible :) – Sergey Kalinichenko Sep 26 '12 at 16:46
@dasblinkenlight: Right, because the lack of a sequence point makes this UB, so anything can happen. There are actually two issues here. – David Schwartz Sep 26 '12 at 16:47

score 3 · Accepted Answer · answered Sep 26 '12 at 16:53

Since you can have 8 or more pairs, I would like to suggest you to use a more readably and easy to extend version:

std::vector<std::string> vec_values = {"foo", "blah", "orange"};
for (size_t index = 0; index < vec_values.size(); ++index)
    std::cout << vec_values[index] << "-" << index;

score 1 · Answer 3 · answered Sep 26 '12 at 16:50

You might think of it in terms of the prefixes, not of the number:

int i = 0;
std::vector<std::string> prefixes { "foo", "bar", "baz", "ham" };

bool first = false;
for (const auto& prefix : prefixes) {
    if (!first)
        std::cout << ',';
    std::cout << prefix << '-' << i++;
    first = false;
}

If you can’t use C++11 range-based for, you can write it out in full:

for (auto prefix = prefixes.begin(); prefix != prefixes.end(); ++prefix) {
//   ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    if (!first)
        std::cout << ',';
    std::cout << *prefix << '-' << i++;
    //           ~~~~~~~
    first = false;
}

Alternatively, use indices:

for (i = 0; i < prefixes.size(); ++i) {
//   ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    if (!first)
        std::cout << ',';
    std::cout << prefixes[i] << '-' << i;
    //           ~~~~~~~~~~~           ~
    first = false;
}

formatting string with incrementing integer c++

3 Answers3