How to Write the Range-based For-Loop With Argv? - c++

Question

From the c++0x Wikipedia site:
int my_array[5] = {1, 2, 3, 4, 5};
for (int &x : my_array) {
x *= 2;
}
So why does this code not work?
int main(int argc, char* argv[])
{
for (char *arg : argv)
{
// Do something.
}
}
Error:
main.cpp:36: error: no matching function for call to ‘begin(char**&)’
I am using Qt with g++ 4.6.1 on Ubuntu 11.10.
Additional Information
Is There a Range Class in C++0x
Range-Based For-Loop Statement Definition Redundance

Answer 1

Usually, the first thing I do with argc and argv is this:
std::vector<std::string> arguments(argv, argv + argc);
Now I have a vector of strings to work with and I can easily use not only the range-based for loops, but also C++ standard library facilities.
for(std::string& s : arguments) {
// do stuff...
}
The wikipedia code works because the type of my_array is a variable of array type.
The original code does not work, because argv is not an array. The syntax char* argv[] may make it look like it is an array, but that's just a sad artifact of C syntax. char* argv[] is exactly the same as char** argv. argv is not an array; it's actually just a pointer.
The range-based for loop works on:
arrays;
any type that has member functions begin() and end() that return iterators;
any type for which exist non-member functions begin and end that can be called like begin(x) and end(x), with x being the thing that you're iterating over.
As you can see, pointers are not part of this list.

Answer 2

You don't, because the system can't tell how long argv is at compile time. Someone can likely find the proper section of the standard to quote you about this.
There is a way around it though, and that's to create a custom class to wrap argv. It's not even that hard.
class argv_range {
public:
argv_range(int argc, const char * const argv[])
: argc_(argc), argv_(argv)
{
}
const char * const *begin() const { return argv_; }
const char * const *end() const { return argv_ + argc_; }
private:
const int argc_;
const char * const *argv_;
};
Here's how you use it:
for (const char *arg: argv_range(argc, argv)) {
// Do something.
}
Yeah, I use a lot of consts. Basically, argv is an array of character pointers, none of which should be modified, each pointing to a string, none of the characters of which should be modified either.

Answer 3

The vector solution proposed copies the array (the pointers only, not the strings1 - but still). Unnessary. The argv_range solution is what I would have tried, too, if I absolutely wanted to enforce a range based loop. But that produces a lot of extra code (admitted, only once, if you write it to a header file and keep it, but still).
The classic loop appears so easy to me that I allow myself just to post it, I don't consider it worth to have all this effort just to have a range based loop...
for (char** a = argv; *a; ++a)
{
// use *a, or if you don't like:
char* arg = *a;
// use arg...
}
Or, if you won't ever need the argv array afterwards any more:
for (++argv; *argv; ++argv)
{
// use *argv, or if you don't like:
char* a = *argv;
// use a...
}
There is a little difference, you might have noticed: In the first variant, I iterate over all the values, in the second, I leave out the first one (which normally is the program name we are not interested in in many cases). The other way round, for each:
for (char** a = argv + 1; *a; ++a);
for (; *argv; ++argv);
Note that all these variants profit from the fact that the strings array itself is null-pointer-terminated as well, just as any of the strings is null-character-terminated, thus the simple checks for *a or *argv.
1This applies only, if using std::vector<char*>; if using std::vector<std::string>, as actually proposed, even the strings themselves are copied!

Answer 4

You can use C++20 std::span() or gsl::span() to make view for argv:
for (auto const arg : std::span(argv, argc)) {
std::cout << arg << '\n';
}
Or similar, using Ranges (std::view::counted()):
for (auto const arg : std::views::counted(argv, argc)) {
std::cout << arg << '\n';
}
or directly using std::ranges::subrange:
for (auto const arg : std::ranges::subrange{argv, argv+argc}) {
std::cout << arg << '\n';
}

Answer 5

argv is a double pointer, argv**
So where &x creates a reference to the element in the array, your *arg does not create a reference but rather is the same type as every element in argv.

Answer 6

I take you'd like the range based loop in order to not have to write that much..
so to annoy c++ purists, while at the same time not exactly answering your question, we'd like to post some nicely unsafe ways to do it without any std::vector or OO-wrapper-crap :)
for (;argc--; argv++) {
printf(" %s \n ", *argv);
}
or
while (argc--) {
printf(" %s \n ", *(argv++));
}
or even
while (*argv) {
printf(" %s \n ", *(argv++));
}