const char * vs const char ** on argv

Question

I want to get a filename from a commandline argument so I can pass it to a file-opening function, however, argv[] seems to be of type const char ** by default, even if it was defined as "const char * argv[]" in the main's arguments. My function requires const char *, is there a way to convert this two, or something?

int main(int argc, const char *argv[]) {
    memory Memory;
    Memory.loadROM(argv); // Error 'argument of type "const char **" is incompatible with parameter of type "const char *"'
}

Note that `const char *var[]` or `const char **var` is rarely what you really mean. If you don't want to change the array elements, use `const char *const var[]` or `const char *const *var` instead. — aschepler, May 18 '18 at 00:28
@aschepler Nope. `const char * const * const var` is right. Without the last `const`, you can do `var = some_other_thing`. — iBug, May 18 '18 at 01:04

score 5 · Accepted Answer · answered May 18 '18 at 01:03

When used in function parameters without being a reference type, const char *argv[] and const char **argv is exactly identical because the former will be adjusted to the latter.

Per N4296, Section 8.3.5p5:

any parameter of type “array of T” or “function returning T” is adjusted to be “pointer to T” or “pointer to function returning T,” respectively

So const char *argv[] is "an array of type const char *" and is thus adjusted to "a pointer to const char *".

And since argv is of type const char **, you want to dereference it to get a const char *:

Memory.loadROM(argv[1])

score 2 · Answer 2 · answered May 18 '18 at 00:24

2

argv contains an array of char *, so if you only want one, simplify specify which you want:

Memory.loadROM(argv[1]);

Read here for more on what'll be in argv.

answered May 18 '18 at 00:24

scohe001

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tadman · Answer 3 · 2018-05-18T00:34:23.873

-2

The difference between char* argv[] and char** argv is actually pretty subtle, so you'll often see them used interchangeably. The typical default is char** argv since that's more conventional, yet the alternative isn't wrong.

In both cases argv[0] is the first argument. Remember that pointers act like arrays and char* argv[] is an array of char*.

edited May 18 '18 at 00:34

answered May 18 '18 at 00:26

tadman

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Subtle? I doubt they have any difference (when used as a function parameter). – iBug May 18 '18 at 00:35
One example of a subtle difference is in the case of `char** argv` you can do `argv++` and `argv = x`, but this is not allowed in the case of `char* argv[]`. These are not things you'd normally do, especially with `argv`, but it's one of those things that's *technically* true and C and C++ are all about technicalities. – tadman May 18 '18 at 00:38
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@tadman: That is false. What you say is true if `argv` was actually an array. But a parameter named `char *argv[]` is exactly `char** argv`. It is in fact a pointer, not an array. Despite the misleading syntax. – Benjamin Lindley May 18 '18 at 00:40
@tadman While your claim looks really intuitive, it's actually *not* defined that way. See my answer above. – iBug May 18 '18 at 03:46

const char * vs const char ** on argv

3 Answers3