cannot convert 'const char **' to 'const char*'

Question

Good morning everyone! I am trying to create a fork/exec call from a parent program via a passed '-e' parameter (e.g. parent -e child key1=val1 ...). As a result I want to copy all the values after the first two in the argv array into a new array child_argv. Something like:

const char *child_argv[10];  // this is actually a global variable
static const char *sExecute;

      int I = 0;
const char *Value = argv[1];
    sExecute = Value;

      for (i=2; i<argc; i++) {
             child_argv[I] = argv[i];
             I++;
      }
    child_argv[I] = NULL;   // terminate the last array index with NULL

This way I can call the exec side via something like:

execl(sExecute, child_argv);

However I get the error message "error: cannot convert 'const char**' to 'const char*' for argument '2' to 'execl(const char*, const char*, ...)'". I have even tried to use an intermediate step:

const char *child_argv[10];  // this is actually a global variable
static const char *sExecute;

      int I = 0;
const char *Value = argv[1];
    sExecute = Value;

      for (i=2; i<argc; i++) {
    const char *Value = argv[i+1];
             child_argv[I] = Value;
             I++;
      }
    child_argv[I] = NULL;   // terminate the last array index with NULL

But I can't figure this out. Any help would be greatly appreciated!

UPDATE

As was pointed out, I should be using 'execv' instead of 'execl' in this situation. Still getting errors though...

UPDATE 2

I ended up copying the array without the desired parameters of argv. See the post here to see the result How to copy portions of an array into another array

Your code doesn't include the `execl` call which is the source of the compile error. But like the error suggests, you are passing in `child_argv` (of type `const char**`) when it wants a `const char*`. — Barry, Aug 26 '15 at 14:42
I recommend you read [the `execl` manual page](http://man7.org/linux/man-pages/man3/exec.3.html), and the reason for the error should be pretty obvious. Actually, if you *read the error message*, you would understand the reason as well. — Some programmer dude, Aug 26 '15 at 14:43
@Barry the 'execl' call was included in the post. How can I be passing a const char** when child_argv was created as an array of const char*? — user1646428, Aug 26 '15 at 14:50

score 2 · Answer 1 · answered Aug 26 '15 at 14:43

2

From here: http://linux.die.net/man/3/exec

I think you mean to call "execv" not "execl". Execl seems to take a variable number of arguments, expecting each const char * to be another argument, while execv takes an array of arguments.

answered Aug 26 '15 at 14:43

Chris Beck

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4
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Oppps, looking back over the reference material, you are right! I updated to use 'execv', but now I'm getting the error "error: invalid conversion from 'const char**' to 'char* const*' [-fpermissive]" – user1646428 Aug 26 '15 at 14:57
you might want to read this I guess, http://stackoverflow.com/questions/7914444/what-are-top-level-const-qualifiers, but it sounds like you need to change `const char *child_argv[10];` to `const char * const child_argv[10];` or something along these lines. – Chris Beck Aug 26 '15 at 15:01
@user the point is that execl expects to be passed an array of constant strings. when you make your type be `const char *child_argv[10]`, that is a constant array of non constant strings. so you need to make it an array of constant strings (in any number of ways) to make the warnings go away. – Chris Beck Aug 26 '15 at 15:09
@Beck I have tried several things but can't figure this out. C++ is not my programming language of choice... Any other thoughts? – user1646428 Aug 26 '15 at 15:28
Really you just need to follow link above and read the answer to that question very carefully. Or, post a new question about this. (But most likely it will be marked as a duplicated.) If you really just want to tap out on this and move on, you could use a `const_cast` to just get the consts right. That will look pretty ugly to C++ programmers though. – Chris Beck Aug 26 '15 at 15:35
@Beck I've been struggling with this for two days. At this point I could care less if it's ugly or not, I just want to get beyond this hurtle. I looked into both links and I'm still in the same position. I will tap out on this and just ask a different question. Thanks for your help though! – user1646428 Aug 26 '15 at 15:42

score 0 · Answer 2 · answered Aug 26 '15 at 15:55

0

You should be using execv. When you convert to execv, for some reason execv expects a array of non-const pointers instead of const pointers, so you either have to just cast the array to (char**) or copy the strings into (char*) pointers.

answered Aug 26 '15 at 15:55

pinkhorror

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1

Thanks for the help pinkhorror! I thought I was copying the strings into (char*) pointers in the second example listed above using the 'Value' variable... apparently not... Any thoughts? – user1646428 Aug 26 '15 at 16:10
Your child_argv is an array of `const char*`. `execv` has an array of (char*) as its expected argument (I don't know why, but it does.) With C types, I'm probably simplifying a bit, but you read right to left. So, `const char* child_argv[10]` is a 10-element array of pointers to chars that are const. execv has this as it's second argument: `char *const argv[]` So, that's an array of const pointers to chars. It promises not to modify the pointers themselves, but the chars they point to are fair game. – pinkhorror Aug 26 '15 at 16:24
thanks for the help! So are you saying I need to change "const char* child_argv[10]" to "char *const argv[]"? Sorry about this... As stated I've worked on this for two days, I'm tired, I'm frustrated, and this problem is holding up development on several fronts. – user1646428 Aug 26 '15 at 18:29
Thanks again for your help, I ended up going at this from a different direction and have updated the post to reflect that. – user1646428 Aug 26 '15 at 20:14

cannot convert 'const char **' to 'const char*'

2 Answers2