I have this simple code that compiles without errors/warnings:
void f(int&, char**&){}
int main(int argc, char* argv[])
{
f(argc, argv);
return 0;
}
And next similar code that doesn't compile:
void f(int&, char**&){}
int main()
{
int argc = 2;
char* argv[] = { "", "", nullptr };
f(argc, argv);
//@VS2013 error: cannot convert argument 2 from 'char *[3]' to 'char **&'
//@GCC error: invalid initialization of non-const reference of type 'char**&' from an rvalue of type 'char**'
return 0;
}
Why char*[]
can be converted to char**&
in the first sample and can't be converted in the second sample? Does it matter if the size is known at compile time?
EDIT: I think there are 2 conversions needed in the second case, and only one implicit conversion can be done by compiler.
This code compiles fine:
void f(int&, char**&){}
int main()
{
int argc = 2;
char* temp[] = { "", "", nullptr };
char** argv = temp;
f(argc, argv);
return 0;
}
Because despite appearances, the second argument to main
has
type char**
. When used as the declaration of a function
argument, a top level array is rewritten to a pointer, so char
*[]
is, in fact, char**
. This only applies to function
parameters, however.
A char*[]
(as in your second case) can convert to a char**
,
but the results of the conversion (as with any conversion) is an
rvalue, and cannot be used to initialize a non-const reference.
Why do you want the reference? If it is to modify the pointer,
modifying the char**
argument to main
is undefined behavior
(formally, in C, at least—I've not checked if C++ is more
liberal here). And of course, there's no way you can possibly
modify the constant address of an array. And if you don't want
to modify it, why use a reference?
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