问题
void f(char* p)
{}
int main()
{
f("Hello"); // OK
auto p = "Hello";
f(p); // error C2664: 'void f(char *)' : cannot convert parameter 1
// from 'const char *' to 'char *'
}
The code was compiled with VC++ Nov 2012 CTP.
§2.14.15 String Literals, Section 7
A narrow string literal has type “array of n const char”, where n is the size of the string as defined below, and has static storage duration.
Why is f("Hello") OK?
回答1:
This behaviour differs between C and C++, at least in theory.
In C: a string literal decays to a non-const pointer. However, that doesn't make it a good idea; attempting to modify the string through that pointer leads to undefined behaviour.
In C++: it's never ok (AFAIK).* However, some compilers may still let you get away with it. GCC, for example, has the -Wwrite-strings flag, which is enabled by default (at least in 4.5.1 onwards).
* In C++11, at least. (I don't have older specs to hand.)
回答2:
The difference between
f("Hello");
and
f(p);
is that the former involves a literal. In C++03 conversion from string literal to char* (note: not const) was supported. It isn't supported any longer in C++11, but few if any compilers have yet caught up with that rule change.
回答3:
f("Hello");
Even this is not okay in C++. The compiler should give diagnostic, or else it needs to be updated.
In C++, "Hello" is convertible to const char*, not char*.
The conversion from "Hello" to char* is allowed in C++03, though it is deprecated. And in C++11, the conversion is invalid, and the code is ill-formed.
回答4:
I think because auto keyword. it's type deduction so compiler doesn't know it how to convert to char* anymore.
来源：https://stackoverflow.com/questions/14415488/why-can-a-string-literal-be-implicitly-converted-to-char-only-in-certain-case