问题
Update
turns out this is just another case of "c++ is not c blues"
What I want
const char hex[16] = "0123456789ABCDEF";
the only thing that works
char hex[16] = "0123456789ABCDE"; hex[15] = "F";
are there any compiler options or something I can do to make strings not null terminated in the gcc compiler. so that I can make a(n) constant array
回答1:
No need for a compiler option, it's already non-NUL terminated. The standard says a NUL should only be added if it can fit, otherwise it would be an overflow. It may just be that the next byte in memory past your array is \0
§ 6.7.8p14
An array of character type may be initialized by a character string literal, optionally enclosed in braces. Successive characters of the character string literal (including the terminating null character if there is room or if the array is of unknown size) initialize the elements of the array.
回答2:
No. NUL-terminated strings are intrinsic to the language. You can have a character array though, and set each character one by one:
char hex [] = {'0', '1', '2', ... 'F'};
回答3:
You answered your own question. If you explicitly give the array a length, as in:
const char hex[16] = "0123456789ABCDEF";
then of course it won't be null-terminated because there is no storage reserved for null termination. (hex[16] is outside the bounds of the object and thus reading or writing it is undefined behavior. If it happens to read as 0, that's UB for ya...)
It's only if you leave the length implicit, as in:
const char hex[] = "0123456789ABCDEF";
or if you use the string literal as an object rather than as an initializer, that it will have null termination.
By the way, why do you care if the null termination is there or not, if you're not planning to use it. Are you trying to shave bytes off your binary? :-)
回答4:
Strings are null terminated in C. If you want to populate a non-null-terminated char array you can use an array initializer.
回答5:
I believe the question is a bit unclear: In C, the qoted initialization:
static const char hex[16] = "0123456789ABCDEF";
is legal. In C++, it is not. So it is one of the pieces of code, that fail (fortunately at compile time), when you move from C to C++.
It would be nice to have a way to force string literals without termination \0 byte. Something like:
static const char hex[16] = "0123456789ABCDEF\!0";
where the \!0 at the end tells the compiler to not zero-terminate the string! \! or even \!0 anywhere else in the string would behave unmodified, so just put out a literal ! or !0 .
来源：https://stackoverflow.com/questions/4348154/non-null-terminated-string-compiler-option-for-gcc