If I do:
const char* const_str = "Some string";
char* str = const_cast<char*>(const_str); // (1)
str[0] = "P"; // (2)
Where (which line) exactly is开发者_开发百科 the undefined behavior ?
I've been searching a lot for this on SO but haven't found any explicit and precise answer (or at least, none that I could understand).
Also related: if I use an external library which provides this kind of function:
// The documentation states that str will never be modified, just read.
void read_string(char* str);
Is it ok to write something like:
std::string str = "My string";
read_string(const_cast<char*>(str.c_str()));
Since I know for sure that read_string()
will never try to write to str
?
Thank you.
Line (2) has undefined behaviour. The compiler is at liberty to place constants in read-only memory (once upon a time in Windows this would have been a "data segment") so writing to it might cause your program to terminate. Or it might not.
Having to cast const-ness away when calling a poorly-defined library function (non-const parameter which should be const) is, alas, not unusual. Do it, but hold your nose.
You are attempting to modify a constant string which the compiler may have put into a read-only section of the process. This is better:
char str[32];
strcpy(str, "Some string");
str[0] = "P";
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