如果我做：
const char* const_str = "Some string";
char* str = const_cast<char*>(const_str); // (1)
str[0] = "P"; // (2)
未定义的行为到底在哪里（哪一行）？
我在 SO 上搜索了很多，但没有找到任何明确和准确的答案（或者至少没有我能理解的答案）。
也相关：如果我使用提供这种功能的外部库：
// The documentation states that str will never be modified, just read.
void read_string(char* str);
可以这样写：
std::string str = "My string";
read_string(const_cast<char*>(str.c_str()));
既然我确定read_string()永远不会尝试写信给str?
谢谢你。
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";