Im a CS student, but I have some experience as a developer as well, which sometimes can make things more difficult, as demonstrated below.In my class my professor asked us to never use the following statement...

    char* str = "Hello world";

According to the Professor...

"This is dangerous (and officially deprecated in the C standard) because you haven't allocated memory for str1 to point at."

I get that its bad, I don't care about that. The statement clearly demonstrates that I am failing to understand the way memory allocation works in regards to the C language. The statement can be ran, even if its deprecated, but I don't what I don't understand, is that if it is not allocating memory to the heap, it wouldn't make sense that it would allocate a variable initialization to the stack, which is where I get confused. Is there another area of memory, or like a cache that C uses specifically for local variable initialization? Where does the string Hello World get placed, and what address is assigned to the pointer in this case?

CodePudding user response：

The problem is in lvalue and rvalue. lvalue defines locator value and it means that it has a specified place in memory and you can easily take it. rvalue has undefined place in memory. For example, any int a = 5 has 5 as rvalue. So you cannot take the address of an rvalue. When you try to access the memory for char* str = "Hello World" with something like str[0] = 'x' you will get an error Segmentation fault which means you tried to get unaviable memory. Btw operators * and & are forbidden for rvalues, it throws compile time error, if you try to use them.

CodePudding user response：

Summary:

• "any strings in double quotes" are const lvalue string literals, stored somewhere in compiled program.

• You can't modify such string, but you can store pointer to this string (of course const) and use it without modifying:

  const char *str = "some string"

  For example:

  int my_strcmp(const char *str1, const char *str2) { ... }
  
  int main() 
  {
      ...
  
      const char *rule2_str= "rule2";
  
      // compare some strings
      if (my_strcmp(my_string, "rule1") == 0)
          std::cout << "Execute rule1" << std::endl;
      else if (my_strcmp(my_string, rule2_str) == 0)
          std::cout << "Execute rule2" << std::endl;
  
      ...
  }
  

• If you want to modify string, you can copy string literal to your own array: char array[] = "12323", then your array will ititialize as string with terminate zero at the end:

  Actually char array[] = "123" is same as char array[] = {'1', '2', '3', '\0'}.

  For example:

  int main()
  {
      char my_string[] = "12345";
      my_string[0] = 5; // correct!
  
      std::cout << my_string << std::endl; // 52345
  }
  

  Remember that then your array will be static, so you can't change it's size, for "dynamic sized" strings use std::string.