To understand what's going on here, two language rules are important:
- Arrays are not assignable.
- An array can be converted to a pointer to its first element.
It's also important to understand what a string literal like "Sunstroke"
is. It's a static array of constant characters, large enough to hold all the characters of a string with a terminator at the end. So in this case, it's a const char[10]
array, containing the nine characters followed by the zero-valued terminator. Being static, the array is stored somewhere in memory for the lifetime of the program.
char a[] = "Sunstroke";
This creates a local array, and initialises it by copying the characters from the string literal.
char *b = "Coldwave";
This creates a pointer, and initialises it to point to the literal itself. Note that this is dangerous: the literal is const
, but the pointer isn't, so you can write code that attempts to modify the literal, giving undefined behaviour. This conversion is deprecated (certainly in C++, I'm not sure about C), so the compiler should give you a warning. You have enabled all the compiler warnings you can, haven't you?
a = "Coldwave";
This attempts to reassign the array, but fails because arrays aren't assignable. There's no particularly good reason why they aren't; that's just the way the languages evolved.
b = "Sunstroke";
This reassigns the pointer to point to a different literal. That's fine (apart from the lack of const
noted above).
If you need to manipulate strings, then:
- in C you'll need to carefully create arrays large enough for your needs, and use the library functions in
<string.h>
(or your own handcrafted code) to manipulate the characters in those arrays;
- in C++, use the
std::string
class to handle memory management, assignment, etc. for you.
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