Why the C string and Char Array and Char pointer differences in C++

Question

Why does this:

const char cwords = "These are even more words";

result in an **error**: cannot initialize a variable of type 'const char' with an lvalue of type 'const char [22]'

but this:

const char * cwordsp = "These are more words"; Work? (Not result in an error)

It seems like the pointer cwordsp should need to point to the memory address of that c string. Am I wrong?

Answer 1

A C-string is nothing more than an array of characters.

So in addition to your working example, you could also do something like this:

const char cString[] = "Hello world";

Which is basically equivalent to this:

const char cString[] = { 'H', 'e', 'l', 'l', 'o', ' ', 'w', 'o', 'r', 'l', 'd' };

Note that this is an array of char s, not a single char .

The reason that you run into problems with this:

const char cString = "Hello world";

is because "Hello world"; can't possibly be interpreted as a char . A char type expects just a single character. Like this:

const char c = 'h';

Answer 2

In const char cwords = "These are even more words"; you are initializing an 8-bit integer ( const char ) with a pointer to const char . This is an illegal conversion.

Answer 3

Because char is 1 byte.

You are trying to put more than 1 byte in to a datatype which holds 1 byte. const char cwords = "These are even more words";

In this example you create an array and get a pointer pointing to the first char in the array const char * cwordsp = "These are more words";

A pointer array and a regular array is almost the same thing, but only almost. You can read more about that: C/C++ int[] vs int* (pointers vs. array notation). What is the difference?