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The char type can only represent a single character What's the difference between char* name which points to a constant string literal, and const char* name When you have a sequence of characters, they are piled next to each other in memory, and the location of the first character in that sequence is returned (assigned to test)

Test is nothing more than a pointer to the memory location of the first character in testing, saying that the type it points to is a char. I want to know what actually happens in terms of storage duration, both at compile and run time. 286 char* and char[] are different types, but it's not immediately apparent in all cases

This is because arrays decay into pointers, meaning that if an expression of type char[] is provided where one of type char* is expected, the compiler automatically converts the array into a pointer to its first element.

} int main() { char *s = malloc(5) // s points to an array of 5 chars modify(&s) // s now points to a new array of 10 chars free(s) } you can also use char ** to store an array of strings

However, if you dynamically allocate everything, remember to keep track of how long the array of strings is so you can loop through each element and free it. Is a pointer to the literal (const) string test The main difference between them is that the first is an array and the other one is a pointer The array owns its contents, which happen to be a copy of test, while the pointer simply refers to the contents of the string (which in this case is immutable).

Technically, the char* is not an array, but a pointer to a char

Similarly, char** is a pointer to a char* Making it a pointer to a pointer to a char Use cout << q to print single character. Char *array = one good thing about music

The difference between char* the pointer and char[] the array is how you interact with them after you create them If you are just printing the two examples, it will perform exactly the same They both generate data in memory, {h, e, l, l, o, /0} The fundamental difference is that in one char* you are assigning it to a pointer, which is a variable

In char[] you are assigning it to an array.

So what is the difference

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