关于c ++中char指针的混淆

Question

I'm new in c++ language and I am trying to understand the pointers concept.

I have a basic question regarding the char pointer,

What I know is that the pointer is a variable that stores an address value, so when I write sth like this:

char * ptr = "hello";

From my basic knowledge, I think that after = there should be an address to be assigned to the pointer, but here we assign "hello" which is set of chars. So what does that mean ?
Is the pointer ptr points to an address that stores "hello"? or does it store the hello itself?
Im so confused, hope you guys can help me..

Thanks in advance.

Answer 1

ptr holds the address to where the literal "hello" is stored at. In this case, it points to a string literal. It's an immutable array of characters located in static (most commonly read-only) memory.

You can make ptr point to something else by re-assigning it, but before you do, modifying the contents is illegal. (its type is actually const char* , the conversion to char* is deprecated (and even illegal in C++11) for C compatibility.

Because of this guarantee, the compiler is free to optimize for space, so

char * ptr = "hello";
char * ptr1 = "hello";

might yield two equal pointers. (ie ptr == ptr1 )

Answer 2

The pointer is pointing to the address where "hello" is stored. More precisely it is pointing the 'h' in the "hello".

Answer 3

"hello" is a string literal : a static array of characters. Like all arrays, it can be converted to a pointer to its first element, if it's used in a context that requires a pointer.

However, the array is constant, so assigning it to char* (rather than const char* ) is a very bad idea. You'll get undefined behaviour (typically an access violation) if you try to use that pointer to modify the string.

Answer 4

编译器将“找到某个地方”，它可以将字符串"hello" ，而ptr将具有“某处”的地址。

Answer 5

When you create a new char* by assigning it a string literal, what happens is char* gets assigned the address of the literal. So the actual value of char* might be 0x87F2F1A6 (some hex-address value). The char* points to the start (in this case the first char) of the string. In C and C++, all strings are terminated with a /0, this is how the system knows it has reached the end of the String.

Answer 6

char* text = "Hello!" can be thought of as the following:

At program start, you create an array of chars, 7 in length: {'H','e','l','l','o','!','\\0'} . The last one is the null character and shows that there aren't any more characters after it. [It's more efficient than keeping a count associated with the string... A count would take up perhaps 4 bytes for a 32-bit integer, while the null character is just a single byte, or two bytes if you're using Unicode strings. Plus it's less confusing to have a single array ending in the null character than to have to manage an array of characters and a counting variable at the same time.]

The difference between creating an array and making a string constant is that an array is editable and a string constant (or 'string literal') is not. Trying to set a value in a string literal causes problems: they are read-only.

Then, whenever you call the statement char* text = "Hello!" , you take the address of that initial array and stick it into the variable text . Note that if you have something like this...

char* text1 = "Hello!";
char* text2 = "Hello!";
char* text3 = "Hello!";

...then it's quite possible that you're creating three separate arrays of {'H','e','l','l','o','!','\\0'} , so it would be more efficient to do this...

char* _text = "Hello!";
char* text1 = _text;
char* text2 = _text;
char* text3 = _text;

Most compilers are smart enough to only initialize one string constant automatically, but some will only do that if you manually turn on certain optimization features.

Another note: from my experience, using delete [] on a pointer to a string literal doesn't cause issues, but it's unnecessary since as far as I know it doesn't actually delete it.