char* 和 char[] 之间的区别

Question

I know this is a very basic question. I am confused as to why and how are the following different.

char str[] = "Test";
char *str = "Test";

Answer 1

char str[] = "Test";

Is an array of chars , initialized with the contents from "Test", while

char *str = "Test";

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).

Answer 2

The diference is the STACK memory used.

For example when programming for microcontrollers where very little memory for the stack is allocated, makes a big difference.

char a[] = "string"; // the compiler puts {'s','t','r','i','n','g', 0} onto STACK 

char *a = "string"; // the compiler puts just the pointer onto STACK 
                    // and {'s','t','r','i','n','g',0} in static memory area.

Answer 3

A pointer can be re-pointed to something else:

char foo[] = "foo";
char bar[] = "bar";

char *str = foo;  // str points to 'f'
str = bar;        // Now str points to 'b'
++str;            // Now str points to 'a'

The last example of incrementing the pointer shows that you can easily iterate over the contents of a string, one element at a time.

Answer 4

One is pointer and one is array. They are different type of data.

int main ()
{
   char str1[] = "Test";
   char *str2 = "Test";
   cout << "sizeof array " << sizeof(str1) << endl;
   cout << "sizeof pointer " << sizeof(str2) << endl;
}

output

sizeof array 5
sizeof pointer 4

Answer 5

The first

char str[] = "Test";

is an array of five characters, initialized with the value "Test" plus the null terminator '\\0' .

The second

char *str = "Test";

is a pointer to the memory location of the literal string "Test" .

Answer 6

Starting from C++11, the second expression is now invalid and must be written:

const char *str = "Test";

The relevant section of the standard is Appendix C section 1.1:

Change : String literals made const

The type of a string literal is changed from “array of char” to “array of const char.” The type of a char16_t string literal is changed from “array of some-integer-type” to “array of const char16_t.” The type of a char32_t string literal is changed from “array of some-integer-type” to “array of const char32_t.” The type of a wide string literal is changed from “array of wchar_t” to “array of const wchar_t.”

Rationale : This avoids calling an inappropriate overloaded function, which might expect to be able to modify its argument.

Effect on original feature : Change to semantics of well-defined feature.

Answer 7

"Test" is an array of five characters (4 letters, plus the null terminator.

char str1[] = "Test"; creates that array of 5 characters, and names it str1 . You can modify the contents of that array as much as you like, eg str1[0] = 'B';

char *str2 = "Test"; creates that array of 5 characters, doesn't name it, and also creates a pointer named str2 . It sets str2 to point at that array of 5 characters. You can follow the pointer to modify the array as much as you like, eg str2[0] = 'B'; or *str2 = 'B'; . You can even reassign that pointer to point someplace else, eg str2 = "other"; .

An array is the text in quotes. The pointer merely points at it. You can do a lot of similar things with each, but they are different:

char str_arr[] = "Test";
char *strp = "Test";

// modify
str_arr[0] = 'B'; // ok, str_arr is now "Best"
strp[0] = 'W';    // ok, strp now points at "West"
*strp = 'L';      // ok, strp now points at "Lest"

// point to another string
char another[] = "another string";
str_arr = another;  // compilation error.  you cannot reassign an array
strp = another;     // ok, strp now points at "another string"

// size
std::cout << sizeof(str_arr) << '\n';  // prints 5, because str_arr is five bytes
std::cout << sizeof(strp) << '\n';     // prints 4, because strp is a pointer

for that last part, note that sizeof(strp) is going to vary based on architecture. On a 32-bit machine, it will be 4 bytes, on a 64-bit machine it will be 8 bytes.

Answer 8

Let's take a look at the following ways to declare a string:

char name0 = 'abcd';           // cannot be anything longer than 4 letters (larger causes error)
cout << sizeof(name0) << endl; // using 1 byte to store

char name1[]="abcdefghijklmnopqrstuvwxyz"; // can represent very long strings
cout << sizeof(name1) << endl;             // use large stack memory

char* name2 = "abcdefghijklmnopqrstuvwxyz"; // can represent very long strings
cout << sizeof(name2) << endl;              // but use only 8 bytes

We could see that declaring string using char* variable_name seems the best way! It does the job with minimum stack memory required.