Pointer arithmetic and const qualifier in C

Question

In the following piece of code, to calculate strlen,

int s(const char* str)
{   
    int count=0;        
    while(*str++) count++;
    return count;
}

You can see that the argument str is const. But, the compiler does not complain when I do a str++. My question is

When passing pointers as arguments to a C function, if is is qualified with const, How can I still perform pointer arithmetic on it? What is const in the above function?

Answer 1

const char* str;

means a non-const pointer to a const data.

char* const str;

means a const pointer to a non-const data.

const char* const str;

means a const pointer to a const data.

The reason for this is that in C++ the variable type declarations are parsed from right to left, which results in that the word "const" always defines the constness of the thing that it's closest to.

Answer 2

It's not declaring the pointer const , it's declaring the thing pointed to as const . Try this:

int s(const char* const str)

With this declaration, you should get a compile error when you modify str .

Answer 3

const char * ch; // non-const pointer to const data
char * const ch; // const pointer to non-constant data.

const char * const ch; // const pointer to const data

Note: Also

const char * ch;

equals to

char  const * ch;

Answer 4

指针指向一个const char ，一个只读字符数组。