char * point to element not the address

Question

void reverse(char * str){
    char * end = str;
    cout << "str" << str << endl;//ABCDE
    cout << "end" << end << endl;//ABCDE
    char tmp;
    if(str){
        while(*end){++end; cout << end << endl;}//ABCDE-->BCDE-->CDE-->DE-->E--> NULL
        --end;//end=E
        cout <<"--end" << end << endl;
        while(str<end){// do swap
            tmp = *str;//*str = str[0] 
            *str++ = *end;//*end = last ele in str[]
            *end-- = tmp;
        }
    }
}

My input is

char test[] = "ABCDE";
cout << test << endl; //ABCDE
reverse(test);
cout << test << endl; //EDCBA

I am feeling not good about the pointer, since c++ primer book says char* pointer to the first element of an array, but when I output the pointer end, it is the content of an array not the address.

Also, reverse(test), I mean to give the address of the first element in an array to the pointer, but it turns out give the whole elements to the pointer.

What is your question exactly ? Don't you have the issue described here ? http://stackoverflow.com/questions/36099466/exc-bad-excess-when-trying-to-reverse-a-c-style-string — Colin Pitrat, Mar 19 '16 at 18:49
Possible duplicate of http://stackoverflow.com/questions/4860788/why-is-address-of-char-data-not-displayed/4860812#4860812 — πάντα ῥεῖ, Mar 19 '16 at 19:16

xvan · Answer 1 · 2016-03-19T19:12:26.840

1

std::cout is overloaded to print strings for char* .

Try:

char *test = "ABCDE";
std::cout << (void *) test << std::endl;

edited Mar 19 '16 at 19:12

answered Mar 19 '16 at 18:56

xvan

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it works, but my question is that char array need to add the void* but when working on the int arrays, I just need to simply cout << test < – flowera Mar 19 '16 at 19:00
So what do you mean by overloaded for char *?? I mean to operate on the address of elements in an array, we need to know the address right? The char array is somehow different from integer array. – flowera Mar 19 '16 at 19:02
`operator<<` has a special definition when the argument is of type `char *` that definition is to print the string instead of the address... See [here](http://en.cppreference.com/w/cpp/io/basic_ostream/operator_ltlt) for all overloads, and [here](http://en.cppreference.com/w/cpp/io/basic_ostream/operator_ltlt2) for string overloads – xvan Mar 19 '16 at 19:09

score 1 · Accepted Answer · answered Mar 19 '16 at 18:57

1

A char* variable is a pointer to a char. A char[] is an array of char. Now, an array of char can be accessed through a pointer, and for char* it is commonly used for string processing (it's used although for other types, but for char it's much more common).

char test[6] = "ABCDE";
char *start = &test[0]; // will point on A

Accessing the array with the pointer can be done with pointer arithmetic:

char *end = start + 5; // equivalent to char *end = &test[5]

Now when you do:

cout << test;

or

cout << start;

It's actually calling an overload of operator<< that takes a const char*. What this operator does is that it print char starting from the pointer passed until it reaches a null char ('\0').

If you want to print the address contained in the pointer and not the string, you have to cast it to void*:

cout << static_cast<void*>(start);

answered Mar 19 '16 at 18:57

Colin Pitrat

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Error 2 error C2679: binary '<<' : no operator found which takes a right-hand operand of type 'overloaded-function' (or there is no acceptable conversion) E:\MyProjects\equilibrium index\string_size.cpp 139 This is the error occurs when I try to cout << &test[0], why? – flowera Mar 19 '16 at 19:08
Your error is strange, don't you have another definition for `test` ? Or is there something else after this on your line (more than just `cout << &test[0];`) ? – Colin Pitrat Mar 19 '16 at 19:15
char test[] = "ABCDEF"; cout << "-- The input add is: --" << (void*)test << endl; cout << "the add of first ele is: " << &test[0] << end; reverse(test); cout << "-- The final result is: --" << test << endl; – flowera Mar 19 '16 at 19:23
end instead of endl is probably your issue – Colin Pitrat Mar 20 '16 at 07:37

char * point to element not the address

2 Answers2