如何使用C ++引用交换两个C样式字符串？

Question

Below code gives me a compilation error, but I don't understand what I am doing wrong. Sorry about asking such a silly question.

$ cat swapcstrings.cc
#include <iostream>

void swap(char*& c, char*& d) {
    char* temp = c;
    c = d;
    d = temp;
}

int main() {
    char c[] = "abcdef";
    char d[] = "ghijkl";
    std::cout << "[" << c << "," << d << "]\n";
    swap(c, d);
    std::cout << "[" << c << "," << d << "]\n";
}
$ g++ swapcstrings.cc
swapcstrings.cc: In function ‘int main()’:
swapcstrings.cc:13: error: invalid initialization of non-const reference of type ‘char*&’ from a temporary of type ‘char*’
swapcstrings.cc:3: error: in passing argument 1 of ‘void swap(char*&, char*&)’
$

Answer 1

Arrays cannot be modified, and they merely decay to temporary pointers, they are not really pointers and cannot be swapped. The address of an array cannot be changed, and the compiler errors when you try to bind the temporary pointer you got from the array to a non- const reference, which is against the rules of the language.

Declare the arrays, then swap two pointers to them.

char a[] = "abcdef";
char b[] = "defghi";

char* aptr = a, *bptr = b;

std::cout << "[" << aptr << "," << bptr << "]\n";
swap(aptr, bptr);
std::cout << "[" << aptr << "," << bptr << "]\n";

Or if you can change the prototype of the function, use const char* in the first place:

void swap(const char*& c, const char*& d) {
    const char* temp = c;
    c = d;
    d = temp;
}

const char* c = "abcdef", // These must be const char* because the arrays are
          * d = "ghijkl"; // const char[N]

std::cout << "[" << c << "," << d << "]\n";
swap(c, d);
std::cout << "[" << c << "," << d << "]\n";

Answer 2

c and d are arrays. They will get autoconverted to pointers where pointers are required. This is the “temporary” your compiler output talks about. Think about it like this:

char c[] = "abcdef", d[] = "ghijkl";
char *cp = (char*)c, *dp = (char*)d;
swap(cp, dp);

The above would compile, but only swap cp and dp , not the original c and d . As the above code gives names to these pointers, you can now have references to them as well. But in your original code, the temporaries had no names, and any modification of them would be an indication of a likely error. So the compiler won't let you do that, and will complain instead.

If you want to exchange C-style strings in their arrays, you'll have to do so one character at a time:

template<size_t n> void swap(char (&a)[n], char (&b)[n]) {
  for (size_t i = 0; i != n; ++i)
    std::swap(a[i], b[i]);
}

This template ensures that both arguments are arrays of the same length.