c++ - 使用 forrange 循环打印 argv

Question

int main(int argc, const char** argv) {

    std::cout << "Hello" << std::endl;

    char arr2d[][4] = {"ABC", "DEF"};

    for (char *i : arr2d)
    {
        std::cout << i << std::endl;
    }

在这里，我评估 forrange 的工作如下:“对于 arr2d 中的每个字符数组，将其打印到控制台”。这是可行的，所以，至少我的理解应该是正确的。上述代码片段的输出是，

muyustan@mint:~/Desktop/C_Files/oop$ g++ main.cpp -o main && ./main
Hello
ABC
DEF

正如预期的那样。

但是，如果我尝试这个，

int main(int argc, const char** argv) {

    std::cout << "Hello" << std::endl;

    char arr2d[][4] = {"ABC", "DEF"};

    for (const char *i : argv)
    {
        std::cout << i << std::endl;
    }

首先 IDE 警告我，

this range-based 'for' statement requires a suitable "begin" function and none was found

如果我尝试编译，我会得到:

muyustan@mint:~/Desktop/C_Files/oop$ g++ main.cpp -o main && ./main
main.cpp: In function ‘int main(int, const char**)’:
main.cpp:30:26: error: ‘begin’ was not declared in this scope
     for (const char *i : argv)
                          ^~~~
main.cpp:30:26: note: suggested alternative:
In file included from /usr/include/c++/7/string:51:0,
                 from /usr/include/c++/7/bits/locale_classes.h:40,
                 from /usr/include/c++/7/bits/ios_base.h:41,
                 from /usr/include/c++/7/ios:42,
                 from /usr/include/c++/7/ostream:38,
                 from /usr/include/c++/7/iostream:39,
                 from main.cpp:1:
/usr/include/c++/7/bits/range_access.h:105:37: note:   ‘std::begin’
   template<typename _Tp> const _Tp* begin(const valarray<_Tp>&);
                                     ^~~~~
main.cpp:30:26: error: ‘end’ was not declared in this scope
     for (const char *i : argv)
                          ^~~~
main.cpp:30:26: note: suggested alternative:
In file included from /usr/include/c++/7/string:51:0,
                 from /usr/include/c++/7/bits/locale_classes.h:40,
                 from /usr/include/c++/7/bits/ios_base.h:41,
                 from /usr/include/c++/7/ios:42,
                 from /usr/include/c++/7/ostream:38,
                 from /usr/include/c++/7/iostream:39,
                 from main.cpp:1:
/usr/include/c++/7/bits/range_access.h:107:37: note:   ‘std::end’
   template<typename _Tp> const _Tp* end(const valarray<_Tp>&);

那么，为什么 argv 的行为与我的 arr2d[][4] 不同？它们不都是 char 指针(char 数组或字符串(？))的指针吗？

如果我的理解有问题，用 forrange 打印 argv 成分的结构应该是什么？

Answer 1

this is something I have been told wrong then, when I was dealing with C back then, at somewhere I have read that "arrays are also pointers!".

关于该声明，我们必须理解一些更细微的要点。

1. 在大多数情况下，数组都会退化为指针，但数组仍然与指针不同。

   • 当用作 sizeof 的参数时，以下两个将导致不同的答案。

     char const* ptr = "Some text.";
     char array[] = "some text.";
     
     std::cout << sizeof(ptr) << std::endl;    // prints sizeof the pointer.
     std::cout << sizeof(array) << std::endl;  // prints sizeof the array.
     

   • 当用作 addressof 运算符的参数时。

     char const* ptr1 = "Some text.";
     char array[] = "some text.";
     
     char const** ptr2 = &ptr1;      // OK.
     char** ptr3 = &array;           // Error. Type mismatch.
     char (*ptr4}[11] = &array;      // OK.
     

2. 二维数组可以衰减为指向一维数组的指针，但它们不会衰减为指向指针的指针。

    int array1[10];
    int* ptr1 = array1;          // OK. Array decays to a pointer
   
   
    int array2[10][20];
    int (*ptr2)[20] = array2;    // OK. 2D array decays to a pointer to 1D array.
    int**  ptr3 = array2;        // Error. 2D array does not decay to a pointer to a pointer.