（C ++）关于语法的非常基本的问题

Question

C++ novice here. I have some basic questions. In int main( int argc, char *argv[] )

1. How is char *argv[] supposed to be read (or spoken out to humans)?
2. Is it possible to clear/erase specific content(s), character(s) in this case, of such array? If yes, how?
3. Can arrays be resized? If yes, how?
4. How can I copy the entire content of argv[] to a single std::string variable?
5. Are there other ways of determining the number of words / parameters in argv[] without argc ? If yes, how? (*)

I'd appreciate explanations (not code) for numbers 2-5. I'll figure out the code myself (I learn faster this way).

Thanks in advance.

(*) I know that main(char *argv[]) is illegal. What I mean is whether there's at least a way that does not involve argc at all, like in the following expressions:

for( int i = 0; i < argc; ++i ) {
    std::cout << argv[i] << std::endl;
}

and

int i = 0;    
while( i < argc ) {
    std::cout << argv[i] << std::endl;
    ++i;
}

Or

int i = 0;
do { 
     std::cout << argv[i] << std::endl;
     ++i; } while( i < argc );

Answer 1

1. It's an array of pointers to char.

2. Sort of - you can overwrite them.

3. Only by copying to a new array.

4. Write a loop and append each argv[i] to a C++ string.

5. Most implementations terminate the array with a NULL pointer. I can't remember if this is standard or not.

Answer 2

char **argv[]

Is wrong. It should be either char **argv or char *argv[] , not a mixture of both. And then it becomes a pointer-to-pointer to characters, or rather a pointer to c-strings, ie, an array of c-strings. :) cdecl.org is also quite helpful at thing like this.
Then, for the access, sure. Just, well, access it. :) argv[0] would be the first string, argv[3] would be the 4th string. But I totally wouldn't recommend replacing stuff in an array that isn't yours or that you know the internals of.
On array resize, since you're writing C++, use std::vector , which does all the complicated allocation stuff for you and is really safe. Generally, it depends on the array type. Dynamically allocated arrays ( int* int_arr = new int[20] ) can, static arrays ( int int_arr[20] ) can't.
To copy everything in argv into a single std::string , loop through the array and append every c-string to your std::string . I wouldn't recommend that though, rather have a std::vector<std::string> , ie, an array of std::strings , each holding one of the arguments.

std::vector<std::string> args;
for(int i=0; i < argc; ++i){
  args.push_back(argv[i]);
}

On your last point, since the standard demands argv to be terminated by a NULL pointer, it's quite easy.

int myargc = 0;
char** argv_copy = argv;
while(++argv_copy)
  ++myargc;

The while(++argv_copy) will first increment the pointer of the array, letting it point to the next element (eg, after the first iteration it will point at c-string #2 ( argv[1] )). After that, if the pointer evaluates to false (if it is NULL ), then the loop brakes and you have your myargc . :)

Answer 3

1. Several options: array of pointer to char OR array of C-string.

2. You can assign to particular characters to clear them, or you can shift the rest of the array forwards to "erase" characters/elements.

3. Normal C-style arrays cannot be resized. If you need a resizable array in C++ you should use std::vector .

4. You'll have to iterate over each of the items and append them to a string. This can be accomplished with C++ algorithms such as copy in conjunction with an ostream_iterator used on an ostringstream .

5. No. If there was such a way, there wouldn't be any need for argc . EDIT: Apparently for argv only the final element of the array is a null pointer.

Answer 4

1. char *argv[] can be read as: "an array of pointers to char"

   char **argv can be read as: "a pointer to a pointer to char"

2. Yes, you may modify the argv array. For example, argv[0][0] = 'H' will modify the first character of the first parameter. If by "erase/clear" you mean remove a character from the array and everything automatically shift over: there is no automatic way to do that - you will need to copy all the characters one-by-one over to the left (including the NULL termination)

3. No, arrays cannot be resized. You will need to create a new one and copy the contents

4. How do you want to represent ALL the parameter strings as 1 std::string? It would make more sense to copy it to an array of std::strings

5. No, there is no special indication of the last entry in the array. you need to use argc

Answer 5

1. It is spoken as "array of pointers to pointers to character" (note that this is not the signature of the main function, which is either int argc, char **argv or int argc, char *argv[] -- which is equivalent).
2. The argv array is modifiable (lack of const). It is illegal to write beyond the end of one of the strings though or extend the array; if you need to extend a string, create a copy of it and store a pointer in the array; if you need to extend the array, create a copy of the array.
3. They cannot be resized per se, but reinterpreted as a smaller array (which sort of explains the answer to the last question).
4. You will be losing information this way -- argv is an array of arrays, because the individual arguments have already been separated for you. You could create a list of strings using std::list<std::string> args(&argv[1], &argv[argc]); .
5. Not really. Most systems have argv NULL terminated, but that is not a guarantee.

Answer 6

1) It is supposed to be char **argv or char *argv[] which is a pointer to an array of characters more commonly known as an array of strings

2) CString is the std library to manipulate C strings (arrays of characters). You cannot resize an array without reallocating, but you can change the contents of elements by referencing it by index:

for(int i = 0; i < argc; ++i)
{
   //set all the strings to have a null character in the
   //first slot so all Cstring operations on this array, 
   //will consider it a null (empty) string
   argv[i] = 0;
}

3) Technically no, however they can be deleted then reallocated:

int *array = new int[15]; //array of size 15
delete[] array;
array = new int[50]; //array of size 50

4) This is one way:

string *myString;
if(argc > 0)
{
   myString = new string(argv[0]);
   for(int i = 1; i < argc; ++i)
      myString->append(argv[i]);
}

5) Yes, according to Cubbi:

POSIX specifies the final null pointer for argv, see for example "The application shall ensure that the last member of this array is a null pointer." at pubs.opengroup.org/onlinepubs/9699919799/functions/exec.html

Which means you can do:

char *val = NULL;
int i = 0;
do
{
   val = argv[i++]; //access argv[i], store it in val, then increment i
   //do something with val
} while(val != NULL); //loop until end of argv array

Answer 7

Array in C/C++ is not an object, but just a pointer to first element of array, so you cannot simply delete or insert values.

Answering your questions:

1. char *argv[] can be read as 'array of pointers to char '
2. It's possible, but involves direct manipulations with data in memory, such as copying and/or moving bytes around.
3. No. But you may allocate new array and copy necesary data
4. By manually copying each element into std::string object
5. No.

As a summary: C++ is much more low-level language that you think.