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# 1  
Old 05-07-2013
Pointer and address
This code is to print out the program name and arguments list one by one:
Code:
  1 #include<stdio.h>
  2 
  3 void main(int argc, char *argv[])
  4 {
  5         int iCount = 0;
  6         while (iCount < argc) {
  7                 printf("argc:%d\t%s\n",iCount, argv[iCount]);
  8                 iCount++;
  9         }
 10 }
 11

I understand it is about pointer and pointer address of the array. Can somebody explain to me why line 7 should not be as :
Code:
printf("argc:%d\t%s\n",iCount, *argv[iCount]);

I thought *argv[iCount] is the correct way as argv[iCount] is one of the array members address, not the content of the pointer pointing to. But my understanding is wrong! What did I miss? Thanks a lot!
Last edited by yifangt; 05-07-2013 at 04:14 PM..
# 2  
Old 05-07-2013
argv is an array of pointers to arrays of char, so first think of it at char**, something that needs one dereferencing ( * or [#] ) to get to char*, which C uses as string. The actual rules for argv is that the last element is a null pointer, so you can size it somewhat analagous to strlen(). The argc is just for your convenience, provided by the loader. char** could be char[][] which is a 2 dimensional heap of characters locally, but the name would still need dereferencing. Since printf's '%s' demands a char*, one layer of dereferencing is right. Another way to think of '*' is '[0]', so **argv is the first char of argv[0], is argv[0][0]. Array names are of type type* even if the array is local, but local arrays have size of the array not size of a pointer. So, 'char *x = "Y" ;' is a local pointer variable initialized with a pointer size 4 to a constant string, two hunks of storage, but 'char x[] = "Y" ;' is a local array size 2 of characters Y and null (sized implicitly by the initializer), one small hunk of storage, and there is no pointer storage, just an array name that is always a pointer.
Last edited by DGPickett; 05-07-2013 at 04:35 PM..
# 3  
Old 05-07-2013
Quote:
Can somebody explain
Suppose you had "int int_array [12]". Then "int_array [n]" would be an int.

argv is declared as "char *argv []" So "argv [n]" is a "char *".

"char *" is what "printf %s" expects and what makes sense, because you are trying to print a string.

"argv [iCount]" is the content of an array slot. The content happens to be an address, because "char *" is an address.

Suppose "argv [1]" is "123456". Then "*argv [1]" is '1', the single character 1. You could do an experiment and change the code to the following and verify this (untested).

Code:
printf("argc:%d\t%c\n",iCount, *argv[iCount]);

The basic answer is that argv[iCount] is compatible with %s and *argv[iCount] is compatible with %c.
This User Gave Thanks to hanson44 For This Post:
yifangt
# 4  
Old 05-07-2013
pointer and address
Actually I did many similar test as you suggested, but I always got error message like:
Code:
program.c: In function ‘main':
program.c:7:3: warning: format ‘%c' expects argument of type ‘int', but argument 3 has type ‘char *' [-Wformat]

I think this is the point that I did not catch
Code:
argv [iCount] , The content  happens to be an address,

My confusion was also with a similar code I was trying:
Code:
1 #include<stdio.h>
  2 
  3 void Print(char *[]);
  4 void main()
  5 {
  6         char * pn[] = {"Fred", "Barney", "Betty", "Wilma", NULL};
  7         Print(pn);
  8 }
  9 void Print(char * arr[])
 10 {
 11         while (*arr != NULL) {
 12                 printf("arr %p \t %s\n", &arr, *arr);
 13                 arr++;
 14         }
 15 }

In this code line 12 was what I expected and printed out correct thing.
and the output is:
Code:
arr 0x7fffd4f3bb58      Fred
arr 0x7fffd4f3bb58      Barney
arr 0x7fffd4f3bb58      Betty
arr 0x7fffd4f3bb58      Wilma

Why this time in line 12 *arr was used instead of arr?
# 5  
Old 05-07-2013
Yes, %s is for char*, a pointer to a null terminated string like "Hi!", a pass by reference

but %c is for just one char, like '!'. A char is 8 bits treated as an unsigned integer and passed here by value, just like int to %d.
# 6  
Old 05-07-2013
Quote:
Why this time in line 12 *arr was used instead of arr
Because arr[0] is exactly the same as *arr

In general, arr[i] is exactly the same as *(arr + 1)

This is one of the most difficult and fundamental concepts of C programming. It took me years of reading and re-reading to "get it". You need to keep at this, because it's so important to understand pointers and arrays.

char *arr[] is an array of pointers. The loop keeps incrementing the address of the start of the array. Each time, it tests the contents of the first element, and if == NULL stops the loop. Each time, it prints the contents of the first element of the array, because *arr and arr[0] mean the same thing.

--------------------------------

Quote:
I always got error message like:
If you can compile, run, and understand this, it will likely help:
Code:
$ cat array.c
#include<stdio.h>

void main(int argc, char *argv[])
{
        int iCount;

        for (iCount = 0; iCount < argc; iCount++) {
                printf("argc:%d\t%s\n",iCount, argv[iCount]);
        }
        for (iCount = 0; iCount < argc; iCount++) {
                printf("argc:%d\t%c\n",iCount, *argv[iCount]);
        }
        for (iCount = 0; iCount < argc; iCount++) {
                printf("argc:%d\t%c\n",iCount, argv[iCount][0]);
        }
}

Code:
$ gcc array.c
$ a.out 123 xyz
argc:0  a.out
argc:1  123
argc:2  xyz
argc:0  a
argc:1  1
argc:2  x
argc:0  a
argc:1  1
argc:2  x

The second two loops are doing exactly the same thing, just written two different ways, *argv[iCount] and argv[iCount][0]
This User Gave Thanks to hanson44 For This Post:
yifangt
# 7  
Old 05-07-2013
Thanks a lot! I had thought I had understood pointer, but actually not at all!
Can I say the address of a pointer is an array of char there in these two cases (strings!)? In the second sample *arr was printed out by looping thru the actual char array, but argv[iCount] was not looped thru, where *argv[iCount] was only the first char of its contents. Is this correct?
Thank you so much,
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