Assigning array to pointer confusion

Question

I was confused by a line of code I found in a tutorial on C. Here is the code:

int main(int argc, char *argv[]){
   ...
   char **inputs = argv+1; // This is the confusing line
   ...
   return 0;
}

I can't understand how can you assign an array to a pointer like that. I would be glad if someone could clarify this for me. Thanks ahead!

Answer 1

Say you execute a program like this

C:\Temp>myprog.exe hello world

the operating system takes these strings and puts them together, in an array of null-terminating strings:

{ "myprog.exe", "hello", "world", NULL } 

Then it calls main() and passes it the number of strings (3) as argc and a pointer to the first string in this array. this pointer is calles argv, and is of type char** (char* argv[] is just a syntactic convenience, semantically equivalent inside function signatures)

but you want inputs to hold only the string "hello" and "world", so you takes this pointer, argc, and point to the next element - add one to it:

char **inputs = argv+1;

now inputs points toward { "hello", "world", NULL } .

Answer 2

argv is an array pointers to strings, last pointer is null. Suppose your executable name is exe and you run it like:

$ exe fist second 

then argv is:

                     +----------+ 
             argv---►| "exe"    | 
                     +----------+ 
         argv + 1---►| "first"  | 
                     +----------+ 
         argv + 2---►| "second" | 
                     +----------+      
         argv + 3---►|  null    |      
                     +----------+      

    * Notice last is null. 

So char** input = argv + 1 points to "first" string that is first input argument.

if your prints argv[0] with %s output will be "exe" that is your executable name and if you prints input[0] with %s output will be "fisrt" string.

Note: even if you don't pass any argument intput will point to NULL (valid address).
(purpose of this is to point to input arguments strings, or say skip program name "exe")

My following code example, and its set of outputs will help you to understand.
code.c:

#include<stdio.h>
int main(int argc, char* argv[]){
    char** input = argv + 1;
    while(*input) /* run untill input-->null */
        printf("%s \n", *input++);
    return 1;
}

outputs:

~$ gcc code.c  -Wall  -pedantic  -o exe 
~$ ./exe 
~$ ./exe first
first 
~$ ./exe first second
first 
second 

Answer 3

Array Name is a constant pointer to the first element of array

MORE:- http://forum.allaboutcircuits.com/showthread.php?t=56256

Answer 4

Arrays used in function arguments are always converted to pointers.

So char *argv[] is the same as char **argv.

argv[0] contains the program name (or the name used to invoke the program, in the case of a multiply-linked file), so argv+1 is just the program's arguments.

Answer 5

argv is not an array in this context; it is a pointer value. In the context of a function parameter declaration, T a[N] and T a[] are equivalent to T *a; in all three declare a as a pointer to T.

However, it would still be possible to make the assignment even if argv were an array. Unless it is the operand of the sizeof or unary & operators, an expression of type "N-element array of T" is converted ("decays") to an expression of type "pointer to T", and the value of the expression is the address of the first element of the array.

Answer 6

Notice that

char *argv[]

Is an array of pointers. An array declaration, is a pointer itself, argv is a pointer. Since here we have an array of pointers, argv is a pointer to a pointer, just like char **inputs, thus

char **inputs = argv+1;

Is just saying inputs is equal to the pointer argv plus one and since argv+1 is also a pointer, then you have a pointer to a pointer.