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.