Different output between argv and normal array of strings

Question

For the program given below, why does the first printf print different values for argv and &argv? The second printf prints the same value for a and &a. What is the mechanism by which argument variables passed to the program are stored?

#include<stdio.h>
    int main(int argc, char * argv[]){
    char * a[10];  
    printf("%d %d\n\n",argv,&argv);
    printf("%d %d",a,&a);  
    return 0;  
}

What is the mechanism of storing command line arguments to an array of strings

Answer 1

The problem is not about argv is special, but about function argument passing. Try this program:

#include<stdio.h>

void foo(int a[])
{
    int b[10];
    printf("foo :%p %p\n",a,&a);
    printf("foo :%p %p\n",b,&b);
}

int main(int argc, char* argv[])
{
    int a[10];
    printf("main:%p %p\n", a, &a);
    foo(a);
    return 0;
}

The output in my machine:

main:0x7fffb4ded680 0x7fffb4ded680
foo :0x7fffb4ded680 0x7fffb4ded628
foo :0x7fffb4ded630 0x7fffb4ded630

The first and third line is not surprising since for an array arr, arr and &arr have the same value. The second line needs to be explained.

As you can see, when a is passed in the function foo, the value of a is not changed, but the address &a is changed since C functions always pass arguments by value, there will be a local copy of the argument inside the function.

It's the same with argv since it's just an argument in the function main.

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EDIT: For those who disagrees with me, my answer is not against the other answer. On the contrary, they complete each other. argv is a pointer exactly because it's an array that's passed as function argument, and it "decayed" into a pointer. Again, argv is not special, it acts no different than other arrays that are passed as function argument, that's the whole point of my example code.