arraysname is a pointer

[carox]

I have understood the concept of pointers and arrays but still one thing is confusing:
If I have one dimensional array for eg.

int x[]={10,20,30};

the x stores the address of is first element i.e x=&x[0][0] but how? This shows that x is a variable that also gets some memory space like x[0],x[1] or any other variable, which contains the address. Am I right? but this does not seems to be true. In C language the array name does not get any space in the memory then how can it has a value..I think anything which has the value gets the space in the memory.

In two dimensional array,

int num[][]={{10,20,30},{40,50,60},{70,80,90}};

num[0] stores the address of the num[0][0],
num stores the address of the num[0],
*num=num[0]
*(*num)=num[0][0]

I have not understood these. I think only num[0][0],num[0][1],num[0][2].....gets the memory. How num, num[0],num[1] gets the memory???

I have gone through some of the topics here but could'nt find them useful.

[laserlight]

the x stores the address of is first element i.e x=&x[0][0] but how?

I do not suggest thinking of it like that. Think of x as an array itself, not a pointer. Also, note that the address of the first element of x is &x[0], not &x[0][0].

This shows that x is a variable that also gets some memory space like x[0],x[1] or any other variable, which contains the address. Am I right?

Yes, there is space allocated for x, but no, that space would not be the space required for a pointer, since x is an array. sizeof(x) would return the size of the entire array.

In two dimensional array,

Your example is incorrect. It should be:

int num[][3]={{10,20,30},{40,50,60},{70,80,90}};

num[0] stores the address of the num[0][0],
num stores the address of the num[0],

Again, do not think of it in that way. num is an array of 3 arrays of 3 ints. Therefore, num[0] is an array of 3 ints.

Now, the point is that in many contexts, an array is converted to a pointer to its first element. Consequently, if you pass num as an argument to a function, you get a pointer to an array of 3 ints, and this pointer will point to num[0] (at least at the start of the function).

[Elysia]

You seem to misunderstand arrays. Arrays are not pointers, and therefore they do not store a memory address.
Memory is allocated for the entire array.
However, when you try to use the array name, it decays to a pointer to its first element.

[BEN10]

I have understood the concept of pointers and arrays but still one thing is confusing:
If I have one dimensional array for eg.

int x[]={10,20,30};

the x stores the address of is first element i.e x=&x[0][0] but how? This shows that x is a variable that also gets some memory space like x[0],x[1] or any other variable, which contains the address. Am I right? but this does not seems to be true. In C language the array name does not get any space in the memory then how can it has a value..I think anything which has the value gets the space in the memory.

In two dimensional array,

int num[][]={{10,20,30},{40,50,60},{70,80,90}};

num[0] stores the address of the num[0][0],
num stores the address of the num[0],
*num=num[0]
*(*num)=num[0][0]

I have not understood these. I think only num[0][0],num[0][1],num[0][2].....gets the memory. How num, num[0],num[1] gets the memory???

I have gone through some of the topics here but could'nt find them useful.

int x[]={10,20,30};

Few points to remember:
x is the base adderss of the array i.e x=&x[0], whereas &x is the address where the whole array is stored. If used with sizeof, x gives the total size of the array.
For a 2D array, think of this way
x[0]=(*(x+0)+0)=&x[0][0]=address of x[0][0]
x[1]=(*(x+1)+0)=&x[1][0]=address of x[1][0]
and so on.