help in pointers

i have read many about pointers and understood very good. but still there are some confusing thinks. please help to understand.

1) int a[];
why *&a doesn't give the first value of array? instead it gives the address. isn't *&a and *a same?

2) int a[n][m], (*p)[n];
for (p = a[0]; p < a[n]; p++) in this case it gives warning "assignment from incompatible pointer type" but it compiles and gives the right answer. if you put address operator before a there will not be problem. why? aren't they same?

thank you

Quote:

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Originally Posted by littletics

1) int a[];
why *&a doesn't give the first value of array? instead it gives the address. isn't *&a and *a same?

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They are not the same. The underpinning of your misunderstanding is that you believe an array and a pointer are the same thing. They are not. It is just that, in some contexts, the compiler can convert the name of an array into a pointer (to the first element of the array). There are lots of contexts where the compiler is forbidden from doing that conversion.

Let's say we've declared a as an actual array.

Code:

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  int a[10];

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&a is of type "pointer to array of ten int". So *&a is of type "array of 10 int". It represents the whole array (all ten elements of it.

In a function (when a is passed as argument to the function) things are a little different. So, if you do this;

Code:

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void func(int a[])
{
  /* body */
}

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a, in the body of the function, is actually of type "pointer to int" So &a is a "pointer to pointer to int" and *&a is of type "pointer to int" (and that pointer has the same value as a).

Quote:

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Originally Posted by littletics

2) int a[n][m], (*p)[n];
for (p = a[0]; p < a[n]; p++) in this case it gives warning "assignment from incompatible pointer type" but it compiles and gives the right answer. if you put address operator before a there will not be problem. why? aren't they same?

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This is one of the many cases where a pointer and an array are actually distinctly different things.

a[0] is of type "array of m ints". p is of type "pointer to an array of n ints". Those types are incompatible.

&a[0] is of type "pointer to array of m ints". If m and n are equal, that is the same type as for p (which is presumably happening in your code, although you haven't shown it). If m and n are not equal, then &a[0] and p have different types - the compiler will generally complain, again, about incompatible pointer types.

@grumpy, I didn't get your explanation. I too, assumed it would work. I mean

Code:

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int a[10] = {10,3,0,} ;

  printf("%p %p", a, &a[0]) ; //This would both give me the address of a
  printf("%p",&a)  //This also would give me the address of a

  printf("%d %d", *a, *&a[0]) //Would give me the number 10, rightfully

  printf("%d", *&a)// doesn't work, however

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Why does *&a not work? It still gives me the address of a.

Quote:

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&a is of type "pointer to array of ten int".

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&a points to the first element in the array of ten ints, yeah?

Quote:

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So *&a is of type "array of 10 int". It represents the whole array (all ten elements of it.

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I don't understand. What do you mean of type "array of 10 int"? So it is equivalent to saying :

Code:

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      int *ptr = &a ;

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&a and a will give you same value(address) but the type is different.
&a + 1 is equal to (char*)&a + sizeof(a). since &a type is pointer to array.
a + 1 gives (char*) a + sizeof(a[0]).

Try to print them out and see for yourself.

Code:

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printf("a is at %p\n",a);
printf("&a is %p\n",&a);
printf("&a + 1 = %p\n",(&a) +  1 );
printf("a + 1 = %p\n",a + 1 );

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Question 6.2

Quote:

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isn't *&a and *a same?

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*& cancel each other..

Code:

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int x;
  *&x = 3;      // same as x = 3!

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oh yes, i see what grumpy means
It also explains, why *&a[0] works. cool!

thank you guys, i have understood the type difference &a and a.
i am sorry i have a mistake in the 2 question in declaration

Quote:

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2) int a[n][m], (*p)[n];

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it should be (*p)[m]. after all, now i can answer my 2 question myself (please correct if wrong)

Answer: however p is type of "pointer to array of m ints" and a[i] is type of "pointer to the "i"s rows first element of 2D array", they are incompatible. for compatibility it must be &a[i], which is type of "pointer to array of m ints" and represents the whole array of m elements of the "i"s row of 2D array

and one more question, as i have read many about that, but never met the explanation of type difference &a and a. where did you know that?

The difference between &a and a is part of C.

The basic rule is: If x is of type "thing", then &x is always of type "pointer to thing" and has a value called "address in memory of x". This rule is always true.

The problem you've had with understanding is that there is special handling for arrays, so the compiler treats "a" and "&a[0]" as equivalent in some contexts. The "in some contexts" is underlined as it is important - arrays and pointers are different things so, sometimes, they can be treated as equivalent and sometimes they can't.



In terms of how I learned that ..... come around the campfire, young brave.

Many many moons ago I, like you, had read an introductory text that told me "a pointer and an array are the same thing".

Eventually, I was writing code, and ran into a case where the compiler complained.

I was arrogant, and insisted the compiler was broken because, after all, "a pointer and an array are the same thing".

However, the stars and the moon were not content with my arrogance.

I had to work with code that could be ported between different operating systems and compilers. All of the compilers complained about my code.

Compilers are like the buffalo - they are wise, they do not change often, and they are unforgiving of your errors. No matter how I insisted that I was right, the compilers kept rejecting my code.

I consulted with other buffaloes. Every one rejected my code.

I consulted with a fellow brave, and he could not explain the way of the buffalo to me.

Faced with the wisdom of multiple buffaloes, I was forced to humbly ask the heart-wrenching question "what am I doing wrong?".

By asking that question, I took the a step towards enlightenment, and realised that the basic texts had taught me incorrectly.

I then consulted with a chieftain - an experienced programmer - who told me "that's the way it is".

I was unsatisfied with the guidance of that chieftain, so I went on a quest to find a wiser chieftain - a programmer who could explain the real world to me.

Eventually, on my hands and knees, I found a chieftain who had new and mighty knowledge - he had buffalo skins branded with the text of the (then) new C standard. After abasing myself to the chieftain, he permitted me to examine the skins.

I paid homage to those skins - and spent much time attempting to understand the strange markings.

Such were my first steps on my never-ending path towards wisdom.....

Quote:

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Originally Posted by grumpy

Compilers are like the buffalo

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I find compilers to be more like a bison or yak.

correct.

Code:

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      int (*p)[2][10] ;

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Is a pointer to an array of 10 integer elements. Notice that it isn't an array of 10 pointers unlike

Code:

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      int *p[2][10]

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Since it is a pointer that has a width of 100, when you increment the pointer, it would skip 100columns.
I will rather prefer this declaration when using strings.