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# Operator [ ]

 P: n/a Until recently I grasp the relationship between pointers and arrays and I have finally understand that [ ] is an operator. But I still can't quite figure it out: What does really does [ ] in a statement such as: int array[5]; int matrix[5][5]; dArray = new int[]; dMatrix = new int*[]; Also, is this valid (useful?): int x; int *array=&x; for(int i=0; i
12 Replies

 P: n/a Gaijinco wrote: Until recently I grasp the relationship between pointers and arrays and I have finally understand that [ ] is an operator. But I still can't quite figure it out: What does really does [ ] in a statement such as: int array[5]; int matrix[5][5]; dArray = new int[]; dMatrix = new int*[]; The brackets here are part of the syntax. However, it the presented form it isn't going to compile, there has to be an integral expression inside the brackets. Also, is this valid (useful?): int x; int *array=&x; for(int i=0; i

 P: n/a > The brackets here are part of the syntax. However, it the presented form it isn't going to compile, there has to be an integral expression inside the brackets. I'm a little lost! When I use something like: int array[5]; I'm saying "look for 5 block of free memory to allocate int's and create a pointer "array" to the first block". True or false? So when I'm saying int matrix[5][5]; It looks for 25 blocks instead of 5, but I cn't understand what [5][5] really means? And if that's the way it works then what it means: Then int array[5] is the same as int* array = new int[5] ? Oct 31 '05 #3

 P: n/a > The brackets here are part of the syntax. However, it the presented form it isn't going to compile, there has to be an integral expression inside the brackets. I'm a little lost! When I use something like: int array[5]; I'm saying "look for 5 block of free memory to allocate int's and create a pointer "array" to the first block". True or false? So when I'm saying int matrix[5][5]; It looks for 25 blocks instead of 5, but I cn't understand what [5][5] really means? And if that's the way it works then int array[5] is the same as int* array = new int[5] ? Oct 31 '05 #4

 P: n/a Gaijinco wrote: The brackets here are part of the syntax. However, it the presented form it isn't going to compile, there has to be an integral expression inside the brackets. I'm a little lost! When I use something like: int array[5]; I'm saying "look for 5 block of free memory to allocate int's and create a pointer "array" to the first block". True or false? No, the program statically allocates an array of 5 ints, either on the stack or in its global memory, depending on the scope in which this declaration appears. There is no dynamic memory allocation invoked by this declaration. So when I'm saying int matrix[5][5]; It looks for 25 blocks instead of 5, but I cn't understand what [5][5] really means? This statement statically allocates an array that contains 5 arrays exactly like the array in first declaration. N And if that's the way it works then int array[5] is the same as int* array = new int[5] ? The "array" variables are interchangeable since an array can be accessed via a pointer and a pointer can be subscripted, but technically they are not the same. An array declaration reserves memory for its elements while a pointer declaration does not reserve memory for the item or items it points to. Greg Oct 31 '05 #5

 P: n/a (a) Don't post twice within minutes: this is not a chat room. Gaijinco wrote:The brackets here are part of the syntax. However, it the presented formit isn't going to compile, there has to be an integral expression insidethe brackets. I'm a little lost! When I use something like: int array[5]; I'm saying "look for 5 block of free memory to allocate int's and create a pointer "array" to the first block". True or false? False. You're saying, 'array' is the name that will in my program identify an array of 5 objects, each of type 'int'. Whether (and where) the memory is allocated, is irrelevant. And 'array' is not a pointer. So when I'm saying int matrix[5][5]; It looks for 25 blocks instead of 5, but I cn't understand what [5][5] really means? It is, again, irrelevant how it looks and for how many "blocks". What you say by this declaration is that 'matrix' is the _identifier_ that is associated with an array of 5 arrays, and each of them is of 5 ints. And if that's the way it works It's not. then int array[5] is the same as int* array = new int[5] ? Of course not. V Oct 31 '05 #6

 P: n/a Gaijinco sade:The brackets here are part of the syntax. However, it the presented formit isn't going to compile, there has to be an integral expression insidethe brackets. I'm a little lost! When I use something like: int array[5]; I'm saying "look for 5 block of free memory to allocate int's and create a pointer "array" to the first block". True or false? No, you declare an array of 5 integers. int array[5] = {1,2,3,4,5}; So when I'm saying int matrix[5][5]; It looks for 25 blocks instead of 5, but I cn't understand what [5][5] really means? You declare an array with 5 arrays of 5 integers. int matrix[5][5] = { {1,2,3,4,5}, // matrix[0][x] {1,2,3,4,5}, // matrix[1][x] {1,2,3,4,5}, // matrix[2][x] {1,2,3,4,5}, // matrix[3][x] {1,2,3,4,5} // matrix[4][x] }; TIT Oct 31 '05 #7

 P: n/a And if that's the way it works then what it means: Then int array[5] is the same as int* array = new int[5] ? True, but with a subtle difference: doing a sizeof(array) will return 5 * sizeof(int) whereas the last one will return sizeof(int *) -M Oct 31 '05 #8

 P: n/a Gaijinco wrote:The brackets here are part of the syntax. However, it the presented formit isn't going to compile, there has to be an integral expression insidethe brackets. I'm a little lost! When I use something like: int array[5]; I'm saying "look for 5 block of free memory to allocate int's and create a pointer "array" to the first block". True or false? Not really, in fact almost completely wrong. '5' is correct and 'int' is correct but nothing else is. There is no looking, there is no allocating and there is no pointer. The above is a declaration which creates an array (helpfully called array) of 5 integers. There's nothing more to be said, you are making it more complex than it is. So when I'm saying int matrix[5][5]; It looks for 25 blocks instead of 5, but I cn't understand what [5][5] really means? 25 is right, I guess you are asking for the difference between int matrix[5][5]; and int array[25]; Both create arrays of 25 integers. The differnce in in the way they are accessed. To get at an individual int in 'matrix' you must subscript twice (e.g. matrix[1][2]) whereas with 'array' you only need to subscript once (e.g. array[12]). I think you probably know this and are again seeing compications where none exist. And if that's the way it works then int array[5] is the same as int* array = new int[5] ? No, different. In that case array is a pointer to a block of memory large engough to hold 5 integers. Here all the stuff about looking and allocating really does happen. The similarity bettwen pointers and array happens in the way they are accessed not the way they are created. int* ptr = new int[5]; int array[5]; The creation of ptr and array are completely different. But once they have been created they way there are accessed is very similar. For instance ptr[1] = 2; and array[1] = 2; are both legal. There still are differences, for instance sizeof ptr != sizeof array and ptr = array; is legal but array = ptr; is not. john Oct 31 '05 #9

 P: n/a OK, lot's of intresting things here! But I don't understand why if: int a[5]; int* b = new int[5]; Then why b = a; is valid but a = b; isn't? Also if this works: int a[5]; for(unsigned i=0; i<5; ++i) *(a+i)=i; It should prove that in the first line we really define a pointer named "a" because the unary operator * is only for pointers. Am I wrong? P.D: I didn't post twice within minutes but I tried to edit the post, I couldn't figure why so I delete and repost edited later. Nov 1 '05 #10

 P: n/a Gaijinco wrote: OK, lot's of intresting things here! But I don't understand why if: int a[5]; int* b = new int[5]; Then why b = a; is valid but a = b; isn't? Because you can assign another value to a pointer but arrays are not assignable. Language limitation. Also if this works: int a[5]; for(unsigned i=0; i<5; ++i) *(a+i)=i; It should prove that in the first line we really define a pointer named "a" because the unary operator * is only for pointers. Am I wrong? No. Using the name of an array in some expressions causes the type to _decay_ from "an array of T" to "a pointer to T". Additive expression is just one of those cases. When you write (a+i), 'a' degrades, decays, to a pointer. Same with a[i] (or i[a]). V Nov 1 '05 #11

 P: n/a >Also if this works:int a[5];for(unsigned i=0; i<5; ++i) *(a+i)=i;It should prove that in the first line we really define a pointernamed "a" because the unary operator * is only for pointers. Am Iwrong? No. Using the name of an array in some expressions causes the type to _decay_ from "an array of T" to "a pointer to T". Additive expression is just one of those cases. When you write (a+i), 'a' degrades, decays, to a pointer. Same with a[i] (or i[a]). This is the essential truth. When books confusingly talk about the 'equivalence' of arrays and pointers they actually mean what Victor has said above. Arrays and pointers are different but in many contexts (but not all) you can use an array as if it were a pointer. john Nov 1 '05 #12

 P: n/a Gaijinco wrote: OK, lot's of intresting things here! But I don't understand why if: int a[5]; int* b = new int[5]; Then why b = a; is valid but a = b; isn't? In C++ the name of an array can be used as the address of that array. The expression, b = a, demonstrates this point. Similarly, the name of a function can be used as the address of that function. But neither arrays nor functionS are themselves pointers, so the expression, a = b, which attempts to assign an address to an array, is invalid. Also if this works: int a[5]; for(unsigned i=0; i<5; ++i) *(a+i)=i; It should prove that in the first line we really define a pointer named "a" because the unary operator * is only for pointers. Am I wrong? Again a is being used as the address of the a array, or &a[0], if you prefer. Greg Nov 1 '05 #13

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