Dynamic Array printing

Question

I am trying to print a dynamic array, but I am having trouble with the bounds for the array. For a simple example, lets say I'm trying to loop through an array of ints. How can I get the size of the array? I was trying to divide the size of the array by the size of the type like this sizeof(list)/sizeof(int) but that was not working correctly. I understand that I was trying to divide the size of the pointer by the type.

int *list

// Populate list

int i;
for(i = 0; i < ????; i++)
  printf("%d", list[i]);

walkingTarget · Accepted Answer · 2011-03-24 14:27:38Z

2

With dynamic arrays you need to maintain a pointer to the beginning address of the array and a value that holds the number of elements in that array. There may be other ways, but this is the easiest way I can think of.

sizeof(list) will also return 4 because the compiler is calculating the size of an integer pointer, not the size of your array, and this will always be four bytes (depending on your compiler).

answered Mar 24, 2011 at 14:27

walkingTarget

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unexpectedvalue · Accepted Answer · 2011-03-24 14:30:16Z

0

YOU should know the size of the array, as you are who allocated it.
sizeof is an operator, which means it does its job at compile time. It will give you the size of an object, but not the length of an array. So, sizeof(int*) is 32/62-bit depending on architecture.

Take a look at std::vector.

answered Mar 24, 2011 at 14:30

unexpectedvalue

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artyom.stv · Accepted Answer · 2011-03-24 14:29:24Z

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There is no standardized method to get the size of allocated memory block. You should keep size of list in unsigned listSize like this:

int *list;
unsigned listSize;

list = malloc(x * sizeof(int));
listSize = x;

If you are coding in C++, then it is better to use STL container like std::vector<>

answered Mar 24, 2011 at 14:29

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MByD · Accepted Answer · 2011-03-24 14:29:37Z

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As you wrote, you really did tried to divide the size of a pointer since list is declared as a pointer, and not an array. In those cases, you should keep the size of the list during the build of it, or finish the list with a special cell, say NULL, or anything else that will not be used in the array.

answered Mar 24, 2011 at 14:29

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Jens Gustedt · Accepted Answer · 2011-03-24 15:04:59Z

Seeing some of the inapropriate links to C++ tools for a C question, here is an answer for modern C.

Your ideas of what went wrong are quite correct, as you did it you only have a pointer, no size information about the allocation.

Modern C has variable length arrays (VLA) that you can either use directly or via malloc. Direcly:

int list[n];

and then your idea with the sizeof works out of the box, even if you changed your n in the mean time. This use is to be taken with a bit of care, since this is allocated on the stack. You shouldn't reserve too much, here. For a use with malloc:

int (list*)[n] = malloc(*list);

Then you'd have to adapt your code a bit basically putting a (*list) everywhere you had just list.

Evan · Accepted Answer · 2011-03-24 15:05:30Z

0

If by size you mean the number of elements then you could keep it in a counter that gets a ++ each time you push an element or if you dont mind the lost cycles you could make a function that gets a copy of the pointer to the first location, runs thru the list keeping a counter until it finds a k.next==null. or you could keep a list that as a next and a prev that way you wouldnt care if you lost the beginning.

answered Mar 24, 2011 at 15:05

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