If I have a an array of ints, how could I directly edit each int?
int i = arr + 1; // Getting the integer in pos 1
i is just a copy, correct? If I set i = 4, then arr + 1 would still be 1.
Would this work?
int *i = &(arr + 1);
*i = 4;
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4 Answers
You should use the array operators:
int i = arr[1];
arr[1] = 4;
Comments
Change your code to this:
int *i = arr + 1;
*i = 4;
and it will work. Arrays in C are just pointers to first element in the array.
So this arr + 0 will give address of first element in array and this arr + 1 is an address of second element.
Comments
You've got:
int arr[4] = {0, 1, 2, 3};
Want to edit it further?
arr[0] = 42;
// arr[] = {42, 1, 2, 3};
Want to change all of them at once? There's:
for(int i = 0; i < 4; ++i)
arr[i] = i * 2;
// arr[] = {0, 2, 4, 6};
And don't forget memset()!
memset(arr, 42, 4);
// arr[] = {42, 42, 42, 42};
Want to change everything but the first element to 7?
memset(&arr[1], 7, 4 - 1);
// arr[] = {42, 7, 7, 7};
Would you like to know somethin' about pointers? (Here's a more useful link.)
See this? (If you can't, please stop reading this. Thanks!)
int *ptr = &arr[1];
It's equivalent to:
int *ptr = arr + 1;
Which is also equivalent to:
int *ptr = arr;
ptr = ptr + 1;
OK, now that we've got that down, let's show you a more efficient for-loop than the one I did above:
int *ptr = arr;
for(int i = 0; i < 4; ++i)
{
*ptr = i << 2;
// i * 2 == i << 2
++ptr;
}
// arr[] = {0, 2, 4, 6};
Not that you should code like that; the compiler will handle it for you, most likely.
Would you like another answer in the form of a series of questions?
Comments
Array indexing operators can do what you need.
arr[3] = 101; //assignment to array
int x = arr[37]; //get value from array
etc.
No need for for that memory arithmetic here..
2 Comments
blubb
I think you meant to say 'address arithmetic' instead of 'memory arithmetic'
jon_darkstar
yes thatd be better way to say it. or maybe pointer arithmetic