Trying to get an array.length to work in for loop

Question

Can someone help me figure out why this is not working.

int main() {
int A[100];

    for(int i=0; i<A.length; i++)
    {

    }

    return 0;
}

Here is my error

[Error] request for member 'length' in 'A', which is of non-class type 'int [100]'

@sje397 `sizeof` won't give the number of elements in the array though. — juanchopanza, Oct 19 '13 at 12:22
See http://stackoverflow.com/questions/37538/how-do-i-determine-the-size-of-my-array-in-c — sje397, Oct 19 '13 at 12:23
Correct is as `for(int i=0; i< sizeof(A) / sizeof(A[0]); i++)`, Read [this answer](http://stackoverflow.com/questions/18009725/weird-behavior-when-printing-array-in-c/18009736#18009736) — Grijesh Chauhan, Oct 19 '13 at 12:25

score 3 · Accepted Answer · answered Oct 19 '13 at 12:23

3

There is no such thing like length attribute for C style arrays. Consider using std::array and it's size() member, or sizeof(A)/sizeof(int), if you insists on C style arrays.

answered Oct 19 '13 at 12:23

erenon

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The requirement for the assignment specifically states not to use vectors so I am stuck doing it this way. – GiBiT Oct 19 '13 at 12:25
1

It's an array, not a vector ; ) – erenon Oct 19 '13 at 12:26
I know I was just stating because you said if you insist on C style arrays. This worked btw thank you. – GiBiT Oct 19 '13 at 12:30

score 2 · Answer 2 · answered Oct 19 '13 at 12:26

Plain arrays do not have any members. If you need to know their length, you either have to keep track of it, or you can write a simple function template to get it:

template<class T, size_t N>
constexpr size_t size(const T (&)[N]) { return N; }

then

int A[100];
....
for(int i=0; i < size(A); i++) { ... }

It would be far easier to use an std::array<int, 100>. This has a size method.

Trying to get an array.length to work in for loop

2 Answers2