Why am I iterating less times than expected?

Question

I have printed out my sizeof(array)/sizeof(*array) and receive the number three, but when I pass my array into my function it prints out 2 when stored into a variable and therefor iterates one less time than it should.

 void printAry(int ary[])
 {
     int size = sizeof(ary)/sizeof(ary[0]);
       cout<<size;//size will print out 2! why?
       for(int i = 0; i < size;i++)
     {
         cout<<ary[i];
     }
 }

 int main() {
     int ary[3] = {1,2,3};
     cout<<sizeof(ary[0]);

     return 0;
 }

Why is it iterating one time less than expected ?

The array is being interpreted as a _pointer_ inside the function, hence the problem. Even worse, you're passing _a single member_ of `ary` to the function, not the whole array. — ForceBru, Jan 11 '17 at 17:27

score 4 · Answer 1 · answered Jan 11 '17 at 17:27

You can't use the sizeof(ary) / sizeof(ary[0]) idiom if ary is a function parameter since it will have already decayed into a pointer type: sizeof(ary) is merely the size of a pointer to int on your platform.

In C, you need to pass the size information explicitly, or adopt a convention that a particular value signals the end of the array (cf. '\0' in the C standard library string functions.)

In C++, you can use a std::vector.

score 4 · Accepted Answer · edited Jan 11 '17 at 17:29

4

The reason is that arrays decay into pointers.

Basically

void printAry(int ary[])

is equivalent to:

void printAry(int* ary)

So sizeof(ary) is equal to the size of pointer to int on your system.

edited Jan 11 '17 at 17:29

Bathsheba

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answered Jan 11 '17 at 17:28

Edgar Rokjān

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If you don't mind: I've added "pointer to `int`" as not all pointers *necessarily* need to be the same size. – Bathsheba Jan 11 '17 at 17:30
@Bathsheba yes, for example pointers to functions hypothetically may have another size. Good point! – Edgar Rokjān Jan 11 '17 at 17:32
It's true for primitives too, e.g. `sizeof(int*)` might be different to `sizeof(long*)`. – Bathsheba Jan 11 '17 at 17:33
@Bathsheba is it guaranteed by the standard? – Edgar Rokjān Jan 11 '17 at 17:36
I wouldn't bet my job on it, but I'm sure the standard allows for `sizeof(int*)` to be different to `sizeof(long*)`. – Bathsheba Jan 11 '17 at 17:36

Why am I iterating less times than expected?

2 Answers2