Arrays and functions in c language

Question

There i something i dont get, if have the following:

double average (double scores[]){

 double sum = 0;
 int n;
 int nscores = sizeof(scores)/sizeof(double);
 for (n=0 ;n<nscores ;++n){
 sum+=scores [n];
 return sum/nscores;
}

and i send this function an array like this:

  double scores[3]={1,2,3};

why will sizeof(scores) will be 0?

If your question is about C, then why did you use the tag `c++`? — Lightness Races in Orbit, Feb 07 '12 at 22:07
And `sizeof(scores)` won't be 0. It'll be the size of a pointer. This has been asked a milion times. — Lightness Races in Orbit, Feb 07 '12 at 22:07
dup of http://stackoverflow.com/questions/1975128/sizeof-an-array-in-the-c-programming-language — RussS, Feb 07 '12 at 22:10
@LightnessRacesinOrbit but sum/nscores = 0/0 meaning nscores is 0 meaning sizeof(scores) is 0? or is it sizeof(void*) ? — Itzik984, Feb 07 '12 at 22:11
@Itzik984 It will be (4/8) or whatever your compiler sizes for a pointer and double. But since they are both ints the answer is also an int. (4/8)=0 in c integer math. — RussS, Feb 07 '12 at 22:13

score 6 · Accepted Answer · answered Feb 07 '12 at 22:09

6

sizeof(scores), inside the function, is equivalent to sizeof(double *): the compiler has no way to tell, at that point, how big the array was.

It won't be zero, but because sizeof(scores) and sizeof(double) are both integer-type expressions, sizeof(scores)/sizeof(double) is integer division, so if sizeof(scores) < sizeof(double), then sizeof(scores)/sizeof(double) == 0.

answered Feb 07 '12 at 22:09

ruakh

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so why do i get a 0? sizeof(scores)=sizeof(void*)=4? and the division is 0? – Itzik984 Feb 07 '12 at 22:14
2

@Itzik984: Yes, exactly. On your platform, `sizeof(double *)` is `4` and `sizeof(double)` is `8`, so `sizeof(scores)/sizeof(double)` is `4/8`, which is `0`. – ruakh Feb 07 '12 at 22:15

score 3 · Answer 2 · answered Feb 07 '12 at 22:10

This is because array as a parameter of the function is treated as pointer. For example,

double average (double scores[]);

... is an equivalent to:

double average (double *scores);

and so sizeof(scores)/sizeof(double) is equal to sizeof(double *)/sizeof(double), and if you happen to have those types to be of the same size, the result is 1, but if size of double is 8 and pointer is just 4, then result is 0.

score 1 · Answer 3 · answered Feb 07 '12 at 23:31

Two things to remember about arrays in C:

Except when it is the operand of a sizeof or unary & operator, or is a string literal being used to initialize another array in a declaration, an expression of type "N-element array of T" will be replaced with ("decay to") an expression of type "pointer to T" whose value is the address of the first element of the array.
In the context of a function parameter declaration, T a[] and T a[N] are identical to T *a; IOW, a is declared as a pointer to T, not as an array. Note that this is only true for function parameter declarations.

When you call average(scores) from your main function, the expression scores will be replaced with another expression of type double *, so what average receives is a pointer value, not an array. Thus, the sizeof trick to get the number of elements won't work within the average function. You will have to pass the number of elements in scores as a separate parameter, like so:

double average(double *scores, size_t count)
{
   ...
}

and call it as

average(scores, sizeof scores / sizeof *scores); // or pass a constant 3, or
                                                 // something similar.

score 0 · Answer 4 · answered Feb 07 '12 at 22:13

0

scores is a double *const, hence sizeof(scores) will be size required to store a pointer

answered Feb 07 '12 at 22:13

Aditya Naidu

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Arrays and functions in c language

4 Answers4