Determining Array Size in C

Question

Possible Duplicate:
Sizeof an array in the C programming language?

In my main function I have the following code:

struct student s1 = {"zack",0,0,82};
struct student s2 = {"bob",0,0,80};
struct student s3 = {"joe",0,0,97};
struct student s4 = {"bill",0,0,100};

struct student c[] = {s1,s2,s3,s4};

printf("%d\n\n", arrayLengths(c));
printf("%d\n\n", sizeof(c)/sizeof(c[0]));

where arrayLength is defined as:

int arrayLengths(struct student g[]) {
    return sizeof(g)/sizeof(g[0]);
}

With duplicate code though I am getting two different results. My console output is:

0
4

What gives? Why is this so and how can I put this in its own function?

score 2 · Answer 1 · answered May 18 '12 at 09:16

When an array is passed to a function it decays to a pointer, pointing to the first element in the array. So:

int arrayLengths(struct student g[]) {
    return sizeof(g)/sizeof(g[0]);
}

is actually:

int arrayLengths(struct student* g) {
    return sizeof(struct student*)/sizeof(struct student);
}

score 1 · Answer 2 · answered May 18 '12 at 09:15

1

How would the code inside arrayLengths know how big the array is?

answered May 18 '12 at 09:15

David Schwartz

179,497
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score 0 · Answer 3 · answered May 18 '12 at 09:18

Array passed to a function decays into a pointer. Pointer size is usually 4 bytes on a 32 bit system.

Whereas the size of the array in the main function will be equal to

sizeof(array element type) * (number of elements in the array)

This should explain the behavior which you are seeing.

Determining Array Size in C

3 Answers3