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say I have:

I am not really clear what int a[] means, does it mean that the function takes elements of an array or it takes an array ?

so say if I do:

func1(b, 'a');

or

func1(b[1], 'a');

which one is correct?
``````int a, b[1];

int func1 (int a[], char b) {
return a[-1] + d[a[1] * 3];
return a[1];
return a[0];
}
``````
0
kuntilanak
• 2
1 Solution

Commented:
in C# lets say we have array called numbers

int[] numbers = {1, 2, 3, 4, 5};
//the first argument is passing complete array in simple words i.e as a reference
//2nd argument is just passing one element of array
func1(numbers,numbers[0]);
}
public void func1(int[] a, int h)
{
int i = a[1];
int j = a[2];
int k = h;
}
0

Commented:
In your example, func1's argument a[] means that the function accepts an array of int;

The array contains elements, so saying "takes elements of an array" is the same as "takes an array", to my thinking.

In your example, the a[-1] is a real problem -- that's saying "the integer just before the first integer in a".  It is a complete crap shoot as to what value is in that location or whether the location is even a legal one to access.  It certainly is not legal to access it in this manner, although C will let you do it.

Here's an example where the output values are given next to the printf statements.
#include <stdlib.h>
int d[5] = { 10, 11, 12, 13, 14 };

int func0 (int a[], char b) {
return a[0];
}

int func1 (int a[], char b) {
return a[1];
}

int func2 (int a[], char b) {
return a[1] + d[a[1] * 3];
}

int main(void) {
int a[5] = { 0, 1, 2, 3, 4 };
int a0 = func0(a,'q');
int a1 = func1(a,'q');
int a2 = func2(a,'q');
printf("a0 = %d\n",a0); // 0
printf("a1 = %d\n",a1); // 1
printf("a2 = %d\n",a2); // 14
}
0

Commented:
Don't feel stupid for asking questions like this.

The only stupid action would be to know that you don't understand something and then be afraid to look stupid by asking it.

Every expert at this site has encountered things they didn't understand.  They became experts by getting answers to their question.

Well done for asking the question!
0

Author Commented:
thank you guys
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