how to pass array as arguments in C?
int a,b,c[10];
void Name1(int x, int y, int *z)
{
a = x;
b = y;
c = z;
}
I try to pass it as argument, but it does not build, how to fix it?
And is the declaration of void Name1(int x, int y, int *z); is the same as void Name1(int x, int y, int z[])? does the void Name1(int x, int y, int z[]) will be treated as void Name1(int x, int y, int *z); by compiler?
When you pass array as an argument to function it decays as an pointer to its first element.
So,
void Name1(int x, int y, int *z)
will work.
But arrays are not assignable so:
c = z;
does not work, you will need to explicitly copy each array element from source to destination.
c in your example is going to hold a copy of the elements of z. In that case, you have to manually copy each element from z to c in a loop. Otherwise, just declare c as a int * and you can set it to z as you do above. But keep in mind then that any changes to the elements of c will be reflected in z.Doing c = z;, you're assigning to a non l-value, which is not allowed. It's like doing &c[0] = &z[0];.
There are a couple of answers here, but I don't think any of them are completely sufficient, so...
basically you can either copy the array or use the pointer, but either way you will need to keep track of the length.
int a,b, *c, len;
void Name1(int x, int y, int *z, int z_len)
{
a = x;
b = y;
c = z;
len = z_len;
}
//usage:
int arr[5];
Name1(1,2,arr /* or &arr[0] ,*/, sizeof(arr )/ sizeof (int));
if you never need to add items this will be sufficient... If you do it gets more complicated...
it is important to keep the length around so that you know how many elements you have, even if you are going to copy them.
c=z; is invalid
c pointer doesnt have memory to save the z address.
you can say:
int a,b,*c;
c=(int *)calloc(10,sizeof(int) );
void Name1(int x, int y, int *z)
{
a = x;
b = y;
c = z;
}
