my function:
int f1(int* r1, int* r2, int* r3, int* r4) {
*r1 = 1;
*r2 = 343;
*r3 = 34;
*r4 = 3;
return c; // c = 1 if success
}
caller:
f1 (&r1, &r2, &r3, &r4);
if I want to simplify my function, should I pass in an array of four int pointers? or should i do it with an array of four int ?
int f1(int* r1[4])?
or
int f1(int r1[4])?
Thank you.
Since arrays decay into pointers when passed to a function, an array would do:
void f(int *p)
{
p[0] = 1;
p[1] = 2;
p[2] = 3;
p[3] = 4;
}
int arr[] = { 0, 0, 0, 0 };
f(arr);
p[4] = 5 -- this little addition to your code will cause buffer overflow. There is no hint to recognize it. Quite the contrary, in the case of `void f(int p[4]) good compiler or code analyzer or just careful programmer will catch buffer overflow
When array is used for function's parameter,it's a pointer. So
int f(int *p) == int f(int r1[])
You don't need to write int f(int r1[4]) .Because the compiler don't care the size of the 1D array when passed to a function.
int f1(int* r1[4]) == int f1( int **p ) // That's not what you want.
The true C99 way to do this is to have:
int f(int p[static 4])
{
p[0] = 1;
p[1] = 2;
p[2] = 3;
p[3] = 4;
return c;
}
The static 4 inside the square brackets indicates to the compiler that the function expects an array of at least 4 elements. clang will issue a warning if it knows at compile-time that the value you are passing to f doesn't contain at least 4 elements. GCC may warn too but I don't use this compiler often.
Array of four int
int f1(int r1[4])
{
p[0] = 1;
p[1] = 2;
p[2] = 3;
p[3] = 4;
}
Since array decay into pointer you can omit the size of array
int f1(int r1[])
or
int f1(int * r1)
But it is a bad way. For example, p[4] = 5 will cause buffer overflow, which compiler can not catch without knowing size of buffer.