How to iterate over a array using a pointer and length unknown? - c++

Question

I've been given a c api to work with and the minimum docs.
Developer is not around at the moment and his code is returning unexpected values (arrays not of expected length)
Im having problems with methods that return pointers to arrays and was wondering am I iterating over them correctly.
Q:does the following always return the correct len of an array?
int len=sizeof(sampleState)/sizeof(short);
int len=sizeof(samplePosition)/sizeof(int);
typedef unsigned char byte;
int len=sizeof(volume)/sizeof(byte);
And I iterate over the array using the pointer and pointer arithmetic (am I doing it correctly for all types below)
And last example below is multidimensional array? Whats the best way to iterate over this?
thanks
//property sampleState returns short[] as short*
short* sampleState = mixerState->sampleState;
if(sampleState != NULL){
int len=sizeof(sampleState)/sizeof(short);
printf("length of short* sampleState=%d\n", len);//OK
for(int j=0;j<len;j++) {
printf(" sampleState[%d]=%u\n",j, *(sampleState+j));
}
}else{
printf(" sampleState is NULL\n");
}
//same with int[] returned as int*
int* samplePosition = mixerState->samplePosition;
if(samplePosition != NULL){
int len=sizeof(samplePosition)/sizeof(int);
printf("length of int* samplePosition=%d\n", len);//OK
for(int j=0;j<len;j++) {
printf(" samplePosition[%d]=%d\n",j, *(samplePosition+j));
}
}else{
printf(" samplePosition is NULL\n");
}
Here byte is type def to
typedef unsigned char byte;
so I used %u
//--------------
byte* volume = mixerState->volume;
if(volume != NULL){
int len=sizeof(volume)/sizeof(byte);
printf("length of [byte* volume = mixerState->volume]=%d\n", len);//OK
for(int j=0;j<len;j++) {
printf(" volume[%d]=%u\n",j, *(volume+j));
}
}else{
printf(" volume is NULL\n");
}
Here is int[][] soundFXStatus.
do I just use same method above and have 2 loops?
//--------------
int** soundFXStatus = mixerState->soundFXStatus;

Answer 1

The sizeof(array)/sizeof(element) trick only works if you have an actual array, not a pointer. There's no way to know the size of an array if all you have is a pointer; you must pass an array length into a function.
Or better use a vector, which has a size function.

Answer 2

sizeof(sampleState)/sizeof(short);
This will only give the length of an array if sampleState is declared as an array, not a pointer:
short array[42];
sizeof(array)/sizeof(short); // GOOD: gives the size of the array
sizeof(array)/sizeof(array[0]); // BETTER: still correct if the type changes
short * pointer = whatever();
sizeof(pointer)/sizeof(short); // BAD: gives a useless value
Also, beware that a function argument is actually pointer even if it looks like an array:
void f(short pointer[]) // equivalent to "short * pointer"
{
sizeof(pointer)/sizeof(short); // BAD: gives a useless value
}
In your code, sampleState is a pointer; there is no way to determine the length of an array given only a pointer to it. Presumably the API provides some way to get the length (since otherwise it would be unusable), and you'll need to use that.
In C++, this is one reason why you would prefer std::vector or std::array to a manually allocated array; although that doesn't help you since, despite the question tags, you are using C here.

Answer 3

int len=sizeof(sampleState)/sizeof(short);
int len=sizeof(samplePosition)/sizeof(int);
sizeof is done at compile time, so this approach doesnt work if the length of the arrays are not known at compile time (for example the memory is reserved using a malloc).

Answer 4

ok ignore the method I used above it was all wrong - though you do need to know the length of the array which I finally got from the API developer.