Reputation: 113
Why last line parameter maybe has no effect outside the function:
void save_last_frame( uint8_t *saveframe, uint8_t *curframe,
int width, int height, int savestride, int curstride )
{
height /= 2;
height--;
while( height-- ) {
blit_packed422_scanline( saveframe, curframe, width );
saveframe += savestride;
interpolate_packed422_scanline( saveframe, curframe, curframe + (curstride*2), width );
saveframe += savestride;
curframe += (curstride*2);
}
blit_packed422_scanline( saveframe, curframe, width );
saveframe += savestride;
blit_packed422_scanline( saveframe, curframe, width );
saveframe += savestride; // <-- Assignment of function parameter has no effect outside the function
}
Thanks
Upvotes: 2
Views: 4910
Reputation: 109
You have passed the variable saveframe
as a pointer; to change the value outside the function, do this:
*saveframe += savestride;
instead. This way, your value will now be retained even after function exits.
Upvotes: 0
Reputation: 27233
In C parameters are essentially local variables which are initialized with values passed in as arguments. This means they exist only for as long as the function is being executed. Your saveframe
variable ceases to exist once the function exists and with it the value you assigned.
In order to modify values existing outside the function you should use a pointer and modify the value pointed to by that pointer.
Since the value you're working with is a pointer already, you should use a pointer to pointer:
void save_last_frame( uint8_t **saveframe, uint8_t **curframe,
int width, int height, int savestride, int curstride )
You should then modify the code accordingly, replacing saveframe
with *saveframe
. Similarly for curframe
if you also wish for it to be updated by the function.
An example of such "output pointer" argument is endptr
used to record the end of parsed numeric string in strtol()
.
Upvotes: 1