for( k = 0; k < CycleCount;k++)
{
//make Org data
int* Data = MakeData(DataCount[i]);
......
the function look like this one. i think this is right. so ...
int* MakeData(int DataCount)
{
//
int* Data=(int*)malloc(DataCount*sizeof(int));
int i;
for( i=0; i<DataCount; i++)
{
//
Data[i] = rand()%DataCount +1;
}
return Data;
}
i dont know why this didn't work.
what shoud i have to do???
When the C compiler finds a function call without having seen a function prototype it assumes a function that returns an int.
You should tell the compiler the correct function signature with the "function prototype":
int* MakeData(int DataCount);
This should be placed in a .h file that will be included in all compilation units that call or define the function.
If you have a static function, that is visible only in the current compilation unit, you can place the prototype (incl. static) before all functions in that file.
Further you should never cast the return from malloc. It returns a void*. In the language C this can be converted to any other pointer type. You get the correct prototype when you #include <stdlib.h>.
DataCount[i] is supposed to be equivalent *(DataCount + i). But if you wrote that it would give you an true error, and not just try to implicitly cast it to a pointer-to-int.
MakeData() was expecting an int, not a pointer. And my (wrong) theory was that it was, indeed getting an int, but that the dereferencing process involved an implicit int-to-pointer conversion.
DataCount?MakeDatait's an integer parameter, 2. Somewhere else whereMakeDatais called it's something else.int* Data = MakeData(DataCount[i]), the other one being defined as a parameter inMakeData. I should have been more explicit.