It doesn't work. I don't understand.Code:#include <stdio.h>#include <stdlib.h> float calculate(int *); int main(void) { float result; int number; scanf("%d",&number); result=calculate(number); printf("%f",result); return 0; } float calculate(int *rakam) { int i; float average=0; int top=0; if(*rakam>1000 && *rakam<9999) { while(rakam!=0) { i=*rakam%10; *rakam=*rakam/10; top=top+i; } average=top/4; return average; } else { return *rakam; } }
Why did rakam a pointer to begin with? It could just have been an int. Anyway, the problem is probably because you wrote:
instead of:Code:while(rakam!=0)
Code:while(*rakam!=0)
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
It's carelessness. Thank you.
Your function 'calculate()' expects an pointer to an integer as argument.
But on line 11 you pass an integer as parameter.
Do you don't receive a warning from your compiler?
You must use the 'address-of' operator to pass the address of that integer.
Now, i see you return the integer 'rakam' on line 35 if the value don't comply your condition, but the function returns a float per definition.Code:… result = calculate(&number); …
I think you should cast the integer to float.
Code:… else return (float) *rakam; …
Last edited by WoodSTokk; 01-15-2015 at 04:59 PM.
Other have classes, we are class
