Thread: Convert seconds to hours and minutes

My first issue is I am receiving a warning that the float* differs in levels of indirection from int. I have tried changing min an hours to int with the same warning but replacing int* in the warning.

The assignment is to write a c function name time () that accepts an interger number of seconds and the addresses of the 3 variable name hours, min, and sec. The function is to conver the passed number os seconds into the equivalent of hours minutes and seconds and directly alter the value of the respective variable using their passed addresses.

Here is the code I have so far

.

Code:

#include
<stdlib.h>
#include
<stdio.h>
#pragma
warning (disable: 4996)
void
 time(int, float*, float*); /*function prototype*/
　
int
 main ()
{
	
int seconds;
	
float hours, min;
		
	printf(
"Enter an amount of seconds passed: ", &seconds);
	scanf(
"%d", &seconds);
	time (seconds, &hours, &min);
	printf(
"The equivalent of %d seconds in hours is %6.2f, minutes is %6.2f, and seconds is %d)", seconds, hours, min, seconds);
	system (
"pause");
	
return 0;
}
void
 time(int seconds, float *hoursaddr, float *minaddr)
{
	hoursaddr = seconds;
	minaddr = seconds/60;
}
　
　

I believe that hours, minutes and seconds variables should all be ints, not floating point variables. 3.15 hours is misleading in the presence of minutes and seconds, for example. Minutes and seconds *are* fractions of hours, and don't correlate directly with base 10 fractions of an hour.

I tried switching everything to int, I receive the same warning instead of saying *float to int it says int* to int differs in level of indirection

To Salem; I looked up the issue I was having with the pasting of the code it said to try to paste to notepad then to the website so hopefully this is right

Code:

void time(int, int*, int*); /*function prototype*/

int main ()
{
 int seconds;
 int hours, min;
  
 printf("Enter an amount of seconds passed: ", &seconds);
 scanf("%d", &seconds);
 time (seconds, &hours, &min);
 printf("The equivalent of %d seconds in hours is %6.2f, minutes is %6.2f, and seconds is %d)", seconds, hours, min, seconds);
 system ("pause");
 return 0;
}
void time(int seconds, int *hoursaddr, int *minaddr)
{
 hoursaddr = seconds;
 minaddr = seconds/60;
}

Okay, that makes sense. I am receive an error in the fuction time that "/" illegal, left opperance includes int*

An int and an int * (pointer to int) are not the same thing. You can not assign from one to the other, and you can't multiply or divide them*. You must dereference the pointer before you can get the the number it points to, and do regular number stuff like assigning, multiplying, etc. You need *minaddr = seconds / 60;

* You are allowed to add and subtract integer values from pointers, but I wouldn't worry about the details of that just yet.

I changed my formulas to *hoursaddr = seconds/360 and same for minutes. When I run m program both min and hours print 0

Careless mistake. I am getting a warning on line with the time function below scanf.
It says function int * differs in levels of indirection from int and that time: different types for formal and actual parameter
Also, when I run my program if I enter for example 120 seconds I receive 0 for hours. I tried switching hours to float but I just received 0.000000, is there any way to make it so the answer is hours as an accurate decimal?

Code:

void time(int, int*, int*); /*function prototype*/

int main ()
{
 int seconds;
 int hours, min;
  
 printf("Enter an amount of seconds passed: ", &seconds);
 scanf("%d", &seconds);
 time (seconds, &hours, &min);
 printf("The equivalent of %d seconds in hours is %d, minutes is %d, and seconds is %d)", seconds, hours, min, seconds);
 system ("pause");
 return 0;
}
void time(int seconds, int *hoursaddr, int *minaddr)
{
 *hoursaddr = seconds/3600;
 *minaddr = seconds/60;
}

Almost everything you need has already been said.

I'll only add that if "seconds/3600" is hours "seconds/60" can't be minutes.

Soma