The variable 'pminutes' is being used without being initialized - c

Question

I keep getting this error message:
The variable 'pminutes' is being used without being initialized.
This is my code:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
void ConvertMinutesToHM(int time_minutes, int *phours, int *pminutes);
void main ()
{
int min, *phours, *pminutes;
printf("Enter a number of minutes:\n");
scanf("%d", &min);
ConvertMinutesToHM(min, phours, pminutes);
printf("Hours=%d", phours);
printf("Minutes=%d", pminutes);
}
void ConvertMinutesToHM (int time_minutes, int *phours, int *pminutes)
{
int hours, minutes;
*phours = time_minutes / 60;
*pminutes = time_minutes % 60 * 60;
}

Answer 1

In the c programming language, variables that are not global or static or explicitly initialized by the programmer are not initialized at all and reading them would cause undefined behavior.
You are passing it to a function that will dereference the pointer and since it was not initialized the function would read from an undetermined location in memory, because the value the pointer is holding is not defined anywhere. This does not mean that the pointer is not holding a value, it means that the value is a random garbage value that cannot be used as an address but will be if you are not careful and dereference it.
You can initialize it by making it point to an actual variable that would be allocated on the stack and make it point to valid memory, the like this
int minutes;
int *pminutes = &minutes;
of course, you can simply pass &minutes directly but this way, the pminutes pointer will be holding the address of a stack variable minutes which is properly allocated, but uninitialized.
The minutes variable is not required to be initialized because it's intended to be used as storage for the value the function will read and store into whatever pminutes is pointing to.
The problem is then that pminutes is a pointer, and not initializing it will cause undefined behavior when the program attempts to read the address it is holding.
It's still good practice to explicitly initialize most variables, but I also consider good practice not to do it when you are certain that they will be initialized somewhere and adding a explicit initialization would not be meaningful, thus making the code harder to understand.

Answer 2

Because you haven't initialised pminutes in main before passing it to ConvertMinutesToHM() function. Remember you need to initialise all parameters before passing them to a function - otherwise how does the function know what values to work on?
In main you should really just need normal int and initialise them properly:
int min = 0, phours = 0, pminutes = 0;
In when calling ConvertMinutesToHM(), you can pass in the address of the corresponding variables:
ConvertMinutesToHM(min, &phours, &pminutes);
Edit: You made other mistakes, too in that ConvertMinutesToHM() function: local hours, minutes are completely not needed, and the calculation of pminutes makes no sense.
Sample fix:
void ConvertMinutesToHM(int time_minutes, int *phours, int *pminutes);
int main()
{
int min = 0, phours = 0, pminutes = 0;
printf("Enter a number of minutes:\n");
scanf("%d",&min);
ConvertMinutesToHM(min,&phours,&pminutes);
printf("Hours=%d\n",phours);
printf("Minutes=%d\n",pminutes);
}
void ConvertMinutesToHM (int time_minutes, int *phours, int *pminutes)
{
*phours=time_minutes/60;
*pminutes=time_minutes%60;
}

Answer 3

The following code:
cleanly compiles
performs the desired operation
incorporates the comments on the OPs question
and now the code
#include <stdio.h> // printf(), scanf(), perror()
#include <stdlib.h> // exit(), EXIT_FAILURE
//#include <math.h>
void ConvertMinutesToHM(int time_minutes, int *phours, int *pminutes);
int main ( void )
{
int min;
int hours;
int minutes;
printf("Enter a number of minutes:\n");
if( 1 != scanf("%d", &min) )
{
perror( "scanf failed to read the minutes from the user" );
exit( EXIT_FAILURE );
}
// implied else, scanf successful
ConvertMinutesToHM(min, &hours, &minutes);
printf("Hours=%d\n", hours);
printf("Minutes=%d\n", minutes);
} // end function: main
void ConvertMinutesToHM (int time_minutes, int *phours, int *pminutes)
{
//int hours, minutes;
*phours = time_minutes / 60;
*pminutes = time_minutes % 60;
} // end function: ConvertMinutestoHM
and here is the results that are displayed on the terminal:
Enter a number of minutes:
100
Hours=1
Minutes=40

Answer 4

You are using pointer that is uninitialised. Pointers with static storage are initialized with null pointers. Pointers with automatic storage duration are not initialized.
From C89 Standard:
C89 §3.5.7 Initialization
If an object that has static storage duration is not initialized explicitly, it is initialized implicitly as if every member that has arithmetic type were assigned 0 and every member that has pointer type were assigned a null pointer constant. If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate.
See Also ISO C 99 6.7.8.10:
If an object that has automatic storage duration is not initialized explicitly, its value is
indeterminate. If an object that has static storage duration is not initialized explicitly,
then:
if it has pointer type, it is initialized to a null pointer;
You are defining int *pminutes and not initialising it so its value is indeterminate and after that doing *pminutes = time_minutes%60*60 try to access that indeterminate location and result in undefined behaviour.