C function program not working

Question

#include <stdio.h>
#include <conio.h>
#include <stdlib.h>


int main()
{
   int ActChoice=0;
   //do
   //{
          printf("What Activity were you doing?");
          printf("\n1. Running" );
          printf("\n2. Swimming");
          printf("\n3. Cycling" );
          scanf("\n%d",ActChoice);

          /*if(ActChoice == 1)
          {
               RunEdit();
          }
          else if(ActChoice == 2)
          {
               SwimEdit();
          }
          else if(ActChoice == 3)
          {
               CyclEdit();
          }*/

   //}
  // while(1==1);
     getch();
}

Here i have a very simple piece of code designed to choose a desired function, however, on running this program, it crashes after i input "ActChoice" .

I don't know whether this is a mistake i have in my code here, or the code further down, But it seems to break at the scan.

Edit: I forgot the & , i'm actually retarded

bitcell · Accepted Answer · 2014-11-10 09:43:30Z

2

Change scanf("\n%d",ActChoice);

to:scanf("\n%d",&ActChoice);

scanf() is expecting a pointer to a variable.

answered Nov 10, 2014 at 9:43

bitcell

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Comments

David Ranieri · Accepted Answer · 2014-11-10 09:44:09Z

2

Your scanf statement should be like this :

scanf("\n%d",&ActChoice);

You are missing the &.

edited Nov 10, 2014 at 9:44

David Ranieri

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answered Nov 10, 2014 at 9:43

Shwetha

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Comments

Sourav Ghosh · Accepted Answer · 2014-11-10 09:47:54Z

2

Change scanf("\n%d",ActChoice); to scanf("\n%d",&ActChoice); scanf() needs the address of the variable where it has to store the input data.

Please check the man page of scanf() for details.

edited Nov 10, 2014 at 9:47

answered Nov 10, 2014 at 9:42

Sourav Ghosh

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Comments

PU. · Accepted Answer · 2014-11-10 10:26:54Z

1

Change scanf("\n%d",ActChoice); to scanf("\n%d",&ActChoice); because the value sent to scanf is address of ActChoice. The second argument (&ActChoice) specifies the variable into which the typed response will be placed. In this case the response will be placed into the memory location associated with the variable ActChoice.

answered Nov 10, 2014 at 10:26

PU.

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Comments

Hojin · Accepted Answer · 2014-11-10 10:31:20Z

1

You need an ampersand in front of the name of the integer variable ActChoice when receiving input via scanf because it requires an address. Your code should look like the following:

scanf("\n%d", &ActChoice);

answered Nov 10, 2014 at 10:31

Hojin

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Comments

ajay kumar · Accepted Answer · 2014-11-10 09:52:36Z

0

your program doesn't crash.It just doesn't know where to store the input you are giving.Just add a & before ActChoice in the scanf statement to resolve it.

your scanf statement should look like this,

scanf("\n%d",&ActChoice);

answered Nov 10, 2014 at 9:52

ajay kumar

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3 Comments

Sourav Ghosh Over a year ago

What do you mean by your program doesn't crash?

ajay kumar Over a year ago

@SouravGhosh He specified in his question that his program crashes.It doesn't actually crash it just closes down.(I am new to c bro correct me if im wrong)

Sourav Ghosh Over a year ago

yes, it does crash. Reason is trying to access memory region which the process is not permitted to access. HInt: ActChoice = 0;

Vignesh K Viki · Accepted Answer · 2014-11-24 14:28:25Z

0

Scanf in c program expects the address where the data to be stored . In your case you have given

scanf("\n%d",ActChoice);

ActChoice = 0 so the scanf will try to store the value in location zero which is actually a program memory, so i would lead to segfault. this same program may work in Harvard architecture where the program memory is different from data memory.

answered Nov 24, 2014 at 14:28

Vignesh K Viki

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