Thread: Simple Program Help

thanks I have made that change and it now gives the following results...


Enter No. For Variable : 3
51 is at 6cffdc

if I pass 4 into it it give result as 52 and so on. Any ideas?

Taking the ; out after the For line has also made the loop run now.

When you use:

Code:

itst = getchar();

to read a character from the input, the computer stores the ascii code of the character in itst. So, when you type '3', the value stored in itst is the ascii value which in decimal notation is 51, '4' is 52, '\n' is 10, 'R' is 82 to quote some examples.

Code:

printf ("%d ",itst);

will printf the value of itst in decimal notation. Which is exactly what your code is doing.

To print the character use:

Code:

printf("%c ", itst)

Hope that helps

Hi Gav,
My suggestions are here. I want to know really what you are doing. Is that while loop neccessary. Is EOF neccessary. EOF is for ctrl-z. I dont know why you used such things.Any way i modified your code.

Code:

#include<stdio.h>
#include<stdlib.h>
int main ()

{
   int itst = 1;
   int cnt;

   for (cnt=0;cnt<5;cnt++)//comma removed.
      {
      printf ("Enter No. For Variable : ");
	  //I got confused why you use while loop here.You are just getting a character and printing it and its 
	  // address.
      while(((itst = getchar()) != '\n') && (itst != EOF));
      printf ("%d is at %p  : ",itst,&itst);
      }
}

ok I rewrote the program and it now works can you just tell me if the following code is how you would have done it? You lot seem to not like the scanf function but getchar was only obviously taking the first character and if I entered 233 it only took the 2 whereas scanf seems to work fine.

all its supposed to do is show the variable contents and pointer 5 times and take input for each one, its just to help me learn and serve no purpose.

Code:

#include<stdio.h>
#include<stdlib.h>
int main ()

{
   int itst = 1;
   int cnt;
   for (cnt=0;cnt<5;cnt++)
      {
      printf ("Enter No. For Variable : ");
      scanf ("%d",&itst);
      printf ("%d is at %p  \n",itst,&itst);
      }
}

>>how you would have done it?
Well, using your code as a basis:

Code:

#include <stdio.h>

int main(void)
{
    int itst, 
        cnt;
    for (cnt = 0; cnt < 5; cnt++)
    {
        printf("Enter No. For Variable : ");
        fflush(stdout);
        if (scanf("%d", &itst) == 1)
          printf("%d is at %p  \n", itst, (void*)&itst);
    }
    return 0;
}

But then I'd recommend you read this re getting numbers from the user.

Code:

#include<stdio.h>
#include<types.h>
int main ()

{
   int itst = 1;
   int cnt;
   char buf[11];
   for (cnt=0;cnt<5;cnt++)
      {
      printf ("Enter No. For Variable : ");
      fgets(buf, 10, stdin);
      itst = atoi(buf);
      printf ("%d is at %p  \n",itst,&itst);
      }
}

Originally posted by WaltP

Code:

#include<types.h>

   char buf[11];

      fgets(buf, 10, stdin);
      itst = atoi(buf);

Why?

atoi() is in 'stdlib.h', and 'fgets' is in 'stdio.h'. What's the purpose of 'types.h'?

Quzah.

Originally posted by WaltP

Code:

#include<stdio.h>
#include<types.h>
int main ()

{
   int itst = 1;
   int cnt;
   char buf[11];
   for (cnt=0;cnt<5;cnt++)
      {
      printf ("Enter No. For Variable : ");
      fgets(buf, 10, stdin);
      itst = atoi(buf);
      printf ("%d is at %p  \n",itst,&itst);
      }
}

You've missed some points I implied in my code.

fflush() is needed after printf to ensure the output is actually written when needed.

The %p argument of printf requires a void* parameter.

>>char buf[11];
>>fgets(buf, 10, stdin);
You're wasting 1 byte of memory, fgets() will write N-1 characters to the array, the Nth will be \0 if too much data is read.

atoi() is not a particularly good choice of a replacement:

From C99
1 The functions atof, atoi, atol, and atoll need not affect the value of the integer
expression errno on an error. If the value of the result cannot be represented, the
behavior is undefined.

Originally posted by quzah
Why?

atoi() is in 'stdlib.h', and 'fgets' is in 'stdio.h'. What's the purpose of 'types.h'?

Quzah.

I forgot
So Solly