CODEWITHSUNDEEP
cscanf
paulakis
paulakis

Reputation: 89

Can scanf() store values?

Hi I am now learning the C language and I have a little problem with a exercise of the book I read. My code is this:

#include<stdio.h>
int main()
{
  unsigned char one=0;
  unsigned char two=0;
  printf("Quantity 1 = ");
  scanf("%d",&one);
  printf("Quantity 2 = ");
  scanf("%d",&two);
  printf("The value is %d",one);
  return 0;
}

Why when I am trying to see the value of one the initial value appears and not the value after the scanf?

Upvotes: 5

Views: 9525

Answers (5)

Nemanja Boric
Nemanja Boric

Reputation: 22157

You need to use int type in conjuction with %d specifier, and char with %c specifier. And %u with unsigned integers.

#include<stdio.h>
int main()
{
    unsigned int one=0; unsigned int two=0;
    printf("Quantity 1 = ");scanf("%u",&one);
    printf("Quantity 2 = ");scanf("%u",&two);
    printf("The value is %u",one);
    return 0;
}

Basicaly, scanf will try to read integer from input and it will try to store it inside memory location that is not large enough, so you will have undefined behavior.

You can find good reference here.

However, if you try to use character for an input type, you may want ask yourself why you won't get a chance to enter a second Quantity (if you type 4 and press enter). This is because second scanf will read enter key as a character. Also, if you try to type 21 (for a twentyone), it will fill the first value with 2 and second with 1 (well, with their ASCII values).

So, be careful - be sure that you always choose the right type for your variables.

Upvotes: 8

abelenky
abelenky

Reputation: 64682

You declared the variable one to be a char:

unsigned char one=0;

But then you told scanf to read an int:

scanf("%d",&one);  /* %d means int */

Int is bigger than char (typically 4-bytes vs. 1-byte), causing the problem you describe.

Change your scanf to:

scanf("%c",&one);  /* %c means char */

Then when you print out the value, also print a char:

printf("The value is %c",one); /* %c means char */

Upvotes: 1

zwol
zwol

Reputation: 140619

Never use scanf. Never use scanf. Seriously, never use scanf.

Use fgets (or getline, if you have it) to read an entire line of input from the user, then convert strings to numbers with strtol or its relatives strtod and strtoul. strsep may also be useful.

Upvotes: 2

haccks
haccks

Reputation: 106012

Change 

unsigned char one=0; unsigned char two=0;

to 

unsigned int one=0; unsigned int two=0;

and also use %u instead of %d then it will print the value after scanf().

Upvotes: 1

0xF1
0xF1

Reputation: 6116

Check if scanf() is working properly by reading its return value. For quickstart, read the details about scanf() at this link.

What you are doing is inputting a integer using "%d" into an unsigned char variable, therefore scanf() may not be working as it should.

Upvotes: 1

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