How can I store and print a character input? - c

I am familiar with storing and printing characters using getchar(); and putchar();.
However, when I use it with my entire code, it does not seem to work. In the command window, it will take the character, but not print it. Thus, I have no idea what the computer is doing. I tried the code for storing and printing a character on its own and it works fine.
int ans;
printf("\n\t Would you like to remove an item from your cart? (Y or N): ");
ans = getchar();
printf("\n\t ");
putchar(ans);
But as soon as I use it with the entire code, it does not work properly.
#include <stdio.h>
void main()
{
float items[6];
float sum;
float taxSum;
printf("\n\n\n\n");
printf("\t Please enter the price for Item 1: ");
scanf_s(" %f", &items[0]);
while (!((items[0] >= 0.001) && (items[0] <= 999.99)))
{
printf("\n\t [ERROR] Please enter number between $0.01 and $999.99: ");
scanf_s(" %f", &items[0]);
}
int ans;
printf("\n\t Would you like to remove an item from your cart? (Y or N): ");
ans = getchar();
printf("\n\t ");
putchar(ans);
I'm extremely curious as to why that is and what I need to do to get it work.

Use
char ans;
printf("\n\t Would you like to remove an item from your cart? (Y or N): ");
scanf( " %c", &ans );
^^^^^
ans = toupper( ( unsigned char )ans );
putchar( ans );
See the leading space in the format string. It allows to skip white space characters as for example the new line character '\n' that corresponds to the pressed Enter key.
Or as #chux - Reinstate Monica wrote in his comment instead of declaring the variable ans as having the type char you can declare it with the type unsigned char. For example
unsigned char ans;
printf("\n\t Would you like to remove an item from your cart? (Y or N): ");
scanf( " %c", &ans );
^^^^^
ans = toupper( ans );
putchar( ans );

Related

C program, getting the integer value of a character. Why everything comes out after line 10?

Hi I am new to programming and I got a trouble when I try to make a little change to the example in the book.
/* Chapter 3 Example, C Prime Plus */
#include <stdio.h>
int main(void)
{
char Letter, ch;
int intValue;
printf("Please enter a letter: \n");
scanf("%c", &Letter); /* user inputs character */
printf("The code for %c is %d.\n", Letter, Letter);
printf("Now is another we to implement the process: \n");
printf("RN, the value of ch is %c, and the value of intValue is %d\n", ch, intValue);
printf("Please enter a letter: \n");
scanf("%c", &ch);
intValue = ch;
printf("The code for %c is %d.\n", ch, intValue);
return 0;
}
When I run it, the outcome would be
Please enter a letter:
M
The code for M is 77.
Now is another we to implement the process:
RN, the value of ch is , and the value of intValue is 0
Please enter a letter:
The code for
is 10.
and the part
"
Now is another we to implement the process:
RN, the value of ch is , and the value of intValue is 0
Please enter a letter:
The code for
is 10. " will all come out without asking me to enter a value.
I want to know why and are there any other way to implement it that is different from examples in the book?
Thank you for your time!
Hi Matt_C and welcome to SO.
First, you don't need the second bloc of printfs and the scanf, it just trying to do the same thing and there is an order error.
Second, it is tricky when you try consecutive scanf, it holds the last key pressed (enter is the last key pressed = \n). This is why it skips the second scanf.
There a little solution for that, add a space at the beginning of the scanfs. Try this:
int main() {
char exit, letter;
while (1) {
printf("Please enter a letter: ");
scanf(" %c", &letter);
printf("\nThe code for '%c' is %d. \n\n", letter, letter);
printf("Exit ? (y/n): ");
scanf(" %c", &exit);
if(exit == 'y')
{
break;
}
system("clear"); // UNIX
//system("cls"); // DOS
}
}
Don't forget to choose one answer that you believe is the best solution to your problem.

How to read three variables using scanf in C?

I am writing a simple calculator. I want to read first integer and save it in el1. Then I want to print "Enter el2: \n" and after this read the second integer and save it in el2. After this I need to print "Choose from ( + , - , / , * )" and read it and save in op[0]. My program is printing Enter el1: and then it waits for me to input 2 integers, then it pronts Enter el2: and waits for me to input 1 integer and then prints Choose from.. and reads nothing.
int el1 = 0;
printf("Enter el1: \n");
scanf(" %d \n", &el1);
int el2 = 0;
printf("Enter el2: \n");
scanf(" %d \n", &el2);
printf("Choose from ( + , - , / , * ):\n");
char op[2];
scanf("%c", &op[0]);
How to make it work properly?
as mentioned in comments remove white spaces from scanf .otherwise it will wait for you to enter a non-white space character.
and add one space here scanf(" %c", &op[0]); , because it prevent scanf to take \n in buffer as input.
look
printf("Enter el1: \n");
scanf("%d", &el1);
int el2 = 0;
printf("Enter el2: \n");
scanf("%d", &el2);
printf("Choose from ( + , - , / , * ):\n");
char op[2];
scanf(" %c", &op[0]);

C language: The Array did not return a correct value

I have an assignment in C language that requires to ask users to enter values to arrays. My idea is createing two different arrays which one contains integer values and the other holds character values. This is my code so far:
#include <stdio.h>
int main()
{
char continued;
int i = 0;
char instrType[10];
int time[10];
printf("\nL-lock a resource");
printf("\nU-unlock a resource");
printf("\nC-compute");
printf("\nPlease Enter The Instruction Type");
printf(" and Time Input:");
scanf("%c", &instrType[0]);
scanf("%d", &time[0]);
printf("\nContinue? (Y/N) ");
scanf("%s", &continued);
i = i + 1;
while (continued == 'Y' || continued == 'y')
{
printf("\nL-lock a resource");
printf("\nU-unlock a resource");
printf("\nC-compute");
printf("\nPlease Enter The Instruction Type ");
printf("Time Input:");
scanf("%c", &instrType[i]);
scanf("%d", &time[i]);
printf("\nContinue? (Y/N) ");
scanf("%s", &continued);
i = i + 1;
}
return 0;
}
The expected value should be: L1 L2 C3 U1
My Screenshot
The loop just stopped when I tried to enter new values and the condition did not check the value even I entered 'Y' meaning 'yes to continue' please help :(
You are comparing a string with a character that is instead of using scanf("%s",&continued) try using "%c"
The main problem is scanf("%c", &char) because scanf() after had read the input print a \n to pass at the next line, this cause that the next scanf() instead of reading your input, go to read \n, causing the failure in the reading of the input.
To avoid this problem put a space before %c ==> scanf(" %c", &char)
#include <stdio.h>
int main()
{
char continued;
int i = 0;
char instrType[10];
int time[10];
do
{
printf("L-lock a resource\n");
printf("U-unlock a resource\n");
printf("C-compute\n");
printf("Please Enter The Instruction Type and Time Input: ");
scanf(" %c%d", &instrType[i], &time[i]);
printf("Continue? (Y/N) ");
scanf(" %c", &continued);
i++;
} while (continued == 'Y' || continued == 'y');
return 0;
}
Other things:
Instead of i = i + 1 you can use i++
Instead of using a while() is better using a do{...}while() for saving some line of code.
You can concatenate more inputs in a single line ==> scanf(" %c%d", &instrType[i], &time[i])

Program that asks the user for a number n and gives him the possibility to choose between computing the sum and computing the factorial of n

The code is supposed to ask the user whether to find the sum of numbers from 1 to x or finding the factorial of x. After taking the user's input for the value of x, the program directly ends without running the if and else if statement. This is my code.
#include <stdio.h>
int sum(int num);
int fact(int num);
int main(void)
{
int x = 0;
char choice;
printf("Enter a number : \n");
scanf("%d", &x);
printf("Enter f for factorial, s for sum \n");
choice = getchar();
//These lines are ignored by C
if (choice == 'f' || choice == 'F')
{
printf("The factorial of %i is %i \n",x, fact(x));
}
else if (choice == 's' || choice == 'S')
{
printf("The sum from 1 to %i is %i \n",x, sum(x));
}
}
int sum (int num)
{
int sum =0;
for (int i =1; i <=num; i++ )
sum = sum+i;
return sum;
}
int fact (int num)
{
int fact =1;
for (int i =1; i <=num; i++ )
fact = fact*i;
return fact;
}
Can anyone please explain to me what is wrong with my code and how can I fix it? Thank you.
I think buffer problem. So, use
scanf(" %d", &x);
^^^
white-space
instead of
scanf("%d", &x);
and also, use
scanf(" %c", &choice);
instead of
choice = getchar();
The problem in this code is in getchar() function.
In first scanning : scanf("%d", &x); when user press enter key, it remain in the input buffer and the integer val is stored in variable x.
In second scanning: choice = getchar();, it reads the enter key in variable choice.
And you have written only two conditions:
if (choice == 'f' || choice == 'F')
else if (choice == 's' || choice == 'S')
That's why it is directly ending the code; as there is no code written for choice = other than 'f' and 's'
if you write 'else' part like this:
else printf("%d", choice);
It will print: 10 which is the ascii value of Enter / New line feed.
To avoid this, try to make following changes in your code:
int x = 0;
char choice;
printf("Enter a number : \n");
scanf("%d", &x); //here the integer is scanned in variable 'x'
choice = getchar(); //here the enter key is scanned in variable 'choice' so now input buffer is free
printf("Entr f for factorial, s for sum \n");
scanf("%c", &choice); //here the character entered by use will be stored in variable 'choice' so it is overwritten.

scanf function not working with characters

I was just trying out a simple program in C, inserting into an array.
I used a scanf function to accept characters but it seems the compiler just skipped that and just went to end of the program.
This was the code that I used :-
#include <stdio.h>
void main()
{
int a[50], i, j, m, n, x;
char ch;
printf("Enter the no. elements of the array :- ");
scanf("%d", &m);
printf("Enter the elements below :- ");
for (i = 0; i < m; i++)
{
scanf("%d", &a[i]);
}
printf("The array is :- \n");
for (i = 0; i < m; i++)
{
printf("%d", a[i]);
}
printf("\nDo you want to enter an element ? (Y/N)\n");
scanf("%c", &ch); // The compiler just skips this along with the
while (ch == 'y' || ch == 'Y') // while loop and goes straight to the printf
{ // statement
printf("The index of the element :- ");
scanf("%d", &n);
printf("\nEnter a number :- ");
scanf("%d", &x);
for (i = m; i > n; i--)
{
a[i] = a[i - 1];
}
a[n] = x;
printf("\nInsert more numbers ? (Y/N)");
scanf("%c", &ch);
m = m + 1;
}
printf("\nThe array is :- ");
for (i = 0; i < m; i++)
{
printf("%d", a[i]);
}
}
I used the variable ch in order to allow the user to have a choice, whether or not to insert elements i.e. Y or N.
But the compiler basically skips the third scanf function, the one that accepts the char, along with the while loop.
I just want to know why the scanf function was skipped ?
Back to the previous scanf which is the last array member.
scanf("%d",&a[i])
In the input file if you entered:
32\n
^^
input will wait just before the newline after reading the decimal number.
In the scanf which causes problem:
scanf("%c", &ch);
It will read the newline character as it is available in input that's why it will skip that line after being executed implicitly.
To ignore the whitespace you have only to add space before the specifier %c as stated in comment by #xing and #WeatherVane.
scanf(" %c",&ch);
C99 7.19.6.2
Input white-space characters (as specified by the isspace function)
are skipped, unless the specification includes a [, c, or n
specifier.250)

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