I was trying out a simple program in C for validating user data. The program is supposed to identify whether a user entered character is a number, alphabet or a special character.
Somehow , the code identifies every kind of input character as a number. I have appended the code below, I'd be grateful if someone could kindly point out where I'm going wrong ?
//Program to take input from the user and determine whether it is character, number, or a special character
#include<stdio.h>
#include<conio.h>
#include<string.h>
char ch;
int main()
{
clrscr();
printf("Enter a character \n");
scanf("%c \n",ch);
if ((ch>='A'&& ch<='Z')||(ch>='a'&& ch<='z') )
{
printf("The character entered is an alphabet \n" );
}
else if ((ch>=0)&&(ch<=9))
{
printf("Character entered is an number \n");
}
else
{
printf("Character entered is a special character");
}
return 0;
}
scanf accepts a pointer as the argument for %c. In other words,
scanf("%c \n",ch);
should be written as:
scanf("%c\n",&ch);
Without the reference operator (&), scanf receives the value of ch. In this case, the value is garbage, because ch is unset.* Referencing ch gives scanf a pointer to ch, not ch itself, so scanf can modify the value of ch by dereferencing the pointer (using the dereference operator, *).
There's also the issue with digit checking that Himadri mentioned.
* This is actually undefined behaviour.
Oh, Arun very silly mistake.
In your second condition in else if you have to right 0 and 9 in single quotation mark.
So, your code will be -
if ((ch>='A'&& ch<='Z')||(ch>='a'&& ch<='z') )
{
printf("The character entered is an alphabet \n" );
}
else if ((ch>='0')&&(ch<='9'))
{
printf("Character entered is an number \n");
}
else
{
printf("Character entered is a special character");
}
May be this is the only mistake. Now, it should work.
A few comments on style:
conio.h and clrscr() are non-standard.
Global variables are bad (char ch). Declaring them non-static is also bad.
Always check the return value of scanf. This will help you catch input format errors.
In this case, as we need to just a single character, getchar is more appropriate.
This is how I would've written this program:
#include <stdio.h>
#include <ctype.h>
int main()
{
int ch; /* We use an int because it lets us check for EOF */
printf("Enter a character: ");
fflush(stdout); /* Remember to flush the output stream */
ch = getchar();
if (ch == EOF)
{
printf("end-of-file or input-error\n");
return 1;
}
if (isalpha(ch))
printf("The character entered is an alphabet\n" );
else if (isdigit(ch))
printf("Character entered is an number\n");
else
printf("Character entered is a special character\n");
return 0;
}
Related
I am trying to take in user input with spaces and store it in an array of characters.
After, I want to take in a single character value and store it as a char.
However, when I run my code, the prompt for the character gets ignored and a space is populated instead. How can I take in an array of chars and still be allowed to prompt for a single character after?
void main()
{
char userIn[30];
char findChar;
printf("Please enter a string: ");
scanf("%[^\n]s", userIn);
printf("Please enter a character to search for: ");
scanf("%c", &findChar);
//this was put here to see why my single char wasnt working in a function I had
printf("%c", findChar);
}
scanf("%c", &findChar); reads the next character pending in the input stream. This character will be the newline entered by the user that stopped the previous conversion, so findChar will be set to the value '\n', without waiting for any user input and printf will output this newline without any other visible effect.
Modify the call as scanf(" %c", &findChar) to ignore pending white space and get the next character from the user, or more reliably write a loop to read the read and ignore of the input line.
Note also that scanf("%[^\n]s", userIn); is incorrect:
scanf() may store bytes beyond the end of userIn if the user types more than 29 bytes of input.
the s after the ] is a bug, the conversion format for character classes is not a variation of the %s conversion.
Other problems:
void is not a proper type for the return value of the main() function.
the <stdio.h> header is required for this code.
Here is a modified version:
#include <stdio.h>
int main() {
char userIn[30];
int c;
char findChar;
int i, found;
printf("Please enter a string: ");
if (scanf("%29[^\n]", userIn) != 1) {
fprintf(stderr, "Input failure\n");
return 1;
}
/* read and ignore the rest of input line */
while ((c = getchar()) != EOF && c != '\n')
continue;
printf("Please enter a character to search for: ");
if (scanf("%c", &findChar) != 1) {
fprintf(stderr, "Input failure\n");
return 1;
}
printf("Searching for '%c'\n", findChar);
found = 0;
for (i = 0; userIn[i] != '\0'; i++) {
if (userIn[i] == findChar) {
found++;
printf("found '%c' at offset %d\n", c, i);
}
}
if (!found) {
printf("character '%c' not found\n", c);
}
return 0;
}
scanf("%[^\n]s", userIn); is a bit weird. The s is guaranteed not to match, since that character will always be \n. Also, you should use a width modifier to avoid a buffer overflow. Use scanf("%29[^\n]", userIn); That alone will not solve the problem, since the next scanf is going to consume the newline. There are a few options. You could consume the newline in the first scanf with:
scanf("%29[^\n]%*c", userIn);
or discard all whitespace in the next call with
scanf(" %c", &findChar);
The behavior will differ on lines of input that exceed 29 characters in length or when the user attempts to assign whitespace to findChar, so which solution you use will depend on how you want to handle those situations.
While trying different things with getchar I figured out that it usually only safes on character in an variable. Somehow when I use a while loop the behaviour changes and it returns more characters if the input is more than one.
Here is my example code for "normal" behaviour. putchar() return only one character even more are put in:
#include <stdio.h>
main() {
char c;
printf("Type a character in here: ");
c = getchar();
printf("You just typed : ");
putchar(c);
}
Somehow when I want to use it in a while loop the putchar() function returns more than one character if more are put in:
Here is the second part:
#include <stdio.h>
void print_input();
main() {
char c;
printf("Type a character in here: ");
c = getchar();
printf("You just typed : ");
putchar(c);
print_input();
}
void print_input() {
char ch = 'x';
while (ch != '#') {
ch = getchar();
putchar(ch);
}
return;
}
Additional question:
While running this in debugger the behaviour somehow is different than if I try this in runtime. Why is that so?
When you enter multiple chars and hit enter, the program will see that whole input (because it's line buffered). So multiple calls to getchar will return subsequent characters and remove them from the stream:
Try to play with this:
char c;
char d;
printf("Type chars in here: ");
c = getchar();
d = getchar();
printf("C: %c \n", c);
printf("D: %c", d);
This is duplicated with:
How is the "getchar()" function able to take multiple characters as input?
You can read more there.
I am trying to make a simple code that will read a char from input and execute "Correct" or "Incorrect input" and run the code again until the correct input is entered. First of all it does not work for capital X. The other issue that I want to fix is that after the incorrect input I have to press enter to get the "Enter x" message, instead of getting in immediately after the incorrect input message.
#include <stdio.h>
int main()
{
do
{
printf("Enter x\n");
if (getchar()=='x'|| getchar()=='X')
{
printf("Entered char is X\n");
return 0;
}
else
{
printf("Input incorrect! Please try again!!!\n");
}
}
while (getchar()!='x' || getchar()!='X');
return 0;
}
You need to store the input in a variable, otherwise you keep asking for input several times in a row, for each getchar call.
For weird historic reasons, getchar actually returns an int, since the value EOF that can be returned from it is an int. So the variable must be int.
And finally, each time the user hits enter, a invisible line feed character \n is appended to the input stream. This character does you no good, so you should discard it with an extra read.
#include <stdio.h>
int main (void)
{
int input;
do
{
printf("Enter x\n");
input = getchar();
getchar(); // extra getchar to chew up line feed from stdin
if (input=='x'|| input=='X')
{
printf("Entered char is X\n");
}
else
{
printf("Input incorrect! Please try again!!!\n");
}
} while (input!='x' && input!='X');
return 0;
}
Please note that the opposite of input=='x'|| input=='X' is input!='x' && input!='X' (De Morgan's laws). "If input is not 'x' and input is not 'X' then loop".
When you hit the ENTER key the newline character \n is placed in input buffer. You need to consume that newline character in order to read the next character.
Also you are reading two time, which is unnecessary in this case. So your code should be like this
#include <stdio.h>
int main()
{
char inp;
do
{
printf("Enter x\n");
inp = getchar();
getchar(); // reading the newline character '\n'
if (inp == 'x'|| inp =='X')
{
printf("Entered char is X\n");
return 0;
}
else
{
printf("Input incorrect! Please try again!!!\n");
}
}
while (inp !='x' || inp !='X');
return 0;
}
p.s There is no need to put condition checking in while loop, since you are returning in if condition. while(true) would work fine. Thanks #bruno for pointing that out.
In your code:
if (getchar()=='x'|| getchar()=='X')
getchar() is called twice.
Instead, you should write it this way:
char c = getchar();
if (c=='x'|| c=='X')
for the second part, if your goal is print the message on a new line, then just simply change your printf to:
printf("\nInput incorrect! Please try again!!!\n");
I'm just a beginner and am trying to make a a program that asks for a number and if a letter is input, it says "that's not a number" and asks for a number again, until a number is input.
However, my program keeps going into an infinite loop with the current code. Any help would be appreciated to fix this. Also, I would also like the program to say "please input something" if nothing is input, but don't know how to do this. Thanks.
#include <stdio.h>
int main()
{
float i;
printf("enter a number");
while(scanf("%f", &i) != 1)
{
puts("That is not a number.");
scanf("%f", &i);
}
}
You need to clear the bad input from stdin after your scanf fails:
#include <stdio.h>
int main()
{
float i;
char trash[1024];
while (1)
{
printf("Please enter a number: ");
fflush(stdout);
if (1 == scanf("%f", &i))
break;
/* scanf failed: clear the bad input from stdin */
if (NULL == fgets(trash, sizeof(trash), stdin)) /* NOTE: assumes 1 entry per line and no line longer than 1023 characters */
exit((fprintf(stderr, "Unexpected EOF or error!\n"), 1));
puts("That is not a number.");
}
printf("You entered: %f!\n", i);
return 0;
}
As an alternative to the fgets() to clear the line, you could call scanf("%1023s", trash), which would only suck in the next whitespace delimited series of characters. This would allow you to handle multiple entries on a single line with mistakes intermixed, for example.
Your program goes into infinite loop because after the invalid input, (scanf("%d", &i) != 1)condition being TRUE, the invalid input which is left in the input buffer is not consumed, it's still in the buffer. So the same invalid input is read over and over again.
To avoid, once scanf() fails, you need to flush out all the input buffer contains before calling next scanf().
Maybe inside the while loop, calling getchar() until a newline or EOF will help. Also, the second scanf() can be removed, IMHO.
After your non numeric[More precisely input which doesn't match the formating of scanf()] input you need to clear the stdin. If not the same input will be read till stdin get cleaned or you terminate the program. A reference answer can be found on this question
Quoted
On success, the function returns the number of items of the argument list successfully filled. This count can match the expected number of items or be less (even zero) due to a matching failure, a reading error, or the reach of the end-of-file.
Reason for infinite loop : Since you don't clear stdin, same values will be read by scanf() and always full fill while condition resulting in a infinite loop.
Use following edited code :
#include <stdio.h>
int main()
{
float i;
char c;
printf("enter a number");
while(scanf("%f", &i) != 1)
{
puts("That is not a number.");
scanf("%f", &i);
while ((c = getchar()) != '\n' && c != EOF); // Flush stdin
}
}
Your code is taking input of a number. Thats why if you even give input a letter, it will take the ASCII value of the letter, which is a valid number. And Further more please take a character as an input. Here I have modified your code which should work
#include <stdio.h>
int main()
{
char i;
printf("enter a number");
while(1)
{
scanf("%c",&i);
if (c >=48 && c <= 57) // here ascii value of numbers between 0-9 is 48-57 respectively
{
puts("That is a number.");
break;
}
else
{
puts("That is not a number.");
scanf("%c",&i);
}
}
}
I created a program to make a diamond out of *'s. I am looking for a way to check if the type of input is an integer in the C language. If the input is not an integer I would like it to print a message.
This is what I have thus far:
if(scanf("%i", &n) != 1)
printf("must enter integer");
However it does not display the message if it's not an integer. Any help/guidance with this issue would be greatly appreciated!
you can scan your input in a string then check its characters one by one, this example displays result :
0 if it's not digit
1 if it is digit
you can play with it to make your desired output
char n[10];
int i=0;
scanf("%s", n);
while(n[i] != '\0')
{
printf("%d", isdigit(n[i]));
i++;
}
Example:
#include <stdio.h>
#include <string.h>
main()
{
char n[10];
int i=0, flag=1;
scanf("%s", n);
while(n[i] != '\0'){
flag = isdigit(n[i]);
if (!flag) break;
i++;
}
if(flag)
{
i=atoi(n);
printf("%d", i);
}
else
{
printf("it's not integer");
}
}
Use fgets() followed by strtol() or sscanf(..."%d"...).
Robust code needs to handle IO and parsing issues. IMO, these are best done separately.
char buf[50];
fgets(buf, sizeof buf, stdin);
int n;
int end = 0; // use to note end of scanning and catch trailing junk
if (sscanf(buf, "%d %n", &n, &end) != 1 || buf[end] != '\0') {
printf("must enter integer");
}
else {
good_input(n);
}
Note:
strtol() is a better approach, but a few more steps are needed. Example
Additional error checks include testing the result of fgets() and insuring the range of n is reasonable for the code.
Note:
Avoid mixing fgets() and scanf() in the same code.
{ I said scanf() here and not sscanf(). }
Recommend not to use scanf() at all.
strtol
The returned endPtr will point past the last character used in the conversion.
Though this does require using something like fgets to retrieve the input string.
Personal preference is that scanf is for machine generated input not human generated.
Try adding
fflush(stdout);
after the printf. Alternatively, have the printf output a string ending in \n.
Assuming this has been done, the code you've posted actually would display the message if and only if an integer was not entered. You don't need to replace this line with fgets or anything.
If it really seems to be not working as you expect, the problem must be elsewhere. For example, perhaps there are characters left in the buffer from input prior to this line. Please post a complete program that shows the problem, along with the input you gave.
Try:
#include <stdio.h>
#define MAX_LEN 64
int main(void)
{ bool act = true;
char input_string[MAX_LEN]; /* character array to store the string */
int i;
printf("Enter a string:\n");
fgets(input_string,sizeof(input_string),stdin); /* read the string */
/* print the string by printing each element of the array */
for(i=0; input_string[i] != 10; i++) // \0 = 10 = new line feed
{ //the number in each digits can be only 0-9.[ASCII 48-57]
if (input_string[i] >= 48 and input_string[i] <= 57)
continue;
else //must include newline feed
{ act = false; //0
break;
}
}
if (act == false)
printf("\nTHIS IS NOT INTEGER!");
else
printf("\nTHIS IS INTEGER");
return 0;
}
[===>] First we received input using fgets.Then it's will start pulling each digits out from input(starting from digits 0) to check whether it's number 0-9 or not[ASCII 48-57],if it successful looping and non is characters -- boolean variable 'act' still remain true.Thus returning it's integer.