Counting Number Of User Input in C Program - c

printf("Enter number of patients:");
int numberOfInputs = scanf("%d", &patients);
if (numberOfInputs != 1) {
printf("ERROR: Wrong number of arguments. Please enter one argument d.\n");
}
I am asking the user to input one number as an argument, but would like to print out a statement if the user does not input anything or puts in more than one input. For example, once prompted with "Enter number of patients:", if the user hits enter without entering anything, I would like to print out a statement. The code above is what I have been specifically tinkering around with it for the past couple hours as a few previous posts on this site have suggested but when I run it in terminal, it does not work. Any suggestions? Thank you in advance, and all advice is greatly appreciated!

If I understand your question right, you want to print an error when the input is anything other than an integer and this includes newline as well. You can do that using a char array and the %[] specifier.
Example:
#include<stdio.h>
#include<stdlib.h>
int main(void)
{
int patients;
char str[10];
printf("Enter number of patients:");
int numberOfInputs = scanf("%[0-9]", str);
if (numberOfInputs != 1) {
printf("ERROR: Wrong number of arguments. Please enter one argument.\n");
}
patients = atoi(str); //This is needed to convert the `str` back to an integer
}
This will print the error when the user just hits ENTER as well.

This looks super over-complicated, but it basically splits the input, checks it to be exactly one and than checks it to be an integer (and converts it). It works fine in loop as well and handles empty input.
I'm sure there are more elegant solutions to this problem, it's just a suggestion.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <stdbool.h>
int getNumberOfInput(char* str);
bool isNumber(char* str);
int main()
{
char str[512];
while(1)
{
printf("Enter text: ");
fgets(str, 512, stdin);
int numberOfInput = getNumberOfInput(str);
if ( numberOfInput == 0 )
printf("You must give an input\n");
else if ( numberOfInput > 1 )
printf("You have to give exactly one input\n");
else
{
if (!isNumber(str))
printf("The input is not an integer\n");
else
{
int input = atoi(str);
printf("input: %d\n", input);
}
}
}
return 0;
}
int getNumberOfInput(char* str)
{
char* word = strtok(str, " \t\n\v\f\r");
int counter = 0;
while(word != NULL)
{
++counter;
word = strtok(NULL, " \t\n\v\f\r");
}
return counter;
}
bool isNumber(char* str)
{
int i, len = strlen(str);
for (i=0; i<len; ++i)
if (!isdigit(str[i]))
return false;
return true;
}

Related

Program fails to receive second input

It's just a code to receive user inputs in C program, but fails to do so and accepts null space as input. I have tried fgets() as well and the same thing keeps happening. Please advice on how to fix.
#include <math.h>
#include <stdio.h>
//#include <string.h>
#define len 16
int main(void)
{
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int n,i=0,j=0;
printf("enter the number of cards:");
n = getchar();
//scanf("%d",&n);
int c1[len][n],card[len][n];
char buf[len];
printf("Enter card number:");
gets(buf);
system("Pause");
return (0);
}
"...code to receive user inputs in c program, but fails to do so and accepts null space as input..."
The reasons your existing code has problems is covered well in the comments under your post.
Consider a different approach: Define the following:
char inBuf[80] = {0};//
int numCards = 0;//Pick variable names that are descriptive (n is not)
int cardNum = 0;
bool isnum;
Then use it in conjunction with printf() etc.
printf("enter the number of cards:");
if(fgets(inBuf, sizeof(inBuf), stdin))//will read more than just a single char, eg. "12345"
{
int len = strlen(inBuf);
isnum = true;
for(int i=0;i<len;i++)
{
if(!isdigit(inBuf[i]))
{
isnum = false;
break;
}
}
if(isnum)
{
numCards = atoi(inBuf);
}
else
{
printf("input is not a number\n"
}
}
printf("Enter card number:");
if(fgets(inBuf, sizeof(inBuf), stdin))
{
...
Repeat variations of these lines as needed to read input from stdin, with modifications to accommodate assignment statements based on user input i.e. an integer (this example is covered), a floating point number, a string (eg. a persons name)
Although there is more that you can do to improve this, it is conceptually viable for your stated purpose...

Encryption in C using a Caesar Cipher

I have been asked to create a program where I have to encrypt multiple pieces of information using a Caesar Cipher. I understand the concept behind it but what I'm having trouble visually is how to enter pieces of data within the function. For example, I have encrypted passwords saved in a file ("hrkk1" meaning "pass1" and so on). I have to create a cipher function to read the input from a scanf and strcmp so it matches what's in the file allowing the user to login.
Whats the best way to validate the user input and make "pass1" turn into "hrkk1" so it matches what's in the file and allows user login?
Thank you
This is the code I have so far:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
#include <windows.h>
void checkValid(void);
void loginDetails(char username[5][6], char password[5][9]);
void encryption(char username[5][6], char password[5][9]);
int main(void)
{
FILE *EP;
FILE *UN;
char username[5][6];
char password [5][9], ch, key;
EP = fopen("encrypted_passwords.txt", "r");
fscanf(EP, "%s %s %s %s %s", password[0], password[1],
password[2], password[3], password[4]);
fclose(EP);
UN = fopen("username.txt", "r");
fscanf(UN, "%s %s %s %s %s", username[0], username[1], username[2],
username[3], username[4]);
fclose(UN);
printf("Welcome.");
loginDetails(username, password);
return 0;
}
void loginDetails(char username[5][6], char password[5][9])
{
int i;
char nurseUsername[6];
char nursePassword[6];
bool useValid = 0;
bool passValid = 0;
printf("Please Enter your username: \n");
scanf("%s", nurseUsername);
for (i = 0; i < 5; i++)
{
if(strcmp(nurseUsername, username[i]) == 0)
{
useValid = 1;
}
}
if(useValid != 1)
{
printf("\nError. Invalid Username. Returning to menu.\n");
Sleep(1000);
system("cls");
main();
}
else
{
printf("\nPlease enter your password: \n");
scanf("%s", nursePassword);
}
for(i = 0; i < 5; i++)
{
if((strcmp(nurseUsername, username[i]) == 0) &&
(strcmp(nursePassword, password[i]) == 0))
{
passValid = 1;
}
if(passValid != 1)
{
printf ("Error. Invalid Password. Returning to menu.\n");
Sleep(1000);
system("cls");
main();
}
else
{
printf("\nLogin Successful. Loading menu.\n");
Sleep(1000);
system("cls");
patientEntry();
}
}
}
You need to use the shifting of the character in c. This is possibile with a simple addition( or subtraction) on a char value.
Pay attention your example doesn't shift the number character and maybe also the character doesn't go out the alphabet, and it take in consideration also the capital letters. So pay attention when you do the addition to not exceed the range of capital or non capital letter. My suggestion is to use ascii table.

C Program - How to deny any non-numerical input

I've just started learning the language of C, and would love your help in cleaning up / simplifying my code if you know a better way to reach the following.
I want a program to ask for a number, and if that is found then proceed to print and end, however if anything else is put in (e.g. a letter key), then I want the program to loop asking for a number until one is given.
I started off by using a simple scanf input command, but this seemed to go into an infinite loop when I tried to check if a valid number (as we define them) was put in.
So instead I have ended up with this, from playing around / looking online, but I would love to know if there is any more efficient way!
//
// Name & Age Program
// Created by Ben Warren on 1/3/18.
//
#include <stdio.h>
int main (void)
{
//Setting up variables
int num;
char line[10]; /* this is for input */
//Collecting input
printf("Please enter any number? \t");
scanf("%d", &num);
//If Invalid input
while (num==0)
{
printf("\nTry again:\t");
fgets(line, 10, stdin); //turning input into line array
sscanf(line, "%d",&num); //scaning for number inside line and storing it as 'num'
if (num==0) printf("\nThat's not an number!");
}
//If Valid input
{
printf("\n%d is nice number, thank you! \n\n", num);
*}*
return 0;
}
Instead of checking if the value is different to 0, check the return value of
sscanf. It returns the number of conversions it made. In your case it should be 1. Unless the return value is 1, keep asking for a number.
#include <stdio.h>
int main(void)
{
int ret, num;
char line[1024];
do {
printf("Enter a number: ");
fflush(stdout);
if(fgets(line, sizeof line, stdin) == NULL)
{
fprintf(stderr, "Cannot read from stdin anymore\n");
return 1;
}
ret = sscanf(line, "%d", &num);
if(ret != 1)
fprintf(stderr, "That was not a number! Try again.\n");
} while(ret != 1);
printf("The number you entered is: %d\n", num);
return 0;
}
That is not a bad approach for someone new to C. One small improvement would be to actually check the return value of scanf(), since it returns the number of arguments successfully retrieved. Then you could get away from relying on num being 0 to indicate the input was valid. Unless you do want to specifically flag 0 as invalid input.
int ret = scanf("%d", &num);
ret == 1 would mean an integer was succesffully read into num, ret == 0 would mean it was not.
Consider using strtol to parse a string for a long int. This also allows you to detect trailing characters. In this example if the trailing character is not a newline, the input can be rejected. strtol can also detect overflow values. Read the documentation to see how that works.
#include <stdio.h>
#include <stdlib.h>
int main (void)
{
//Setting up variables
long int num = 0;
char line[40] = ""; /* this is for input */
char *parsed = NULL;
printf("Please enter any number? \t");
fflush ( stdout);
while ( fgets(line, 40, stdin))
{
parsed = line;//set parsed to point to start of line
num = strtol ( line, &parsed, 10);
if ( parsed == line) {//if parsed equals start of line there was no integer
printf("Please enter a number? \t");
printf("\nTry again:\t");
fflush ( stdout);
continue;
}
if ( '\n' != *parsed) {//if the last character is not a newline reject the input
printf("Please enter only a number? \t");
printf("\nTry again:\t");
fflush ( stdout);
}
else {
break;
}
}
if ( !parsed || '\n' != *parsed) {
fprintf ( stderr, "problem fgets\n");
return 0;
}
printf("\n%ld is nice number, thank you! \n\n", num);
return 0;
}
0 (zero) is a number...
But I see what you want to do...
You can check for a valid number, using isdigit or a combination of similar functions
I think its also important to follow the advice of other answers to use the return value from scanf using code such as:
int ret = scanf("%d", &num);
and examining ret for success or failure of scanf.

check input program gets stuck in an infinte loop

I'm trying to create a program that asks to type something and check if it is an integer. If it is an integer, then print "the integer is ...". Else, print "try again" and waits for another input. However, the program prints an infinite number of "try again" if you type in a character. Here's the source code:
#include <stdio.h>
#include <stdbool.h>
int main()
{
int inp;
bool t = 1;
printf("type an integer\n");
while (t) {
if (scanf("%i", &inp) == 1) {
printf("The integer is %i", inp);
t = 0;
} else {
printf("try again");
scanf("%i", &inp);
}
}
}
OP's code fail to consume the offending non-numeric input. It remains in stdin, for the next input function. As it is unfortunately just another scanf("%i", &inp) which fails the same way - infinite loop.
After attempting to read an int, read the rest of the line.
#include <stdio.h>
#include <stdbool.h>
int main() {
int inp;
int scan_count;
printf("Type an integer\n");
do {
scan_count = scanf("%i", &inp); // 1, 0, or EOF
// consume rest of line
int ch;
while ((ch == fgetchar()) != '\n' && ch != EOF) {
;
}
} while (scan_count == 0);
if (scan_count == 1) {
printf("The integer is %i\n", inp);
} else {
puts("End of file or error");
}
}
An even better approach would read the line of user input with fgets(). Example
When you entered a char, the variable inp in scanf("%d", &inp) would get null, since the input that doesn't match the format string. And the character you input would remain in the buffer, so that's the reason both your scanf would not stop.
A simplest way to fix this is modify your second scanf("%i", &inp); to scanf("%c", &c); (don't forget to declare a char c in your main function).
check here while(t) its in an infinite loop because you have to set a condition for t something like while(t==1) or while(t>1) or (t<1) something like that. saying while(t) means that t can be anything and it will continue to run.
There is nothing in to break the while loop.
consider getting rid of the boolean, and simply using a while (1) loop with a break. Also you should be using "%d" to indicate an integer in scanf/printf. And there is no need for the scanf call in the else, since your program would loop back and call scanf again anyway.
#include <stdio.h>
int main() {
int inp = 0;
printf("type an integer\n");
while (1) {
if (scanf("%d", &inp) == 1) {
printf("The integer is %d", inp);
break;
}
else {
printf("try again");
}
}
return 0;
}
I hope this helped.

Anagram problems

I'm new to this forum and would like to seek help. I'm trying to modify an anagram program based on code from http://www.sanfoundry.com/c-program-...ings-anagrams/.
This time, however, I have used array pointers to obtain input from the user. I have also created a function "check_input" to ensure that the input consists of ONLY characters and excludes symbols(!, #, $). However, when I ran the program, it still accepts those symbols and does not break like I wanted it to. Please help.
Plus, I intend to make the program treat upper-case letters the same way as lower-case letters. Can this be achieved by using the "stricmp" function? If so, where should I place that function? Alternative methods are also appreciated.
Update: Sorry. I've added the check_input code at the bottom.
#include <stdio.h>
#include <ctype.h>
#include <string.h>
int test_anagram(char *ptrArray1, char *ptrArray2);
int check_input(char array1[], char array2[]);
int main()
{
char array1[100], array2[100];
char *pArray1, *pArray2;
int flag;
pArray1 = array1;
pArray2 = array2;
printf("Enter the first word: \n");
gets(pArray1);
printf("Enter the second word: \n");
gets(pArray2);
check_input(pArray1, pArray2);
flag = test_anagram(pArray1, pArray2);
if(flag == 1){
printf("\"%s\" and \"%s\" are anagrams.\n", pArray1, pArray2);
}else{
printf("\"%s\" and \"%s\" are not anagrams.\n", pArray1, pArray2);
}
return 0;
}
int test_anagram(char array1[], char array2[])
{
int num1[26] = {0}, num2[26] = {0}, i = 0;
while(array1[i] != '\0')
{
num1[array1[i] - 'a']++;
i++;
}
i = 0;
while(array2[i] != '\0')
{
num2[array2[i] - 'a']++;
i++;
}
for(i=0;i<26;i++)
{
if(num1[i] != num2[i]){
return 0;
}
return 1;
}
}
int check_input(char array1[], char array2[])
{
while(isalpha((int)array1) != 1){
break;
}
while(isalpha((int)array2) != 1){
break;
}
}
You haven't (yet) posted the full code of the check_input() function but one advice would be to validate the input when the user inputs every character.
You can do this using f.e. the getchar() function and checking if the inputted character is a letter, as well as converting it to the lowercase (or uppercase if you will).
You can do lowercase convertion like this:
#include <ctype.h>
// ...
tolower('A');

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