I have a simple problem;
Here is the code :
#include<stdio.h>
main(){
int input;
printf("Choose a numeric value");
scanf("%d",&input);
}
I want the user to only enter numbers ...
So it has to be something like this :
#include<stdio.h>
main(){
int input;
printf("Choose a numeric value");
do{
scanf("%d",&input);
}while(input!= 'something');
}
My problem is that I dont know what to replace in 'something' ... How can I prevent users from inputting alphabetic characters ?
EDIT
I just got something interesting :
#include<stdio.h>
main(){
int input;
printf("Choose a numeric value");
do{
scanf("%d",&input);
}while(input!= 'int');
}
Adding 'int' will keep looping as long as I enter numbers, I tried 'char' but that didnt work .. surely there is something for alphabets right ? :S
Please reply !
Thanks for your help !
The strtol library function will convert a string representation of a number to its equivalent integer value, and will also set a pointer to the first character that does not match a valid number.
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
...
int value;
char buffer[SOME_SIZE];
char *chk;
do
{
printf("Enter a number: ");
fflush(stdout);
if (fgets(buffer, sizeof buffer, stdin) != NULL)
{
value = (int) strtol(buffer, &chk, 10); /* assume decimal format */
}
} while (!isspace(*chk) && *chk != 0);
If chk points to something other than whitespace or a string terminator (0), then the string was not a valid integer constant. For floating-point input, use strtod.
You can't prevent the user from entering anything he wants -- you can only ignore anything s/he enters that you don't "like".
A typical pattern is to read a string with fgets, then scan through the string and check that all the input was digits with isdigit. If it was all digits, then convert to an integer; otherwise, throw it away and get the input again.
Alternatively, use strtol to do the conversion. It sets a pointer to the end of the data it could convert to a number; in this case you (apparently) want it to point to the end of the string.
If you don't mind some non-portable code, you can read one character at a time, and throw away anything but digits (e.g. with getch on Windows).
You should not use scanf to read in numbers - see http://www.gidnetwork.com/b-63.html
Use fgets instead.
However, if you must use scanf, you can do this:
#include <stdio.h>
int main() {
char text[20];
fputs("enter some number: ", stdout);
fflush(stdout);
if ( fgets(text, sizeof text, stdin) ) {
int number;
if ( sscanf(text, "%d", &number) == 1 ) {
printf("number = %d\n", number);
}
}
return 0;
}
Personally, I would read the input into a buffer and scan that string for my number.
char buffer[100];
float value;
do {
scanf("%s", buffer);
} while ( sscanf(buffer,"%f", &value) != 1 )
This will loop until the first thing the user enters on the line is a number. The input could be anything but will only get past this block when the first thing entered is a number.
example input:
43289 (value is 43289)
43.33 (value is 43.44)
392gifah (value is 392)
ajfgds432 (continues to loop)
Related
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...
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.
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.
I want the program to ask the user for a number, and if the user does not enter a number the program will say "input not a integer."
Thx for the help guys!
I propose this (it doesn't handle integer overflow):
#include <stdio.h>
int main(void)
{
char buffer[20] = {0}; // 20 is arbitrary;
int n; char c;
while (fgets(buffer, sizeof buffer, stdin) != NULL)
{
if (sscanf(buffer, "%d %c", &n, &c) == 1)
break;
else
printf("Input not integer. Retry: ");
}
printf("Integer chosen: %d\n", n);
return 0;
}
EDIT: Agreed with chux suggestions below!
One possible way: use scanf() function to read the input. It returns the number of items it successfully read.
Another way: read the input as string with scanf() of fgets() and then try to parse it as integer.
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.