Related
Friends how can I make Scanf to take 1 or 2 or 3 numbers depending on input data I give?
sample data 1: "1 2 5"
sample data 2: "1 4"
sample data 3: "4"
if(scanf("%lf",&a)==1 )
{
printf("1 input num\n");
}
else if(scanf(" %lf %lf",&a, &b)==2 )
{
printf("2 input num\n");
}
else if(scanf("%lf %lf %lf",&a, &b, &c)==3 )
{
printf("3 input num\n");
}else
{
printf("Error message.\n");
return 1;
}
You might consider this an answer:
int InputNums=0;
InputNums = scanf("%lf %lf %lf",&a, &b, &c);
if(InputNums!=0)
printf("%d input num\n");
else
printf("Error message.\n");
It works by NOT eating one number and then trying whether instead more numbers could have been read, like your shown code does.
Instead try to read three numbers and then let scanf() tell you how many worked.
But actually I am with the commenters. If you do not have guaranteed syntax in your input (which scanf() is for) then use something else.
This is nicely describing which alternative in which situation AND how to get scanf to work in the same situation:
http://sekrit.de/webdocs/c/beginners-guide-away-from-scanf.html
Scanf already does this for you, and indeed you used the same scanf function with a variable number of arguments. You can look here: How do vararg work in C?
However you don't need to overload scanf, but rather pass to it a string telling what you need to scan. You can do this dynamically by changing the string at runtime.
The code to try is the following:
#include <stdio.h>
int main()
{
char one[] = "%d";
char two[] = "%d%d";
int o1;
int t1,t2;
scanf(one,&o1);
scanf(two,&t1,&t2);
printf("%d %d %d",o1,t1,t2);
return 0;
}
If you must use scanf() ....
"%lf" is a problem as it consumes leading white-space including '\n', so we lost where a line of input might have ended.
Instead first look for leading white-space and see if an '\n' occurs.
#define N 3
double a[N];
count = 0;
while (count < N) {
// Consume leading white-spaces except \n
unsigned char ch = 0;
while (scanf("%c", &ch) == 1 && isspace(ch) && ch != '\n') {
;
}
if (ch == '\n') {
break;
}
// put it back into stdin
ungetc(ch, stdin);
if (scanf("%lf", &a[count]) != 1)) {
break; // EOF or non-numeric text
}
count++;
}
printf("%d values read\n", count);
for (int i=0; i<count; i++) {
printf("%g\n", a[i]);
}
Alterantive to consume various multiple leading whitespaces that only uses scanf() with no ungetc():
// Consume the usual white spaces except \n
scanf("%*[ \t\r\f\v]");
char eol[2];
if (scanf("%1[\n]", eol) == 1) {
break;
}
If the line contains more than N numbers or non-numeric text, some more code needed to report and handle that.
The best solution to problems with scanf and fscanf is usually to use something other than scanf or fscanf. These are remarkably powerful functions, really, but also very difficult to use successfully to handle non-uniform data, including not only variable data but data that may be erronious. They also have numerous quirks and gotchas that, though well documented, regularly trip people up.
Although sscanf() shares many of the characteristics of the other two, it turns out often to be easier to work with in practice. In particular, combining fgets() to read one line at a time with sscanf() to scan the contents of the resulting line is often a convenient workaround for line-based inputs.
For example, if the question is about reading one, two, or three inputs appearing on the same line, then one might approach it this way:
char line[1024]; // 1024 may be overkill, but see below
if (fgets(line, sizeof line, stdin) != NULL) { // else I/O error or end-of-file
double a, b, c;
int n = sscanf(line, "%lf%lf%lf", &a, &b, &c);
if (n < 0) {
puts("Empty or invalid line");
} else {
printf("%d input num\n", n);
}
}
Beware, however, that the above may behave surprisingly if any input line is longer than 1023 characters. It is possible to deal with that, but more complicated code is required.
here is an example of using fgets, strtok and atof to achieve same:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main() {
char input[256];
double inputsf[256];
while(1) {
printf(">>> "); fflush(stdout);
char *s = fgets(input, 255, stdin);
if (!s)
break;
int count = 0;
char *t;
while (t = strtok(s, " \n")) {
s = NULL;
inputsf[count++] = atof(t);
}
printf("Found %d inputs\n", count);
for (int i = 0; i < count; i++)
printf(" %lf\n", inputsf[i]);
if (count == 0)
break;
}
return 0;
}
Bases on the #chux-ReinstateMonica comment, here is a piece of code which uses strtod. It skips leading spaces, but has an issue with the tailing spaces at the end of the string. So, some extra checking is needed there, which could be used for error checking as well. The following loop can replace the strtok loop from above.
while(*s) {
char *e;
double val = strtod(s, &e);
if (e == s)
break; // not possible to parse, break the loop
inputsf[count++] = val;
s = e;
}
You can solve your problem using those line of code
#include <stdio.h>
int main(){
int a[100];
int n;
printf("How many data you want to input: ");
scanf("%d", &n);
printf("Sample %d data Input: ", n);
for (int i=0; i <n; i++) {
scanf("%d", &a[i]);
}
printf("Sample data %d: ", n);
for (int i=0; i <n; i++) {
printf("%d ", a[i]);
}
if(n == 1){
printf("\n 1 input num\n");
}else if(n==2){
printf("2 input num\n");
}else if(n==3){
printf("3 input num\n");
}else{
printf("Error");
}
return 0;
}
if you want to take multiple input in single line use this line
int arr[100];
scanf ("%lf %lf %lf", &arr[0], &arr[1], &arr[2]);
I read chars until '\n', convert them to int and sum the numbers until the result is only one digit.
I can't use mod or .
The first run went well, but the second one keep running and not waiting to \n.
any reason for keeping the '\n'?
#include<stdio.h>
int main(){
char str[8], conv_str[8],c;
int i,val,ans = 0;
while(1){
printf("Enter 8 values(0-9) :\n");
scanf("%[^\n]", str); // Scan values to str untill \n
for(i = 0;i < 8;i++){
val = str[i]-48; //convert from asci to int
ans += val;
}
while(ans > 9){
// itoa convert int to string, str(the input) is the buffer and 10 is the base
itoa(ans,conv_str,10);
ans = (conv_str[0]-48) + (conv_str[1]-48) ;
}
printf("the digit is: %d", ans);
printf("\ncontinue? (y/n)\n");
scanf("%s", &c);
if (c == 'n')
break;
memset(str, 0, sizeof(str));
}
return 0;
}
TIA
You have multiple problems in the code. Some of them are
scanf("%s", &c); is wrong. c is a char, you must use %c conversion specifier for that.
You never checked for the return value of scanf() calls to ensure success.
While scanning for character input, you did not clear the buffer of any existing inputs. Any existing character, including a newline ('\n') already present in the buffer will be considered as a valid input for %c. You need to clear the buffer before you read a character input.
I have no idea why isdigit() and isalpha() keep doing this. No matter how I use them they always return a 0. Am I using them incorrectly or is my compiler acting up?
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
//example of isdigit not working, it ALWAYS returns 0 no matter what the input is.
int main()
{
int num=0;
int check=0;
printf("Enter a number: ");
fflush(stdin);
scanf("%d",&num);
//Just checking what value isdigit has returned
check=isdigit(num);
printf("%d\n",check);
if(check!=0)
{
printf("Something something");
system("pause");
return 0;
}
else
{
system("pause");
return 0;
}
}
isdigit acts on an int, but it is supposed to be char extended to an int. You are reading a number and convert it to a binary representation, so unless your inout happens to be a value matching 0 - ยด9` (i.e. 0x30 - 0x39) you will get false as a result.
If you want to use isdigit() you must use it on the individual characters from the string the user enters as the number, so you should use %s, or %c for single digits, and loop through the string.
Example:
char c;
scanf("%c",&c);
check=isdigit(c);
or (quick and dirty example)
char buffer[200];
if (scanf("%s", buffer) > 1)
{
int i;
for(i = 0; buffer[i] != 0; i++)
{
check=isdigit(buffer[i]);
}
}
Change your scanf ("%d"...) with scanf ("%c"...) and the definition of num from int to char. You may also change the variable name, as isalpha() and isdigit() don't work with "numbers" but with "ASCII codes of characters" (which are indeed numbers), and the prompt message from "Enter a number" to "Enter a character"
You should pass the number character by character to isdigit. Replace %d specifier by %c in scanf. And also remove the line
fflush(stdin);
it may invokes undefined behavior (if you are not on MS-DOS).
You want to check a character not number. 0 and '0' are not the same. The first one is integer, second one is character.
Change on this:
char character =0;
int check = 0;
printf("Enter a number: ");
scanf("%c", &character);
printf("%d\n",character);
check = isdigit(character);
if(check != 0) {
printf("Something something\n");
return 0;
}
else{
return 0;
}
In action:
Enter a number: 5
53 # ASCII code of 5
2048
Something something
Enter a number: a
97 #ASCII code if a
0
I'm writing a program in C that is suppose to ask the user for a number.
The number has to be greater than zero and cannot have letters before or after the number. (ie: 400 is valid but abc or 400abc or abc400 is not). I can make my program invalidate everything besides 400abc. How would I make it invalidate an input if it starts valid then turns invalid? (I'm about 2 months into an intro to c class so my knowledge is very limited.)
#include<stdio.h>
int check(void);
void clear_input(void);
main()
{
int num;
printf("Please enter a number: ");
num = check();
printf("In Main %d\n", num);
}
int check(void){
int c;
scanf("%d", &c);
while (c < 0){
clear_input();
printf("Invalid, please enter an integer: ");
scanf("%d", &c);
}
return c;
}
void clear_input(void){
char junk;
do{
scanf("%c", &junk);
}while (junk != '\n');
}
You can also check whether ascii value of each char scanned from user input should lie in range 48-57, It will only then be integer value.
strtol can be used to do it, but it takes some extra work.
After running this code:
char *endptr;
int n = strtol(num_text, &endptr, 10);
n will contain the number. But you still have to check that:
1. *endptr=='\0' - this means strtol didn't stop in the middle. In 400abc, endptr will point to abc. You may want to allow trailing whitespace (in this case, check that endptr points to an all-whitespace string.
2. num_text isn't empty. In this case, strtol will return 0, but an empty string isn't a valid number.
Read the input as a line, using fgets.
Check if all characters are numeric.
If not, it's invalid. If yes, use sscanf to get the line into an int.
Check if the int is in the range; you're done.
Scanf with %d will treat the "400abc" as 400, all the trailing characters would be ignored, so there is nothing to worry about.
If you definitely want to treat "400abc" as an invalid input, then maybe you shouldn't use %d in scanf, use %s instead?
One way is to read the whole line as a string and check by yourself if it contains any non-digits.
The other way is reading the integer and then looking into the input using fgetc() to see if the next character after the detected input is valid. Or you could even use the same scanf() for this:
char delim;
if(scanf("%d%c", &c, &delim) == 2 && !isspace(delim))
// the input is invalid
You can read the number in a character array and validate it by checking if all the characters lie in the ascii range 48 to 57 ( '0' to '9' ) .If so your no. is valid otherwise you can safely regard it as invalid input.Here the the demonstration :
#include<stdio.h>
#include<string.h>
int conv( char * word )
{
int ans=0;
int res=0;
for(int i=0;i<strlen(word);i++)
if(word[i]>='0' && word[i]<='9')
ans=(ans*10) + (word[i] - '0');
else
res=-999;
if(res==-999)
return res;
else
return ans;
}
int main()
{
char a[10];
gets(a);
int b=conv(a);
if(b==-999)
printf("Invalid Entry.\n");
else
printf("Success.No is %d.\n",b);
return 0;
}
You can adjust for negatives as well by checking the first character in the word array to be '-' and adjusting the sign accordingly.
This is C99, so compile with -std=c99
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <stdbool.h>
bool getNum(int *n) {
char c, s[10];
if (!scanf("%9s", s))
return false;
for (int i=0; c=s[i]; i++)
if (!isdigit(c))
return false;
*n = atoi(s);
return true;
}
int main() {
int n;
if (getNum(&n))
printf("you entered %d\n", n);
else
printf("you did not enter a number\n");
}
The following is your check function rewritten to fix your problem, so try this:
int check(void){
int n;
char c;
while (EOF==scanf("%d%c", &n,&c) || n < 0 || !isspace(c)){
clear_input();
printf("Invalid, please enter an integer: ");
}
return n;
}
The catch is that I cannot use atoi or any other function like that (I'm pretty sure we're supposed to rely on mathematical operations).
int num;
scanf("%d",&num);
if(/* num is not integer */) {
printf("enter integer");
return;
}
I've tried:
(num*2)/2 == num
num%1==0
if(scanf("%d",&num)!=1)
but none of these worked.
Any ideas?
num will always contain an integer because it's an int. The real problem with your code is that you don't check the scanf return value. scanf returns the number of successfully read items, so in this case it must return 1 for valid values. If not, an invalid integer value was entered and the num variable did probably not get changed (i.e. still has an arbitrary value because you didn't initialize it).
As of your comment, you only want to allow the user to enter an integer followed by the enter key. Unfortunately, this can't be simply achieved by scanf("%d\n"), but here's a trick to do it:
int num;
char term;
if(scanf("%d%c", &num, &term) != 2 || term != '\n')
printf("failure\n");
else
printf("valid integer followed by enter key\n");
You need to read your input as a string first, then parse the string to see if it contains valid numeric characters. If it does then you can convert it to an integer.
char s[MAX_LINE];
valid = FALSE;
fgets(s, sizeof(s), stdin);
len = strlen(s);
while (len > 0 && isspace(s[len - 1]))
len--; // strip trailing newline or other white space
if (len > 0)
{
valid = TRUE;
for (i = 0; i < len; ++i)
{
if (!isdigit(s[i]))
{
valid = FALSE;
break;
}
}
}
There are several problems with using scanf with the %d conversion specifier to do this:
If the input string starts with a valid integer (such as "12abc"), then the "12" will be read from the input stream and converted and assigned to num, and scanf will return 1, so you'll indicate success when you (probably) shouldn't;
If the input string doesn't start with a digit, then scanf will not read any characters from the input stream, num will not be changed, and the return value will be 0;
You don't specify if you need to handle non-decimal formats, but this won't work if you have to handle integer values in octal or hexadecimal formats (0x1a). The %i conversion specifier handles decimal, octal, and hexadecimal formats, but you still have the first two problems.
First of all, you'll need to read the input as a string (preferably using fgets). If you aren't allowed to use atoi, you probably aren't allowed to use strtol either. So you'll need to examine each character in the string. The safe way to check for digit values is to use the isdigit library function (there are also the isodigit and isxdigit functions for checking octal and hexadecimal digits, respectively), such as
while (*input && isdigit(*input))
input++;
(if you're not even allowed to use isdigit, isodigit, or isxdigit, then slap your teacher/professor for making the assignment harder than it really needs to be).
If you need to be able to handle octal or hex formats, then it gets a little more complicated. The C convention is for octal formats to have a leading 0 digit and for hex formats to have a leading 0x. So, if the first non-whitespace character is a 0, you have to check the next character before you can know which non-decimal format to use.
The basic outline is
If the first non-whitespace character is not a '-', '+', '0', or non-zero decimal digit, then this is not a valid integer string;
If the first non-whitespace character is '-', then this is a negative value, otherwise we assume a positive value;
If the first character is '+', then this is a positive value;
If the first non-whitespace and non-sign character is a non-zero decimal digit, then the input is in decimal format, and you will use isdigit to check the remaining characters;
If the first non-whitespace and non-sign character is a '0', then the input is in either octal or hexadecimal format;
If the first non-whitespace and non-sign character was a '0' and the next character is a digit from '0' to '7', then the input is in octal format, and you will use isodigit to check the remaining characters;
If the first non-whitespace and non-sign character was a 0 and the second character is x or X, then the input is in hexadecimal format and you will use isxdigit to check the remaining characters;
If any of the remaining characters do not satisfy the check function specified above, then this is not a valid integer string.
First ask yourself how you would ever expect this code to NOT return an integer:
int num;
scanf("%d",&num);
You specified the variable as type integer, then you scanf, but only for an integer (%d).
What else could it possibly contain at this point?
If anyone else comes up with this question, i've written a program, that keeps asking to input a number, if user's input is not integer, and finishes when an integer number is accepted
#include<stdlib.h>
#include<stdio.h>
#include<stdbool.h>
bool digit_check(char key[])
{
for(int i = 0; i < strlen(key); i++)
{
if(isdigit(key[i])==0)
{
return false;
}
}
return true;
}
void main()
{
char stroka[10];
do{
printf("Input a number: ");
scanf("%s",stroka);}
while (!digit_check(stroka));
printf("Number is accepted, input finished!\n");
system("pause");
}
I looked over everyone's input above, which was very useful, and made a function which was appropriate for my own application. The function is really only evaluating that the user's input is not a "0", but it was good enough for my purpose. Hope this helps!
#include<stdio.h>
int iFunctErrorCheck(int iLowerBound, int iUpperBound){
int iUserInput=0;
while (iUserInput==0){
scanf("%i", &iUserInput);
if (iUserInput==0){
printf("Please enter an integer (%i-%i).\n", iLowerBound, iUpperBound);
getchar();
}
if ((iUserInput!=0) && (iUserInput<iLowerBound || iUserInput>iUpperBound)){
printf("Please make a valid selection (%i-%i).\n", iLowerBound, iUpperBound);
iUserInput=0;
}
}
return iUserInput;
}
Try this...
#include <stdio.h>
int main (void)
{
float a;
int q;
printf("\nInsert number\t");
scanf("%f",&a);
q=(int)a;
++q;
if((q - a) != 1)
printf("\nThe number is not an integer\n\n");
else
printf("\nThe number is an integer\n\n");
return 0;
}
This is a more user-friendly one I guess :
#include<stdio.h>
/* This program checks if the entered input is an integer
* or provides an option for the user to re-enter.
*/
int getint()
{
int x;
char c;
printf("\nEnter an integer (say -1 or 26 or so ): ");
while( scanf("%d",&x) != 1 )
{
c=getchar();
printf("You have entered ");
putchar(c);
printf(" in the input which is not an integer");
while ( getchar() != '\n' )
; //wasting the buffer till the next new line
printf("\nEnter an integer (say -1 or 26 or so ): ");
}
return x;
}
int main(void)
{
int x;
x=getint();
printf("Main Function =>\n");
printf("Integer : %d\n",x);
return 0;
}
I developed this logic using gets and away from scanf hassle:
void readValidateInput() {
char str[10] = { '\0' };
readStdin: fgets(str, 10, stdin);
//printf("fgets is returning %s\n", str);
int numerical = 1;
int i = 0;
for (i = 0; i < 10; i++) {
//printf("Digit at str[%d] is %c\n", i, str[i]);
//printf("numerical = %d\n", numerical);
if (isdigit(str[i]) == 0) {
if (str[i] == '\n')break;
numerical = 0;
//printf("numerical changed= %d\n", numerical);
break;
}
}
if (!numerical) {
printf("This is not a valid number of tasks, you need to enter at least 1 task\n");
goto readStdin;
}
else if (str[i] == '\n') {
str[i] = '\0';
numOfTasks = atoi(str);
//printf("Captured Number of tasks from stdin is %d\n", numOfTasks);
}
}
printf("type a number ");
int converted = scanf("%d", &a);
printf("\n");
if( converted == 0)
{
printf("enter integer");
system("PAUSE \n");
return 0;
}
scanf() returns the number of format specifiers that match, so will return zero if the text entered cannot be interpreted as a decimal integer
The way I worked around this question was using cs50.h library. So, the header goes:
#include <cs50.h>
There you have get_int function and you simply use it for variable initiation:
int a = get_int("Your number is: ");
If a user inputs anything but integer, output repeats the line "Your number is: "; and so on until the integer is being written.
I've been searching for a simpler solution using only loops and if statements, and this is what I came up with. The program also works with negative integers and correctly rejects any mixed inputs that may contain both integers and other characters.
#include <stdio.h>
#include <stdlib.h> // Used for atoi() function
#include <string.h> // Used for strlen() function
#define TRUE 1
#define FALSE 0
int main(void)
{
char n[10]; // Limits characters to the equivalent of the 32 bits integers limit (10 digits)
int intTest;
printf("Give me an int: ");
do
{
scanf(" %s", n);
intTest = TRUE; // Sets the default for the integer test variable to TRUE
int i = 0, l = strlen(n);
if (n[0] == '-') // Tests for the negative sign to correctly handle negative integer values
i++;
while (i < l)
{
if (n[i] < '0' || n[i] > '9') // Tests the string characters for non-integer values
{
intTest = FALSE; // Changes intTest variable from TRUE to FALSE and breaks the loop early
break;
}
i++;
}
if (intTest == TRUE)
printf("%i\n", atoi(n)); // Converts the string to an integer and prints the integer value
else
printf("Retry: "); // Prints "Retry:" if tested FALSE
}
while (intTest == FALSE); // Continues to ask the user to input a valid integer value
return 0;
}
Just check is your number has any difference with float version of it, or not.
float num;
scanf("%f",&num);
if(num != (int)num) {
printf("it's not an integer");
return;
}
This method works for everything (integers and even doubles) except zero (it calls it invalid):
The while loop is just for the repetitive user input. Basically it checks if the integer x/x = 1. If it does (as it would with a number), its an integer/double. If it doesn't, it obviously it isn't. Zero fails the test though.
#include <stdio.h>
#include <math.h>
void main () {
double x;
int notDouble;
int true = 1;
while(true) {
printf("Input an integer: \n");
scanf("%lf", &x);
if (x/x != 1) {
notDouble = 1;
fflush(stdin);
}
if (notDouble != 1) {
printf("Input is valid\n");
}
else {
printf("Input is invalid\n");
}
notDouble = 0;
}
}
I was having the same problem, finally figured out what to do:
#include <stdio.h>
#include <conio.h>
int main ()
{
int x;
float check;
reprocess:
printf ("enter a integer number:");
scanf ("%f", &check);
x=check;
if (x==check)
printf("\nYour number is %d", x);
else
{
printf("\nThis is not an integer number, please insert an integer!\n\n");
goto reprocess;
}
_getch();
return 0;
}
I found a way to check whether the input given is an integer or not using atoi() function .
Read the input as a string, and use atoi() function to convert the string in to an integer.
atoi() function returns the integer number if the input string contains integer, else it will return 0. You can check the return value of the atoi() function to know whether the input given is an integer or not.
There are lot more functions to convert a string into long, double etc., Check the standard library "stdlib.h" for more.
Note : It works only for non-zero numbers.
#include<stdio.h>
#include<stdlib.h>
int main() {
char *string;
int number;
printf("Enter a number :");
string = scanf("%s", string);
number = atoi(string);
if(number != 0)
printf("The number is %d\n", number);
else
printf("Not a number !!!\n");
return 0;
}