I need to get a user input and check if it is a valid input.
The input must:
Start with space(may be multiple) or a number.
(Assuming condition one satisfied) After the number there may be any kind of characters as long as I manage to extract the number
Clarification for number valid values:
Can be multiple digit
Can't be negative number
Can't contain decimal point
Can't be scientific notation
Can't be hexadecimal
So I wrote this basic code which simply gets the input but I have no clue on where to start on applying these conditions
printf("Enter size of input:\n");
int c;
while((c=getchar())!='\n' && c!=EOF){
printf("%c",c);
}
For example :
Input - 4##2311413sadokalda ; expected output - 4
Input - !4a ; expected output - Invalid Size
You can have state machine as below.
printf("Enter size of input:\n");
int c;
int state = 0; //0 = space, 1 = number, 2 = number read
int number = 0;
while((c=getchar())!='\n' && c!=EOF){
switch(state)
{
case 0:
if (isdigit(c))
state = 1;
else if (c == ' ')
break;
else
//error
break;
case 1:
if (isdigit(c))
{
number = number*10 + (c-'0');
break;
}
else {
state = 2;
}
case 2:
printf ("%d\n",number);
}
}
You probably want something like this:
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
int main(int argc, char* argv[]) {
printf("Enter size of input:\n");
char input[100];
fgets(input, sizeof input, stdin);
if (!isdigit(input[0]))
{
printf("Invalid Size\n");
}
else
{
int inputsize = strtol(input, NULL, 10);
printf("%d\n", inputsize);
}
}
Based on the problem statement you have given, i think this should give you your desired output
EDITED (After a few clarifications):
int main()
{
int c;
int i=0;
while((c=getchar())!='\n' && c!=EOF)
{
if(isdigit(c) || (char)c==' ') //isdigit() function check if given variable is a digit
{ printf("%c",c);
i+=1;
}
else
{
break;
}
}
if(i==0)
{ printf("Invalid size"); }
}
Related
First, I apologize if the question doesn't make sense as my English isn't that good...
My question is, how do we print out different things depending on the user input?
What I'm trying to do is: when user inputs integer, the program prints out the inputted number. When the user inputs something that's not integer (like symbols and characters), the program prints out "not integer".
my current idea (pseudo-code) is as follows:
`int main(){
int value;
printf("Enter numbers");
scanf("%d", &value);
if(value is integer){
printf("%d", value);
} else {
printf("not integer");
}
return 0;
}`
what gets me is the scanf; by using %d, I'm assuming that the user will input an integer values, but the user can input values that are not integers so I can't make a comparison using the if statement if( value is integer). How can I make a comparison that will determine whether the inputted value is integer or not?
I don't know if this is a good thing or not.
You can use ASCII to check if the input type is an integer or not
(between 48 - 57 in ASCII)
it will be like this
char value;
int flag = 0; //to check true or false (0 means false, and 1 means true)
printf("Enter numbers");
scanf("%c", &value);
for(int i = 48; i <= 57; i++){
if(value == i){
flag = 1;
break;
}
}
if(flag == 1){
printf("%c", value);
} else {
printf("not integer");
}
How do you print different things depending the user input?
Step 1: Read the line of user input
char buf[100];
if (fget(buf, sizeof buf, stdin)) {
// something was entered
Step 2: test the string
char *end;
long value = strtol(buf, *end);
// If the end is the same as the beginning, no conversion occurred.
if (end == buf) {
puts("not integer");
}
printf("%ld\n", value);
}
}
Additional code could look for input that occurred after the integer. Also code could test for a large number that overflowed the long range.
The code is as follows. It caters for different situations like inputting negative numbers and decimal numbers:
#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main() {
char input[20];
int wrongFlag = 0;
scanf("%s", input);
if (input[0] == '0' && strlen(input) > 1) {
wrongFlag = 1;
//for number starts with 0, and string length>1 eg: 010
}
for (int i = 0; i < strlen(input); i++) {
if (i == 0 && (input[i] == '-' && strlen(input) > 2 && input[i + 1] == '0')) {
//check first round only: negative number with length >2 and starts with 0 eg: -010.
wrongFlag = 1;
continue;
}
if (i != 0 && !isdigit(input[i])) {
//check following rounds, check if it is not digit
wrongFlag = 1;
break;
}
}
if (wrongFlag) {
printf("Not integer");
}
else {
printf("integer");
}
return 0;
}
Try this it works for me.
#include<stdio.h>
#include<string.h>
int main()
{
int i;
char value[50];
int len;
printf("Enter maximum 50 digits\n");
/* enter the values you wanted*/
printf("Enter the value: ");
gets(value);
len = strlen(value);
/*it will iterate upto the end of the user input*/
for(i=0;i<len;i++)
{
if(48<value[i] && value[i]<=57)
{
if(i==(len-1))
printf("It's an integer");
}
else{
printf(" Not an integer");
break;
}
}
return 0;
}
Hi I need to prompt a user for some input and then validate it. The input must only be validated if it is a positive integer and not greater then 23. The only problem I am having with this is when the user enters a non-numerical input like "hello." The code below does not successfully detect that any input is non-numerical and though I have tried many methods to do this, none of them seem to work. Below is the closest I seem to have gotten by taking the input as a string then converting it to an integer, however it still does not work. Any help would be appreciated.
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
int main(void) {
int height;
char input[50];
int cont = 0;
while (cont == 0) {
printf("Please provide a non-negative integer no greater than 23.\n");
scanf("%s", &input);
height = atoi(input);
if (height <= 23 && height >= 0) {
cont = 1;
} else {
//do nothing
}
}
printf("Valid Input.\n");
return 0;
}
The atoi() function has no provision for returning an error indicator. Instead, you can use the strtol() function:
char *end;
height = strtol(input, &end, 10);
if (end == input) {
// no digits were entered
puts("Invalid input.");
continue;
}
#include <stdio.h>
int main(void) {
int height;
while(1){
printf("Please provide a non-negative integer no greater than 23.\n");
//if(2==scanf("%d%c", &height, &nl) && nl == '\n' && 0<= height && height <= 23)//more limited for "number\n"
if(1==scanf("%d", &height) && 0<= height && height <= 23)
break;
//Clear of invalid input
while(getchar()!='\n')
;
}
printf("Valid Input(%d).\n", height);
return 0;
}
I am assuming that you have to consider the whole input into the consideration rather than only certain parts like "12jjj" and "23h" as invalid.
In my opinion, since 23 is only 2 char, so there is no harm in checking the length of the string and the individual characters.
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <string.h>
bool ValidateInput (char * input, int &output)
{
if (strlen(input) > 2)
return false;
for (int index = 0; index < strlen (input); index++)
{
if ((input[index] < '0') || input[index] > '9')
return false;
}
output = atoi(input);
return true;
}
int main(void) {
int height;
char input[50];
int cont = 0;
while (cont == 0) {
printf("Please provide a non-negative integer no greater than 23.\n");
scanf("%s", input);
if (ValidateInput (input, height))
break;
}
printf("Valid Input.\n");
return 0;
}
I hope this helps.
I would appreciate some help with this. I'm trying to create this simple program that repeatedly loops asking for the user to enter in an int. If he enters an int, it exits but if he enters something else or bigger than int (ex.4gs4r33) it will loop again asking to enter an int. This is what I have tried, and it's not working. It says it's an int even if it's not.
#include<stdio.h>
unsigned int findInt();
int main() {
printf("Please input an int.\n");
findInt();
}
unsigned int findInt() {
char input;
long num = 0;
int b = 0;
do {
scanf("%c", &input);
if (isdigit(input)){
num = num*10+input+'0';
b = 1;
}
else if (input == '\n')
b = 1;
else
b = 0;
} while(input != '\n');
if (b == 1)
printf("Great!\n");
else{
printf("Not an int \n");
findInt();
}
return 0;
}
Two possible approaches. One would be to modify your code:
b = 1; // start off with good intentions…
do {
scanf("%c", &input);
if (isdigit(input)){
num = num*10+input -'0'; // *** SUBTRACT '0', don't add it!
}
else if (input != '\n') {
b = 0;
break; // *** break when you find non-digit
}
} while (input != '\n');
Two changes: getting the math right as you compute the integer, and fixing the logic (so you break out of your loop when you find a non digit character)
Second approach:
char buf[100];
char intAsString[100];
fgets(buf, 100, stdin);
sscanf(buf, "%d", num);
sprintf(intAsString, "%d\n", num);;
if(strcmp(buf, intAsString) == 0 ) {
printf("yay - you entered an integer!\n");
}
I'm sure you can figure out how that works.
update a complete code snippet that solves the issue of "loop logic" as well: you call the findInt function once from the top level, and it keeps going until you get the int. Note - in order for this to work properly, I read the entire input at once (rather than one at a time), then pick off the characters one by one using sscanf (and updating the pointer manually). It has a number of advantages - not least of which is that you start with a fresh input every time you call findInt, instead of having the rest of the input buffer that still needs reading (and which was giving rise to "no,no,no,great!" - as you would keep reading the bad input until you got to the newline, and accept that...)
#include<stdio.h>
#include <ctype.h>
unsigned int findInt();
int main() {
findInt();
}
unsigned int findInt() {
char input;
char buf[100];
char *temp;
long num = 0;
int b = 0;
printf("please enter an int:\n");
fgets(buf, 100, stdin);
temp = buf;
do {
sscanf(temp++, "%c", &input);
if (isdigit(input)){
num = num*10+input-'0';
b = 1;
}
else if (input == '\n')
{
b = 1;
break;
}
else {
b = 0;
break;
}
} while(input != '\n');
if (b == 1)
printf("Great! %d is an integer!\n", num);
else{
printf("Not an int \n");
findInt();
}
return 0;
}
In the else branch - i.e. not a digit or a newline - you set b to 0. Now if a digit DOES follow you reset that to 1.
You'll probably want to break or somehow record the permanent failure instead of just continuing.
#include <stdlib.h>
#include <limits.h>
#include <stdbool.h>
void findInt(){
int x;
bool ok;
do{
char buff[32], *endp;
long long num;
ok = true;//start true(OK)
printf("Enter a number: ");
fgets(buff, sizeof(buff), stdin);
//strtoll : C99
x=(int)(num=strtoll(buff, &endp, 0));//0: number literal of C. 10 : decimal number.
if(*endp != '\n'){
if(*endp == '\0'){
printf("Too large!\n");//buffer over
while('\n'!=getchar());
} else {
printf("Character that can't be interpreted as a number has been entered.\n");
printf("%s", buff);
printf("%*s^\n", (int)(endp - buff), "");
}
ok = false;
} else if(num > INT_MAX){
printf("Too large!\n");
ok = false;
} else if(num < INT_MIN){
printf("Too small!\n");
ok = false;
}
}while(!ok);
}
,
My aim is to accept 4-digit numbers, and 4-character strings (string should not contain digits or special characters)
If an invalid input is given the program should not terminate and it must allow the user to enter the details and continue until he wish to terminate.
I am able to find whether the input is a digit.
if(scanf("%d",&input)!=1)
{
printf("enter the number please");
... // I have option to re enter using while and flags
}
else
{
// I continue my work
...
}
To check it is four digits I have tried using the commands
i=0;
num = input;
while(num>0)
{
i = i+1;
num = num/10;
}
if(i==4){
...//I continue
}
else
printf("please enter four digit");
I have no idea of checking the same for characters. (I know how to check its length using strlen())
Please help me with the code in C. (Also help me to reduce/optimize the above logic to check whether the input is a 4-digit number)
I believe you want 2 inputs a number and a string. You can do that as
int number= 0;
char string[10] = { 0 };
do {
printf("please enter four digit");
scanf("%d", &number);
if(number >=1000 && number<= 9999)
break;
} while(1);
do {
printf("please enter four character string");
fgets(string, sizeof(string), stdin);
if(strlen(string) == 4)
break;
} while(1);
To check it is four digit number you can simply put a check whether the number lies between 1000 and 9999. (I am assuming you don't want the number to start with 0.)
strtol can help:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
char s[32], *p;
int x;
fgets(s, sizeof(s), stdin);
if ((p = strchr(s, '\n')) != NULL)
*p = '\0';
x = (int)strtol(s, &p, 10);
if ((p - s) == 4) {
printf("%d\n", x);
} else {
printf("Please enter four digit\n");
}
return 0;
}
char input[16];
int ok = 1, k = 0;
if (scanf("%s", input) > 0 && strlen(input) == 4) {
// check if it's a word
for (; k < 4; k++)
if (!isalpha(input[k])) {
// check if it's a number
for (int k = 0; k < 4; k++)
if (!isdigit(input[k]))
ok = 0;
break;
}
}
else ok = 0;
if (!ok)
printf("invalid input, please enter a 4-digit number or 4-letter word");
else {
printf("valid input");
...
}
You can use gets()1 fgets() to get the whole line and check line length. If the first character is between '0' and '9' then check the remaining if they are 3 numbers too. If the first character is a valid character in string then check the 3 remaining chars if it's also valid in string.
1See Why is the gets function so dangerous that it should not be used?
I've searched in and out of these forums but am still having trouble. My understanding of C is still very basic. I'm creating a small program that takes 3 numerical values entered by the user and then calculates the highest. I nailed that.
I now want to ensure that the user enters only integer values. I managed to get the prog to keep prompting the user to re-enter the value until it is within the specified numerical range (for example, any number between 1 and 1000 inclusive, piece of cake) but that's not good enough. I used 3 while loops to test each value, but this only works as long as the input is of type integer.
The trick is I cant use built in functions. It needs to be manual (sorry, poor choice of words) I tried to use char variables and x = getchar(); to get the ASCII value and test it in a condition but I can't get it working in a loop. (while/ do-while)
I also tried using a "for loop" and array variables but once again am struggling to get it to keep prompting the user.
I've also tried to test the value returned by scanf to see if its integer but my knowledge level of correct C syntax is level: noob. Either my loops don't loop or they loop infinitely.
Here is some sample code:
int x, y, z =0;
printf("Enter the first number:\n");
scanf("d", &x);
while (condition) /* Here is where I need to determine that the entered val is false */
{
printf("Wrong input. Re-enter a valid value.\n");
x =0;
scanf("%d", &x); /*user re-prompted */
}
I'm getting the idea that I'll have to use ASCII and a loop, but I just can't get to it. Also, the values entered get sent to a function for comparing and are then returned.
Could someone give me some advice and a few tips please?
Much thanks
You would have to use something like fgets, and strtol:
long someValue;
char *bufEnd = NULL;
char buf[128]; // max line size
do {
printf("enter a value: ");
fgets(buf, 128, stdin);
someValue = strtol(buf, &bufEnd, 10); // base 10
} while (bufEnd == buf || *bufEnd != '\n');
printf("got value: %li", someValue);
What we are doing here is we are tapping into strtol's capability to tell us where it stopped parsing, by passing in bufEnd.
Then, we are making sure that bufEnd doesn't point to the beginning of buf (in which case, it didn't start with a number), and also checking to make sure that bufEnd points to \n, or the end of the line (making sure that the user didn't enter something like 123abc, which strtol would interpret as 123). You may wish to trim buf of whitespace characters first, however.
You're absolutely on the right track with "scanf()". Just check the return value. If you don't get the expected #/values, then you got invalid input:
char found = FALSE;
int ival;
double x;
while (!found)
{
printf("Please enter a valid integer: ");
if (scanf("%d", &ival) !=1) {
printf ("Invalid! Please re-enter!\n");
continue;
}
printf("Please enter a valid floating point number: ");
if (scanf("%lf", &x) !=1) {
printf ("Invalid! Please re-enter!\n");
continue;
}
found = TRUE;
}
Here's my solution. It safe against buffer overflow and straightforward .
#include <stdio.h>
#define LEN 10
int main() {
int a;
char str[LEN];
fgets( str, LEN, stdin );
while ( !sscanf( str, "%d", &a ) )
fgets( str, 10, stdin );
printf("Num is : %d\n", a);
return 0;
}
#include <stdio.h>
#include <limits.h>
#include <ctype.h>
int getInteger(int* err){
int ch;
int n;//int32
int takeNum, sign;
long long int wk;//long long int as int64
wk=0LL;
*err = 0;
takeNum = 0;//flag
sign = 1;//minus:-1, other:1
/* //skip space character
while(EOF!=(ch=getchar()) && (ch == ' ' || ch == '\t' || ch == '\n'));
ungetc(ch, stdin);
*/
while(EOF!=(ch=getchar())){
if(ch == '-'){
if(takeNum != 0){//in input number
*err = 1;
break;
}
if(sign == -1){//already sign
*err = 2;
break;
}
sign = -1;
continue;
}
if(ch >= '0' && ch <= '9'){//isdigit(ch) in ctype.h
if(takeNum == 0)
takeNum = 1;
wk = wk * 10 + (ch - '0')*sign;
if(INT_MAX < wk || INT_MIN > wk){//overflow
*err = 3;
break;
}
continue;
}
if(ch != '\n'){//input other [-0-9]
*err = 4;
}
break;
}
if(takeNum == 0){//not input number
*err = 5;
} else {
n=wk;
}
while(ch != '\n' && EOF!=(ch=getchar()));//skip to newline
return n;
}
int getValue(const char* redoprompt, int low, int high){
int num, err=0;
while(1){
num = getInteger(&err);
if(err || low > num || high < num)
printf("%s", redoprompt);
else
break;
}
return num;
}
#define max(x,y) ((x)>(y))? (x) : (y)
int main(){
const char *error_message = "Wrong input. Re-enter a valid value.\n";
int x, y, z, max;
x = getValue(error_message, 1, 1000);
y = getValue(error_message, 1, 1000);
z = getValue(error_message, 1, 1000);
max = max(max(x,y), z);
printf("max:%d\n", max);
return 0;
}