How do I break out of this while loop if the user just presses enter without typing anything.
int main()
{
while(1){
int integer;
printf("enter integer:");
scanf("%d",&integer);
}
return 0;
}
You can't. scanf is a horrible function and should be avoided (in particular, you should never use it for user input).
The easiest way to get user input working is to make sure all input is through fgets (which reads a whole line). You can then analyze that line, convert it to a number (e.g. with strtol, strtod, sscanf, ...), or do whatever you want with it.
Here's an example:
#include <stdio.h>
#include <string.h>
int main(void) {
while (1) {
char buf[200];
int integer;
printf("enter integer: ");
fflush(stdout);
if (!fgets(buf, sizeof buf, stdin)) {
/* input error or end of file reached */
break;
}
buf[strcspn(buf, "\n")] = '\0'; /* remove trailing newline, if any */
if (buf[0] == '\0') {
/* empty line */
break;
}
sscanf(buf, "%d", &integer);
}
return 0;
}
According to me...
Take input as a string
IF inputString equals to lineBreak
THEN break
OTHERWISE convert it into integer
int main()
{
while(1)
{
int integer;
char input[1024];
printf("enter integer:");
fgets (input, sizeof (input), stdin);
if (strcmp (input, "\n") == 0)
break;
integer = atoi(input);
}
return 0;
}
It is possible,
Initialise the value of integer with something that is not allowed in input
For eg if the input takes positive integers ,I initialise integer with -1
int main()
{
while(1){
int integer=-1;
printf("enter integer:");
scanf("%d",&integer);
if(integer==-1)
break;
}
return 0;
}
Related
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.
Given the following code:
#include <stdio.h>
int main()
{
int testcase;
char arr[30];
int f,F,m;
scanf("%d",&testcase);
while(testcase--)
{
printf("Enter the string\n");
fgets(arr,20,stdin);
printf("Enter a character\n");
F=getchar();
while((f=getchar())!=EOF && f!='\n')
;
putchar(F);
printf("\n");
printf("Enter a number\n");
scanf("%d",&m);
}
return 0;
}
I want a user to enter a string, a character and a number until the testcase becomes zero.
My doubts / questions:
1.User is unable to enter a string. It seems fgets is not working. Why?
2.If i use scanf instead of fgets,then getchar is not working properly, i.e whatever character I input in it just putchar as a new line. Why?
Thanks for the help.
Mixing functions like fgets(), scanf(), and getchar() is error-prone. The scanf() function usually leaves a \n character behind in the input stream, while fgets() usually does not, meaning that the next call to an I/O function may or may not need to cope with what the previous call has left in the input stream.
A better solution is to use one style of I/O function for all user input. fgets() used in conjunction with sscanf() works well for this. Return values from functions should be checked, and fgets() returns a null pointer in the event of an error; sscanf() returns the number of successful assignments made, which can be used to validate that input is as expected.
Here is a modified version of the posted code. fgets() stores input in a generously allocated buffer; note that this function stores input up to and including the \n character if there is enough room. If the input string is not expected to contain spaces, sscanf() can be used to extract the string, leaving no need to worry about the newline character; similarly, using sscanf() to extract character or numeric input relieves code of the burden of further handling of the \n.
#include <stdio.h>
int main(void)
{
int testcase;
char arr[30];
char F;
int m;
char buffer[1000];
do {
puts("Enter number of test cases:");
if (fgets(buffer, sizeof buffer, stdin) == NULL) {
/* handle error */
}
} while (sscanf(buffer, "%d", &testcase) != 1 || testcase < 0);
while(testcase--)
{
puts("Enter the string");
/* if string should not contain spaces... */
if (fgets(buffer, sizeof buffer, stdin) == NULL) {
/* handle error */
}
sscanf(buffer, "%29s", arr);
printf("You entered: %s\n", arr);
putchar('\n');
puts("Enter a character");
if (fgets(buffer, sizeof buffer, stdin) == NULL) {
/* handle error */
}
sscanf(buffer, "%c", &F);
printf("You entered: %c\n", F);
putchar('\n');
do {
puts("Enter a number");
if (fgets(buffer, sizeof buffer, stdin) == NULL) {
/* handle error */
}
} while (sscanf(buffer, "%d", &m) != 1);
printf("You entered: %d\n", m);
putchar('\n');
}
return 0;
}
On the other hand, if the input string may contain spaces, fgets() can read input directly into arr, but then the stored string will contain a \n character, which should probably be removed. One way of doing this is to use the strcspn() function to find the index of the \n:
#include <string.h> // for strcspn()
/* ... */
puts("Enter the string");
/* or, if string may contain spaces */
if (fgets(arr, sizeof arr, stdin) == NULL) {
/* handle error */
}
/* replace newline */
arr[strcspn(arr, "\r\n")] = '\0';
printf("You entered: %s\n", arr);
putchar('\n');
/* ... */
Note that a maximum width should always be specified when using %s with the scanf() functions to avoid buffer overflow. Here, it is %29s when reading into arr, since arr can hold 30 chars, and space must be reserved for the null terminator (\0). Return values from sscanf() are checked to see if user input is invalid, in which case the input is asked for again. If the number of test cases is less than 0, input must be entered again.
Finally got the solution how can we use scanf and fgets together safely.
#include <stdio.h>
int main()
{
int testcase,f,F,m;
char arr[30];
scanf("%d",&testcase);
while((f=getchar())!=EOF && f!='\n')
;
while(testcase--)
{
printf("Enter the string\n");
fgets(arr,30,stdin);
printf("Enter a character\n");
F=getchar();
while((f=getchar())!=EOF && f!='\n')
;
putchar(F);
printf("\n");
printf("Enter a number\n");
scanf("%d",&m);
while((f=getchar())!=EOF && f!='\n')
;
}
}
We need to make sure that before fgets read anything,flushout the buffer with simple while loop.
Thanks to all for the help.
A simple hack is to write a function to interpret the newline character. Call clear() after each scanf's
void clear (void){
int c = 0;
while ((c = getchar()) != '\n' && c != EOF);
}
Refer to this question for further explaination: C: Multiple scanf's, when I enter in a value for one scanf it skips the second scanf
This is a school assignment but I cannot get my loop to work. We have to use scanf and terminate the loop when an EOF is inputted. This is the part of the code that is the problem:
{int
main(void){
char str[MAX];
while(scanf("%s", str) != EOF)
{
printf("\nEnter a String: ");
scanf("%s", str);
two_ele_subs(str);
}
return 0;
}
The sscanf returns the number of read variables, that in your case is only one, or EOF (ie: -1) in case of end of file. So I suggest you to use a different approach, like in the following:
#include <stdio.h>
#define MAX 100
int main(void){
char str[MAX];
int retVal;
printf("\nEnter a String: ");
while((retVal = scanf("%s", str)) == 1 || retVal != EOF)
{
printf("\nEnter a String: ");
two_ele_subs(str);
}
return 0;
}
My teacher has asked me to "Fool proof" my code from any sort of misuse, So I have come up with an
program that can remove any empty values (by disallowing them entirely)
Here is the Un-foolproofed code
#include <stdio.h>
#include <conio.h>
int main()
{
char text[16];
printf("Type something: ");
fgets(text,16, stdin);
printf("You typed: %s",text);
getch();
}
I have made some simple adjustments to ensure there is no error, however, i cannot get the if filter to work properly, as it still allows the NULL input
#include <stdio.h>
#include <conio.h>
int main()
{
char text[16];
int loop;
do
{
printf("Type something: ");
fgets(text,16, stdin);
if( text[0] == '\0')
{
printf("Try again");
system("cls");
loop=1;
}
else
{
loop = -1;
}
}
while(loop > 0);
printf("You typed: %s",text);
getch();
}
I've tried google and i cannot get a solid answer, this probably is some very simple line of code, but sadly i have no idea what it is.
Edit: it's fixed, the if statement should be:
if (text[0] == '\n')
Using the return value from fgets() is the best first step to fool-proofing user I/O.
char text[16];
printf("Type something: ");
if (fgets(text, sizeof text, stdin) == NULL) {
if (feof(stdin)) Handle_stdin_is_closed(); // no more input
if (ferror(stdin) Handle_IOerror(): // very rare event, more common with files
}
// Test is input is is only a '\n'
if (text[0] == '\n')
printf("Try again");
// Look for long line.
size_t len = strlen(text);
if (len + 1 == sizeof text && text[len - 2] != '\n') HandleLongLine();
The next step is to look for scan errors. Let's assume code is to read a long.
errno = 0;
char *endptr;
long = strtol(text, &endptr, 10);
if (errno) Handle_NumericOverflow();
if (text == endptr) Handle_InputIsNotNumeric();
while (isspace((unsigned char) *endptr)) endptr++;
if (*endptr != '\0') Handle_ExtraTextAfterNumber();
Although this is a lot of code, robust handling of hostle user input is best spun off to a helper function where lots of tests can be had.
char * prompt = "Type something: ";
long number;
int stat = GetLong(stdin, prompt, &number); // put all tests in here.
if (stat > 0) Handle_SomeFailure();
if (stat < 0) Handle_EOF();
printf("%ld\n", number);
fgets reads a whole line including the newline into the buffer and 0-terminates it.
If it reads something and then the stream ends, the read line will not have a newline.
If the line does not fit, it won't contain a newline.
If an error occurs before it successfully reads the first character, it returns NULL.
Please read the man-page for fgets: http://man7.org/linux/man-pages/man3/fgets.3.html
According to the fgets() man page
char *fgets(char *s, int size, FILE *stream);
//fgets() returns s on success, and NULL on error or when end of file
//occurs while no characters have been read.
so, you can check the return value of fgets()
n = fgets(text,16, stdin);
if that value is NULL, then nothing have been read.
you can do this by checking the value of n in a for loop,
if( n == NULL)
{
printf("Try again");
system("cls");
loop=1;
}
else
{
loop = -1;
}
I was working on this sample exercise, and everything works as I would like it to, but there is one behavior I don't understand.
When providing input: if I make consecutive invalid entries everything seems to work great. But if I enter a number different from 1,2,3 in the case of the first question, or 1,2 in the case of the second question, the program just sits there until a new input is given. If another invalid entry is made, it goes back to the error "invalid entry" message, and if an appropriate number is entered, everything moves along fine.
I do not understand why it stops to wait for a second input...anyone?
Thanks guys.
#include <stdio.h>
static int getInt(const char *prompt)
{
int value;
printf("%s",prompt);
while (scanf("%d", &value) !=1)
{
printf("Your entry is invalid.\nGive it another try: %s", prompt);
getchar();
scanf("%d", &value);
}
return value;
}
int main() {
int wood_type, table_size, table_price;
printf("Please enter " );
wood_type = getInt("1 for Pine, 2 for Oak, and 3 for Mahogany: ");
printf("Please enter ");
table_size = getInt("1 for large, 2 for small: ");
printf("\n");
switch (wood_type) {
case 1:
table_price = (table_size == 1)? 135:100;
printf("The cost of for your new table is: $%i", table_price);
break;
case 2:
table_price = (table_size == 1)? 260:225;
printf("The cost of for your new table is: $%i", table_price);
break;
case 3:
table_price = (table_size == 1)? 345:310;
printf("The cost of for your new table is: $%i", table_price);
break;
default:
table_price = 0;
printf("The cost of for your new table is: $%i", table_price);
break;
}
}
You most likely need to flush your input buffer (especially with multiple scanf calls in a function). After scanf, a newline '\n' remains in the input buffer. fflush does NOT do this, so you need to do it manually. A simple do...while loop works. Give it a try:
edit:
static int getInt(const char *prompt)
{
int value;
int c;
while (printf (prompt) && scanf("%d", &value) != 1)
{
do { c = getchar(); } while ( c != '\n' && c != EOF ); // flush input
printf ("Invalid Entry, Try Again...");
}
return value;
}
The blank line you get if you enter nothing is the normal behavior of scanf. It is waiting for input (some input). If you want your routine to immediately prompt again in the case the [Enter] key is pressed, then you need to use another routine to read stdin like (getline or fgets). getline is preferred as it returns the number of characters read (which you can test). You can then use atoi (in <stdlib.h>) to convert the string value to an integer. This will give you the flexibility you need.
example:
int newgetInt (char *prompt)
{
char *line = NULL; /* pointer to use with getline () */
ssize_t read = 0; /* number of characters read */
size_t n = 0; /* numer of chars to read, 0 no limit */
static int num = 0; /* number result */
while (printf ("\n %s ", prompt) && (read = getline (&line, &n, stdin)) != -1)
{
if ((num = atoi (line)))
break;
else
printf ("Invalid Input, Try Again...\n");
}
return num;
}
If some invalid input is entered, it stays in the input buffer.
The invalid input must be extracted before the scanf function is completed.
A better method is to get the whole line of input then work on that line.
First, put that input line into a temporary array using fgets(),
then use sscanf() (safer than scanf because it guards against overflow).
#include <stdio.h>
int main(int argc, const char * argv[]) {
char tempbuff[50];
int result, d , value;
do
{
printf("Give me a number: ");
fgets( tempbuff, sizeof(tempbuff), stdin ); //gets string, puts it into tempbuff via stdin
result = sscanf(tempbuff, "%d", &value); //result of taking buffer scanning it into value
if (result < 1){ //scanf can return 0, # of matched conversions,
//(1 in this case), or EOF.
printf("You didn't type a number!\n");
}
}while (result < 1);
//some code
return 0;
}
Knowledge from: http://www.giannistsakiris.com/2008/02/07/scanf-and-why-you-should-avoid-using-it/