I Have written a simple while program in C
// Online C compiler to run C online.
// Write C code in this online editor and run it.
#include <stdio.h>
typedef enum
{
false = 0,
true = 1
} bool;
int main() {
bool res = true;
while (res)
{
char choice;
printf("Success! \n");
printf("Do you want to continue(y/n): ");
scanf("%c", &choice);
if (choice == 'N' || choice == 'n')
{
res = false;
}
}
printf("Good Bye");
}
The Program run successfully but when I type Y in Do you want to continue the loop is executed twice. How Can I Solve this?
Add a space before %c in scanf it skip all whitspaces(newline) like
scanf(" %c", &choice);
^
when I type Y in Do you want to continue the loop is executed twice ?
When you type Y+enter this will go like "Y\n" and this new line will taked by next iteration, so use above meathod to discard it.
Similiar Questions :
1)scanf() leaves the new line char in the buffer
Thanks.
Related
I am making a program that reads input from a user then check the validity,
but for the while statement in the validateInput function I am getting an warning
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
int ValidateUser(char *);
int ValidateInput(int);
FILE *database;
struct{
int year;
int unit;
float gpa;
char semester;
char grade;
char name[40];
}student;
int ValidateInput(int x)
{
while (x != 1 || x != 2 || x != 3) // code that causing a warning
return x;
}
int ValidateUser(char *input)
{
int result;
result = strcmp(input, student.name);
if (result == 0)
{
return 1;
}
else
{
while (result != 0)
{
printf("The Username you entered does not exit. Please enter correct name.");
}
}
return 1;
}
int main()
{
printf("\t----------------------------------------------------------\n");
printf("\t|\t\t\t\t Santa Monica College\t\t\t\t\t |\n");
printf("\t----------------------------------------------------------\n\n");
printf("\t\tWelcome to Santa Monica Student Record System.\n\n");
printf("Please Enter Your Option\n");
printf("\t1.View GPA/GRADE\n");
printf("\t2.Add New GPA\n");
printf("\t3.Modify Information\n");
int choice;
scanf("%d", &choice);
char UserName[40];
if(choice == 1)
{
printf("Please enter your name\n");
scanf("%s", UserName);
ValidateUser(UserName);
printf("The GPA of %s is %f , %c ", student.name, student.gpa, student.grade);
}
else if (choice == 2)
{
printf("Please enter your name\n");
scanf("%s", UserName);
ValidateUser(UserName);
}
else if (choice == 3)
{
printf("Please enter your name\n");
scanf("%s", UserName);
ValidateUser(UserName);
}
else
{
ValidateInput(choice);
}
return 0;
}
The ide was suggesting to add parentheses around x-values in the argument of while statement in the ValidateUser to make it silence
I've done some research on this warning and I found that the condition I made is not true so that's why I am getting it, but I am not quite sure what the problem is.
Can someone help me out with this???
In validateUser(), your program will reach the
while (result != 0)
only in the event that the while condition is initially true, so if it reaches the loop, it will enter it. The body of the loop does not modify result, so if it enters the loop, it will loop indefinitely.
Ultimately, if the first branch of the if / else is taken, then the function returns from within that branch, and if the second branch is taken then control never exits that branch. Either way, the
return 1;
at the end of that function cannot be reached. Of course, that's just a symptom. The infinite loop is the main problem.
It's unclear what behavior you actually want here, but what would be most in keeping with the name of the function would be for it to only evaluate whether the specified user name is valid, returning a result that conveys either "yes" or "no". There is no particular reason why such a function would need to loop at all.
You put while (result != 0) inside function ValidateUser(char *input) and never put a way to get out of the loop, resulting in the return value below it never running.
To fix this, you would need to add a break; somewhere in the loop to indicate that you are ready to leave the current loop, or make result = 0;
The difference between the two is that break; will indicate that the rest of the loop doesn't need to run and will jump out of the loop, or use result=0; to run the rest of the code in the loop and then jump out of it
I'm new to C and this is my TicTacToe for first C project. For that, I setup a simple process for user where to choose X or O. But it doesn't seem to work for reason. Here it continues to the if statements and goes into infinite loop cause it didn't wait for user input.
I've gone through similar forums about this exact question but I was unable to get an answer that fixed my problem. Also, feedbacks about the code are much appreciated because I do want to improve my code.
#include <stdio.h>
#include <stdlib.h>
void main()
{
char checks_player;
int is_input_valid = 0;
while(is_input_valid == 0)
{
printf("What do you want to choose? (X/O) ");
scanf(" %c",checks_player);
if(checks_player == 'x')
{
checks_player = 'X';
is_input_valid = 1;
}
else if(checks_player == 'o')
{
checks_player = 'O';
is_input_valid=1;
}
else if((checks_player == 'O')|| (checks_player == 'X'))
{
is_input_valid = 1;
}
else
{
printf("Invalid Input!!\nTry Again.\n\n");
}
}
}
You need to pass a pointer to scanf. Instead of this:
scanf(" %c",checks_player);
Use this:
scanf(" %c", &checks_player);
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 have a problem with my input for my program:
#include <stdio.h>
#include <stdlib.h>
int confirm()
{
char c;
printf("Confirm (y/n): ");
scanf("%c", &c);
while (scanf("%c", &c))
{
if (c == 'Y' || c == 'y' || c == 'N' || c == 'n')
{
printf("\nThank you. \n");
break;
}
else
{
printf("\nInput not recognised, ry again. \n");
printf("Confirm (y/n): ");
}
}
}
int main(int argc, char* argv[])
{
confirm();
return 0;
}
When it executes, it asks the first question and inputting the answer is fine. However after entering the character (either y or n) the program prints the second question and stops. The whole program is not running. I don't know what I'm doing wrong.
Loose the first scanf at line 9 and (for me) it then seem to work correctly: if ynYN is entered then the confirm function exits, otherwise it continues looping
This is the first question I've posted about C programming on here as I just started learning C just a few weeks ago. Ill write up my code and ask what my problem is :) If Anyone please knows how I can fix my mistake or whatever I should replace for my code please reply:)!
The problem I am having, is that if you run the code for yourself, you will see that everything works fine, except for the 'else' part in the statement. The issue I am having is that when someone types more than one letter, it will run the last printf statement more than once, and will printf as many times as the user inputs a character other than y or n.
The first part with the Y or N is working fine, yet if they type any number of other chars, it doesnt just state "Please select again", one time and then re-scanf, it types out at least 2 printfs, just for even one character entered, "Please select again" "Please select again", and then, if you type more chars for the answer, it will just type even more "please select again"'s.
Please help me understand what I am doing wrong as I'm so keen on learning to program properly, but I am just stuck here atm :)
#include <stdio.h>
#include <conio.h>
int main()
{
char answer;
int loop = 0;
printf("Please select. [Y/N]:\n");
while (loop == 0)
{
scanf("%c", &answer);
if (answer == 'y' || answer == 'Y')
{
printf("Seeyou Later Aligator.\n");
break;
return 0;
}
else if (answer == 'n' || answer == 'N')
{
printf("Mmkay.\n");
break;
return 0;
}
else
{
printf("Please select again [Y/N]:\n");
loop = 0;
}
}
getch();
return 0;
}
scanf reads the required number of characters each time. If there are more characters, they are not ignored. They are read next time you call scanf. Hence you see multiple prints for every character. Inorder to explicitly ignore pending input, call fflush(stdin) after scanf. Which means to flush out any data in standard input stream.
Update:
fflush should not be used on input streams as said in comments. Use the accepted solution for ignoring output. However I recommend using toupper or tolower instead of bit hack.
The reason as many have pointed out is that your scanf is reading the extra newline character left in the input buffer after the user presses ENTER. So here is an alternative way to read input to avoid that whole mess:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main() {
char answer;
printf("Please select. [Y/N]:\n");
while (1)
{
scanf("%1s%*[^\n]", &answer);
answer |= 0x20;
if (answer == 'y')
{
puts("Seeyou Later Aligator.");
break;
}
else if (answer == 'n')
{
puts("Mmkay.");
break;
}
else
{
puts("Please select again [Y/N]:");
}
}
getchar();
return 0;
}
This will read just the first character found on stdin and ignore everything else after that and at the same time clear the input buffer of the newline character
break; is enough ... return will never be executed as you will break out of the while
Its printing more than once because scanf is taking in '\n' and extra inputs from previous entry
also the variable loop is pointless in your code
here is the fixed code:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main() {
char answer;
int loop = 0;
printf("Please select. [Y/N]:\n");
while (1)
{
scanf("%c", &answer);
if (answer == 'y' || answer == 'Y')
{
printf("Seeyou Later Aligator.\n");
break;
//return 0;
}
else if (answer == 'n' || answer == 'N')
{
printf("Mmkay.\n");
break;
// return 0;
}
else
{
printf("Please select again [Y/N]:\n");
while(getchar()!='\n'){
getchar();
if(getchar() == '\n'){
break;
}
}
}
}
getchar();
return 0;
}
Output:
$ ./test
Please select. [Y/N]:
dddd
Please select again [Y/N]:
ffffff
Please select again [Y/N]:
y
Seeyou Later Aligator.