C - Reading user input - c

I have a program where user input is required, a user types in a number 1-8 to determine how to sort some data, but if the user just hits enter a different function is performed. I get the general idea of how to do this and I thought what I had would work just fine but I'm having some issues when it comes to when the user just hits the enter key. Currently my code looks as follows:
//User input needed for sorting.
fputs("Enter an option (1-8 or Return): ", stdout);
fflush(stdout);
fgets(input, sizeof input, stdin);
printf("%s entered\n", input); //DEBUGGING PURPOSES
//If no option was entered:
if(input == "\n")
{
printf("Performing alternate function.");
}
//An option was entered.
else
{
//Convert input string to an integer value to compare in switch statment.
sscanf(input, "%d", &option);
//Determine how data will be sorted based on option entered.
switch(option)
{
case 1:
printf("Option 1.\n");
break;
case 2:
printf("Option 2.\n");
break;
case 3:
printf("Option 3.\n");
break;
case 4:
printf("Option 4.\n");
break;
case 5:
printf("Option 5.\n");
break;
case 6:
printf("Option 6.\n");
break;
case 7:
printf("Option 7.\n");
break;
case 8:
printf("Option 8.\n");
break;
default:
printf("Error! Invalid option selected!\n");
break;
}
}
Now I've changed the if statement to try input == "", input == " ", and input == "\n" but none of these seems to work. Any advice would be greatly appreciated. Currently from what I can see, the initial if statement fails and the code jumps to the else portion and then prints the default case.
Just to be clear the variables I declared for this code are as follows:
char input[2]; //Used to read user input.
int option = 0; //Convert user input to an integer (Used in switch statement).

The problem is in how you're doing the string comparison (if (input == "\n")). C doesn't have a "native" string type, so to compare strings, you need to use strcmp() instead of ==. Alternatively, you could just compare to the first character of the input: if (input[0] == '\n') .... Since you're then comparing char's instead of strings, the comparison doesn't require a function.

Try:
#include <string.h>
at the top and
if(strcmp(input, "\n") == 0)
in place if your if ( input == ... )
Basically, you have to use string comparison functions in C, you can't use comparison operators.

Try:
input[0] == '\n'
(or *input == '\n')

You need to use single quotes rather than double quotes
if(input == "\n")
compares the input address to the address of the string "\n",
What you want to do is to compare the first character of the input buffer
to the character literal \n like this
if(input[0] == '\n')
Note the use of single quotes around '\n'

You need to capture return code from sscanf, it will tell you how many of the field are "assigned", which in "Enter" key case, return code of 0
edit:
you should use strcmp when comparing string, not the operator "==".

The issue is with strings, you are comparing pointers, i.e. memory addresses. Since the input and "\n" aren't the same exact memory, it always fails (I assume input is a char *). Since you're looking for a single character, you can instead dereference input and compare to a char using single quotes instead of double.
(*input == '\n')
Should work as you intend.

Related

Compare strings correctly in C [duplicate]

This question already has answers here:
How do I properly compare strings in C?
(10 answers)
Closed 8 years ago.
I have to make a program where the input must be one of 2 values, 'M' or 'F'.
Here is the program segment that I have created to demonstrate
do
{
printf("What is your Gender? (M for Male - F for Female) ");
fgets(GenderValue, 16, stdin);
if (GenderValue[0] == '\n')
{
printf("Try again");
getch();
system("cls");
loop=1;
}
else
{
loop = -1;
}
if (GenderValue == "M");
{
loop =-1;
}
else if(GenderValue == "F");
{
loop=-1;
}
else
{
printf("Try again");
getch();
system("cls");
loop=1;
}
}
while(loop > 0); //Checks for NULL input
printf("Gender: %s",GenderValue);
I know I could have done an integer choice input, but I would like to recycle this later if possible.
So far the program calls my Gender value comparisons "Errors", Which is fine, except I have no idea what to do next, I could find a function that compares strings, but I really don't want to over complicate my code.
Edit: Duplicate?, Really?, The example given doesn't even come close to helping me.
Edit 2: Found the problem, i had ;'s next to my if statements, that's fixed now. But seriously, no one spotted this ?
You use strcmp to compare strings. Using the comparison operator == just compares the pointers.
But note that fgets will leave the ending newline in the string, so you need to compare to e.g. "M\n", or remove the newline if it's there.
Also note that scrcmp is case sensitive, the string "M" is not equal to the string "m".
You could use e.g. scanf to read the input as a single character instead, and use e.g. tolower (or toupper) to get case insensitive comparison.
However, using scanf brings other problems, especially if you use it again afterwards to read another single character. The reason for this is that Scanf leaves the newline in the input buffer, causeing the possible next scanf call with the "%c" format to read that newline. There are very simple solution for this: Tell scanf to read and discart leading whitespace by using " %c".
If GenderValue is a string, GenderValue == "M" is wrong . strings are not supposed to be comapred with ==. use strcmp().
[In this case, IMO its suitable] If GenderValue is a single character, comparison should be done as GenderValue == 'M'
For your case, [considering one character input] you can do something like
fgets(GenderValue, 16, stdin);
switch(GenderValue[0])
{
case '\n':
//your code
break;
case 'M':
case 'm':
//your code
break;
case 'F':
case 'f':
//your code
break;
default:
}
In above code, three advantages
You don't need to bother about removing trailing \n caused by
fgets()
Exactly one character input is considered
No need to worry about the upper/lower case of input.
replace (GenderValue == "M") with (strcmp(GenderValue, "M") == 0 )
and
replace (GenderValue == "F") with (strcmp(GenderValue, "F") == 0 )
try replacing GenderValue == "F" with !strcmp(GenderValue, "F\n") and GenderValue == "M" with !strcmp(GenderValue, "M\n")
you might also need to include "string.h"
Use string compare function (strcmp("")) whenever you need to compare strings.

Loop showing menu never ends

I'm trying to write a simple menu in C.
This is my proposed code:
while (end == 0) {
printf("Main menu\n");
printf("=========\n");
printf("1) Option 1.\n");
printf("2) Option 2.\n");
printf("3) Option 3.\n");
printf("4) Option 4.\n");
printf("0) Exit");
printf("\n");
printf("Choose an option: ");
fgets(&menu_option,1,stdin);
switch (menu_option) {
case '0': end = 1; break;
case '1': option1(); break;
case '2': option2(); break;
case '3': option3(); break;
case '4': option4();
}
}
However, when executing this code, the loop never ends.
What's wrong?
Thank you.
EDIT: when I say "the loop never ends", I want to say that I can't write any option because the menu appears again and again.
From fgets
The fgets() function shall read bytes from stream into the array pointed to by s, until n-1 bytes are read, or a is read and transferred to s, or an end-of-file condition is encountered. The string is then terminated with a null byte.
Since you give a length of 1, n-1 will be 0 and so your input is never stored in menu_option. Change it to a char array
char menu_option[10];
// ...
fgets(menu_option, sizeof(menu_option), stdin);
switch (menu_option[0]) {
case '0': end = 1; break;
case '1': option1(); break;
case '2': option2(); break;
case '3': option3(); break;
case '4': option4();
}
Update:
It is necessary to use char arrays, because you need enough space to store the input. The minimum size for a buffer is two characters, e.g. char buf[2], because fgets stores your input plus a terminating NUL character.
The loop iterates twice or even more times, because when you input e.g. 1 return or 1 3 2 return, it will first return 1, then 3, 2 and finally return. If you want to read the whole line and use the first digit only, your buffer must be larger than just 2, like e.g. 10 or even 256.
fgets reads a string instead of character. The statement
fgets(&menu_option,1,stdin);
Will store a string to menu_option but to store a string you need a size at least 2 bytes.
char menu_option[2];
...
fgets(menu_option,2,stdin);
int c;
while((c = getchar()) != '\n' && c != EOF); // This will be needed to flush your input buffer
switch (menu_option[0]) {
...
...
}
From: http://www.cplusplus.com/reference/cstdio/fgets/
fgets
char * fgets ( char * str, int num, FILE * stream );
Get string from stream.
Reads characters from stream and stores them as a C
string into str until (num-1) characters have been read or either a
newline or the end-of-file is reached, whichever happens first.
A newline character makes fgets stop reading, but it is considered a
valid character by the function and included in the string copied to
str.
A terminating null character is automatically appended after the
characters copied to str.
Since you have specified 1 for the number, 0 bytes are read, and the only thing that happens is that a terminating null character is appended. The result is an infinite loop in which nothing is read.
Since in the comments you expressed dislike for using strings, here's alternative implementation with minimal changes to the question code:
while (end == 0) {
printf("...menu print snipped...\n");
int ch = getchar();
switch (ch) {
case EOF: // end of file or error, fall through to exit option
case '0': end = 1; break;
case '1': option1(); break;
case '2': option2(); break;
case '3': option3(); break;
case '4': option4();
// default: // add error message?
}
// ignore any extra chars in the same line
while ((ch = getchar() != '\n') {
// test for end of file and set end to true
if (ch == EOF) {
end = 1;
break;
}
}
}
Note the extra loop, which will read and discard rest of the input line. Without it, user can enter several options on one line, which is probably not desirable. Also note, if the option functions read more user input, consider how you handle long lines then. For example move the extra loop to be before switch (and add another variable to read discarded chars). Actually you should probably put the extra loop to it's own function discard_until_newline.

C while loop - code won't work

I've been writing a simple program to check if input letter is a vowel, and my code doesn't work.
The program should take characters as input one by one until % is entered, which will make it exit. It checks if input chars are vowels, and prints the result. Also it reports an error if input is not a letter.
The problem is, it breaks out of the loop on the second step.
Thank you for help, in advance.
PS Sorry, didn't write that there's no error message, it just breaks out of the loop.
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
int main(void)
{
char processed='q';
while(processed != '%')
{
printf("Enter letter to check if it's a vowel, %% to quit.\n");
char input = getchar();
processed = tolower(input);
printf("%c\n", processed);
if (processed == '%')
break;
if (processed < 'a' || processed > 'z')
{
fprintf(stderr, "Input should be a letter\n");
exit(1);
}
switch(processed)
{
case 'a':
case 'e':
case 'i':
case 'o':
case 'u':
case 'y':
printf ("Vowel\n");
break;
default:
printf ("Non-vowel\n");
}
}
exit(0);
}
Presumably you're entering a character and then hitting [ENTER]. So, in actuality you are entering two characters -- the letter you typed and a line feed (\n). The second time through the loop you get the line feed and find that it's not a letter, so you hit the error case. Perhaps you want to add something like:
if (processed == '\n') {
continue;
}
Someone else mentioned that you're hitting enter after each letter of input, and thus sending a newline ('\n') into your program. Since your program doesn't have a case to handle that, it isn't working right.
You could add code to handle the newline, but using scanf would be easier. Specifically, if you replaced
char indent = getchar();
with
char indent;
scanf("%c\n", &indent);
scanf() would handle the newline and just return back the letters you're interested in.
And you should check scanf()'s return value for errors, of course.

Menu that accepts single character in C language

I want to create a simple menu in C program that accepts single character. The menu will be like this:
[S]how
[E]xit
If the user enter '1','s' or 'S' the program will print "Hello" and prompt again for the input
else if the user enters '2',E or 'E' the program ends.
else it should print "invalid input" and prompt again.
I am able to create the program but the problem is that when user enters 12, 13, 14, 15, 16.....so on starting with 1, it shows Hello and same for other options.
My code is:
#include <stdio.h>
void clearBuffer();
int main() {
int i = 0;
char selection;
do
{
printf("\t1. [S]how\n");
printf("\t2. [E]xit\n");
printf("Enter your selection from the number or character noted above: ");
scanf("%s", &selection);
clearBuffer();
if (selection == '1' || selection == 's' || selection == 'S')
printf("Hello");
else if (selection == '2' || selection == 'E' || selection == 'x')
i = 0;
} while(i != 0);
}
void clearBuffer()
{
while(getchar() != '\n');
}
If you are going to receive only one character consider replacing the scanf() function for getchar() function:
printf("Enter your selection from the number or character noted above: ");
selection = getchar();
You could use strlen, which is part of the standard C library, to check the length of the string returned by scanf and reject entries longer than one character:
if (strlen(selection) > 1)
{
printf("Invalid selection.");
}
Alternatively, I think you could use getchar() to accept just a single character from the user, which means they wouldn't have to press enter.
As already mentioned, you should use getchar() if you only want one character. If you still want to use scanf() for whatever reason you may have, the correct format is "%c", not "%s".
I would also suggest that if you are looking for a single character, the if block looks a little "busy" (read, awkward) ... a switch would be a cleaner, more elegant way to do it (IMHO).
/* something like this ... */
switch ( selection ) {
case '1':
case 's':
case 'S':
printf ( "Hello\n" );
break;
case '2':
case 'e':
case 'E':
i = 0;
break;
}
Other couple of things ... if you don't care about the case of the character being read (that is, 's' and 'S' will do the same thing), you can convert selection to uppercase before your if-block or switch-block using toupper(). Also, and this is just a style suggestion, don't use i for your exit flag. General practice is to use things like i and j for counters or indexes - you could use something like quit_now or user_done which would convey more precisely what the variable means.

While scanf!=EOF or scanf==1?

Ceteris paribus (well formed data, good buffering practices and what not), is there a reason why I prefer to loop while the return of scanf is 1, rather than !EOF? I may have read this somewhere, or whatever, but I may have it wrong as well. What do other people think?
scanf returns the number of items succesfully converted ... or EOF on error. So code the condition the way it makes sense.
scanfresult = scanf(...);
while (scanfresult != EOF) /* while scanf didn't error */
while (scanfresult == 1) /* while scanf performed 1 assignment */
while (scanfresult > 2) /* while scanf performed 3 or more assignments */
Contrived example
scanfresult = scanf("%d", &a);
/* type "forty two" */
if (scanfresult != EOF) /* not scanf error; runs, but `a` hasn't been assigned */;
if (scanfresult != 1) /* `a` hasn't been assigned */;
Edit: added another more contrived example
int a[5], b[5];
printf("Enter up to 5 pairs of numbers\n");
scanfresult = scanf("%d%d%d%d%d%d%d%d%d%d", a+0,b+0,a+1,b+1,a+2,b+2,a+3,b+3,a+4,b+4);
switch (scanfresult) {
case EOF: assert(0 && "this didn't happen"); break;
case 1: case 3: case 5: case 7: case 9:
printf("I said **pairs of numbers**\n");
break;
case 0:
printf("What am I supposed to do with no numbers?\n");
break;
default:
pairs = scanfresult / 2;
dealwithpairs(a, b, pairs);
break;
}
Depends what you want to do with malformed input - if your scan pattern isn't matched, you can get 0 returned. So if you handle that case outside the loop (for example if you treat it the same as an input error), then compare with 1 (or however many items there are in your scanf call).
From http://www.cplusplus.com/reference/clibrary/cstdio/scanf/
On success, the function returns the
number of items succesfully read. This
count can match the expected number of
readings or fewer, even zero, if a
matching failure happens. In the case
of an input failure before any data
could be successfully read, EOF is
returned.
The only way to be sure that you read the number of items intended is to compare the return value to that number.

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