I'm a bit rusty on C. I'm in a school assignment that is asking me to make a program that gets user input, following that print the output. If the character count is higher than 50, to not print anything out and reprompt them. Quit should not print a count (in this case, 4).
This is what I have so far:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main ()
{ /*Start of main*/
printf("Type any word you'd like <= 50 characters.\nType quit to exit:\n");
char word[50];
int wordLength = 0;
while (strcmp(word, "quit") != 0)
{/*open of while loop (strcmp ... != 0)*/
scanf("%s\n", word);
wordLength = strlen(word);
if (wordLength > 50)
{
printf("Try again, >= 50 characters!:\n");
scanf("%s", word);
wordLength = strlen(word);
}
printf("%d\n", wordLength);
}/*End of while loop (strcmp... != 0)*/
return 0;
}/*End of main*/
I can't seem to get the length immediately following the submitted word. It appears to get wonky. It shows the number for the LAST submittted word after putting in a new word. It doesn't matter where I put the printf, it takes its time.
Can someone explain to me why it's so slow, and suggest a method I could implement to make it faster? This is a school assignment, please don't share direct answers, but guide me into the way I should be thinking? :)
I'm almost thinking this is inefficient coding with the delay.
You're all life savers!!!
Remove \n from first scanf()
scanf("%s\n", word);
#---------^
Thanks to #remyable, \n has different meaning in scanf() - not the one you are expecting here to read newline. Refer C-faq 12.17
Also, checking for input for more than 50 chars that is not correct. You would get into buffer overrun. Look for different way to limit that.
The code has multiple problems
the while statement is using word before word is initialized
the scanf doesn't limit the number of characters that get written
into word
strlen can only return a number >50 if you overran the buffer
the code only reprompts once, it should reprompt until the user gets it right
Related
The assignment asks to print out the number of times a chosen character appears in an input (no length limit) string. I wanted to solve it by only using do or do-while loops, and after a bit of googling I found this code (source: https://www.tutorialgateway.org/c-program-to-count-all-occurrence-of-a-character-in-a-string/.).
I get the gist of it, but there are many things I still haven't covered, such as the meaning of str[i], the meaning of the variable ch, and kind of how the structure works. How can I interpret it piece by piece? Or if there's any simpler way, how can I work on it? I'm a beginner and I fear this is much easier than expected but I don't have the base to move on, thank you
#include <stdio.h>
#include <string.h>
int main() {
char str[10], ch;
int i, Count;
i = Count = 0;
printf("\n Please Enter any String : ");
gets(str);
printf("\n Please Enter the Character that you want to Search for : ");
scanf("%c", &ch);
while (str[i] != '\0') {
if (str[i] == ch) {
Count++;
}
i++;
}
printf("\n The Total Number of times '%c' has Occurred = %d ", ch, Count);
return 0;
}
Well i am giving an easy example regarding that problem with a proper explanation. Hope you might understand.
char is a datatype which will accepts character type of variable. Here str[100] will be an array of length 100, where we will store our search example. ch is a character type variable where we will store the character for which we will find the concurrence.
i and count are integer variables where i will be loop variable and the count will keep count of the concurrence.
after taking the text string using puts function we are storing it in the str[100] array.
then we are taking the search letter and stored it in ch.
we are now running for loop from 0 to the length of the string we have given.
strlen() function returning us the length.
now str[i] will search from i=0 to the length size of the string. each time loop will go forward one by one letter and compare the letter with the letter inside ch.
if match found then we will increase the count value.
after the ending of the loop count will be the result of the concurrency.
reference: Concurrency of a letter in a string
#include <stdio.h>
#include <string.h>
int main()
{
char str[100], ch;
int i, Count;
Count = 0;
printf("\n Please Enter any String : ");
gets(str);
printf("\n Please Enter the Character that you want to Search for : ");
scanf("%c", &ch);
for(i = 0; i <= strlen(str); i++)
{
if(str[i] == ch)
{
Count++;
}
}
printf("\n The Total Number of times '%c' has Occured = %d ", ch, Count);
return 0;
}
The code has the following parts:
1)The header files: These contain predefined functions like scanf() which you use in your program
2)The main function: Here you are performing your character count. Usually, this function contains the driver code for the program
ch is a variable for the character you want to count in your string. You are taking this as input from the user using scanf()
str is the string you are performing your count operation in. It is also taken as input.
str[i] is used to denote the index i in the string. For example in "test", the index of 's' will be 2 as it is the 3rd character and your index starts from 0.
Final note: I recommend going through the basic syntax and usage of arrays if you do not know indexing. Also, as someone commented, do not use gets(). It causes security issues in programs as user can access stack values by giving malicious inputs containing format specifiers.
Instead, use a scanf with proper format specifiers or use fgets() when accessing content from files.
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main(){
int n=1,i,cont;
char string[50];
scanf("%d",&n);
while(n!=0){
gets(string);
cont=0;
for(i=0;i<strlen(string);i++){
if(string[i]=='.'){
cont++;
}
}
if(cont%2==0){
printf("S\n");
}else{
printf("N\n");
}
scanf("%d",&n);
}
return 0;
}
My problem is quite simple but troublesome, I want to read an integer value n, and then read a string, after that read n again, but whenever I run the program, it only reads the string value... but if I digit 0 the program ends... it's like my scanf is within the gets function.
Mixing scanf with gets or fgets is troublesome because they each handle newlines differently.
Get rid of the gets call (which is unsafe anyway) and replace it with the following scanf call:
scanf("%49s", string);
This will read at most 49 characters into string (i.e. one less that its size).
From OP's comments, it sounds like the goal is to be able to read strings containing spaces. While there are ways to accomplish this using scanf(), it would be better to use fgets(), which is at the least less error-prone.
The fgets() function can be used to read input for the number into a buffer, and this buffer can then be processed by sscanf() to extract the number. Since fgets() keeps the newline character, it is not left behind to interfere with the next I/O operation.
But, when fgets() is used to get the string, since the newline is retained, it may be desirable to remove it. This can be accomplished in a number of ways, but here strcspn() is used to provide the index of the first \r or \n character encountered; a \0 character is then written to this location, removing the terminating newline from the string.
The code below illustrates these suggestions. Note that both buffer[] and string[] are generously allocated to accommodate reasonably large inputs. If a user enters a large number of characters (more than 999 in this case), the extra characters are left behind in the input stream for the next I/O function call. Also note that the main loop has been streamlined a bit; now there is a for(;;) loop that never terminates, broken out of when the user enters 0 for the number. And, there is a nested loop within the main loop that prompts the user to enter a number until a valid number is entered. Since the #include <stdlib.h> was unnecessary, it was removed. Better code would check the values returned from the calls to fgets() for possible errors.
#include<stdio.h>
#include<string.h>
int main(void)
{
int n = 1, cont;
char buffer[1000];
char string[1000];
for (;;) {
/* Loop until user enters a number */
do {
printf("Please enter a number: ");
fgets(buffer, sizeof buffer, stdin);
} while (sscanf(buffer, "%d", &n) != 1);
/* Break on 0 */
if (n == 0) break;
/* Get a string, and remove trailing newline */
printf("Please enter a string\n");
fgets(string, sizeof string, stdin);
string[strcspn(string, "\r\n")] = '\0';
cont = 0;
for (size_t i = 0; i < strlen(string); i++) {
if (string[i] == '.') {
cont++;
}
}
if (cont % 2 == 0){
printf("S\n");
} else {
printf("N\n");
}
}
return 0;
}
When you enter 5 for an example, you hit a new line character afterwards.
So you are entering 2 characters: 5 and a new line character.
That new line character is causing your headache.
The new line character is also considered an input.
In order to ignore this new line char, simply add a new line that acts as a garbage collection:
char garbage[50];
scanf( "%d", &n);
fgets(garbage, sizeof(garbage), stdin);
C beginner here. For the program below, whenever the user inputs a character or a string it enters an infinite loop. How would you fix this while still using scanf? And what would be the better methods of writing this program rather than using scanf? Thanks to those who will answer.
#include <stdio.h>
#include <ctype.h>
int main() {
int rounds = 5;
do {
printf("Preferred number of rounds per game. ENTER NUMBERS ONLY: ");
scanf("%d", &rounds);
} while(isdigit(rounds) == 0);
return 0;
}
Using 'scanf' require the input to be formatted. Scanf has very limited ability to handle bad input. The common solution will be to use fgets/sscanf, following the structure below:
char buff[256] ;
int rounds = 0 ;
while ( fgets(buff, sizeof(buff), stdin) ) {
if ( sscanf(buff, "%d", &rounds) == 1 ) {
// additional verification here
break ;
} ;
} ;
// Use rounds here ...
The fgets/sscanf will allow recovery from parsing error - the bad input line will be ignored. Depending on requirement, this might be accepted solution.
I'd say there are just two "fixes".
Retain the scanf call(s), warts and all. Carefully refrain from typing non-digits when scanf is expecting digits.
Abandon scanf and use something else. We've just been discussing this tactic over at this new question.
Once you're using scanf, it's always tempting to try to "fix" it, so there's potentially a third answer lurking here, explaining how to do better, more user-friendly, more error-tolerant input while still using scanf. In my opinion, however, this is a fool's errand, a waste of time. The easy, obvious fixes for scanf's many deficiencies are themselves imperfect and have further deficiencies. You will probably spend more time trying to fix a scanf-using program than you would have spent rewriting it to not use scanf -- and you'll get overall better results (not to mention a cleaner program) with the non-scanf-using rewrite anyway.
Change
scanf("%d", &rounds);
To
int ret;
if ((ret = scanf(" %d", &rounds)) != 1) { // Skip over white space as well
if (ret == EOF) break;
scanf("%*[^\n\r]"); // Consume the rest of the line
}
If you really like scanf, you can use getch to discard non-numeric input:
int rounds = MIN_INT;
while (scanf("%d", &rounds)) != 1)
if (getc() == EOF) /* discard a rubbish character */
break; // or other error-handling return
// rounds is only valid if we did not break, when its value should be MIN_INT.
// but you might need another indicator
C beginner here as well. Like you, I use scanf, and it can be problematic sometimes.
I've had your same problem and tried to solve it with scanf and basic stuff before finding a better solution.
I've tried different solution from here but I continue to have the same problems again and again, like if I type:
a number followed by a character (e.g. 123a), the result is a valid number (which i don't want); the result is '123'.
a string of numbers and chars that begin with a number (e.g. 1a2b3), the result is still a valid number which is '1'.
a char at the beginning (e.g. a123) can generate infinite loop.
... and so on... I've tried do...while, only while, for... nothing.
The only solution I have found to prompt the user until he/she writes only numbers is the following, but...
NOTE: if the user type a space, the program considers only the part before it, e.g. '12 3', only 12 is considered, 3 doesn't exist... unless you want use an infinite loop like I did so, in this case, you can enter multiple numbers, check them and run your program on them all at once. e.g.: '12 23 34 45' ...
NOTE 2: this is a very basic beginner solution, I am learning, and this is just what I found with what I know. Can't do any better right now and, as I said, I didn't find any other solution that I liked the output.
NOTE 3: I use the counter to sum up all the inputs that are not numbers and store the value if it finds one. If I don't use this solution I'll end up in the case where if the first character is a number but the rest aren't, it's still valid (e.g.: '12w3' is 12, which I don't want)
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
int main (void)
{
while (1) // try also multiple inputs separated by space
{
char str[10]; // should be enough for short strings/numbers (?!)
int strlength, num, counter;
do
{
printf("Enter a number: ");
scanf("%s", str);
strlength = strlen(str);
counter = 0;
for (int i = 0; i < strlength; i++)
{
if (!isdigit(str[i]))
counter++;
}
if (counter != 0)
printf("%s is not a number.\n", str);
} while (counter != 0);
num = atoi(str);
printf("%d is a number. Well done!\n\n", num);
}
}
You can also put it in a function and away from the main().
the assignment is to get an input string, and using no string library functions to be able to handle the string. this code at the moment doesn't even print out the string i get in. when I remove the functions from main it magically starts to print. any help would be greatly appreciated
#include <stdio.h>
#include <string.h>
#define SIZE 32
int isQuit(char str[]);
void isPalindrome(char str[]);
int main (){
int cont = 0;
char str[SIZE];
fflush(stdin);
printf("please enter a word:\n");
scanf("%s\n", str);
printf("%s\n", str);
while(cont == 0)
{
scanf("%s\n", str);
printf("%s\n", str);
cont = isQuit(str);
isPalindrome(str);
}
return 0;
}
You most likely are suffering from line buffering in your terminal. Until you write a newline character, any characters written are not displayed.
Try adding a newline when displaying your input:
printf("%s\n", str);
The same goes for any other printf calls you do that you want to ensure are displayed.
By the way, your null-termination test is incorrect. The escape character is \, not /. Change your loop to:
while (str[h] != '\0')
Or simply:
while (str[h])
There are a few things wrong with your code here:
while(isQuit(str) == 0)
{
isPalindrome(str);
return 0 ;
}
Since you have the return keyword in your loop body (unconditionally), the loop will execute at most one time.
Also, neither isQuit nor isPalindrome take input from the user. This means that even if you were to fix the loop by removing the return statement, it still wouldn't be right; you'd have an infinite loop of isQuit and isPalindrome being passed the same str that the user got asked for on line 15.
What you have to do is change your while loop to continually poll the user for input and act upon it, in addition to the issues pointed out in #paddy's answer.
I am trying to get a single digit number from stdin.
Using scanf("%d",&choice); is not good because if something like 3fjios or fjaifdj is entered then it keeps everything after the digit (if there is one), so if later I have scanf("%s",name); it takes the other chars and messing up. And also using scanf is bad (or so it seems from Google).
After a lot of digging I understand that we should use fgets, to read input into a string and then parse through it.
But! Nowhere is explained how to properly clear the buffer afterwards.
So if I do something like:
char choice[3];
do {
fgets(choice, 3, stdin);
scanf("%*[^\n]");
scanf("%*c");//clear upto newline
} while (choice[1] != '\n');
this works only if I enter a string longer than 2 chars.
When I enter a single char for fgets then the scanf actually waits for another input... which is bad.
The other big problem is if I enter more than 2 chars (a digit and '\n') then the first 2 chars go into choice, the rest are stuck in the buffer. All the approaches to clearing it seems like they require one to build a nuclear power plant first...
Also, what happens if the user enters an infinitely (a really long) long string?
Can you please show a simple way that will allow the user to enter some string of some (unknown) length, and then to properly check if it contains exactly a single digit at the start, followed by '\n'?
Any other input should loop back to get a new input from the user again.
please don't use complex solutions, only standard simple C please.
I can't believe I wasted 6 hours on this supposedly simple technical thing, just getting an input from the user. Solving the actual problem was easier...
Do not use scanf. It is making things overly complicated. Just use getchar to read and discard the line. eg:
int read_input(void) {
int n;
n = getchar();
if( getchar() == '\n' || n == EOF)
return n;
else
do n = getchar(); while ( n != '\n' && n != EOF);
fputs("invalid entry: ", stderr);
return read_input();
}
int main(void) {
int input;
input = read_input();
printf("user entered: %c\n", input);
return EXIT_SUCCESS;
}