This is a word guessing game. For example, hello is given as h___o and the user must guess the letters.
I set a condition on my loop but don't know why it is not breaking the while loop.
#include <stdio.h>
#include <string.h>
int main()
{
char word[] = "hello";
int length = strlen(word);
int check;
char spaceLetters[length];
int i, j;
spaceLetters[0] = word[0];
char *dash = "_";
for (i = 1; i < length; i++)
{
strncat(spaceLetters, dash, 1);
}
int attemptLeft = length;
printf("\n %s\n", spaceLetters);
printf("\t\t\tAttempt Left: %d\n", attemptLeft);
boolean start = T;
int userInput;
while (1)
{
printf("\n");
printf("Enter Letter:");
scanf("%c", &userInput);
for loop for checking entered letter is true or not
for (j = 1; j < length;j++)
{
if (word[j] == userInput)
{
spaceLetters[j] = word[j];
printf("%s\n", spaceLetters);
printf("\t\t\tAttempt Left: %d\n", attemptLeft);
printf("\n");
}
}
this is my break loop condition when hello == hello break loop
if(word == spaceLetters){
break;
}
}
}
Strings are represented by arrays/pointers. They need to be compared using string library. Replace
if ( word == spaceLetters )
with
if ( strcmp ( word, spaceLetters ) == 0 )
You'll also need to add #include <string.h>.
For starters instead of this call
scanf("%c", &userInput);
^^^
you have to use this call
char userInput;
//...
scanf(" %c", &userInput);
^^^^
Otherwise the call will read also white spaces as the new line character '\n' that is stored in the input buffer after pressing the Enter key by the user.
Arrays do not have the comparison operator ==. You have to compare arrays element by element. To compare strings stored in character arrays you should use the standard function strcmp
#include <string.h>
//...
if( strcmp( word, spaceLetters ) == 0 ){
break;
}
But before using this function the array specialLetters must be declared like
char spaceLetters[sizeof( word )] = ""`;
That is the array shall contain a string.
Otherwise this for loop in your program invokes undefined behavior.
for (i = 1; i < length; i++)
{
strncat(spaceLetters, dash, 1);
}
And it does not make a great sense.
You could just write
char dash = '_';
memset( spaceLetters + 1, dash, sizeof( spaceLetters ) - 2 );
Related
I'm trying to make a program in C where it prints from 0 to 100 then asks the user to print it again. It's not working.
#include <stdio.h>
#include <string.h>
int main()
{
int num;
char resposta[0];
resposta[0] = 's';
num = 0;
do
{
for (int i = 0; i < 100; i++)
{
num += 1;
printf("%i\n", num);
printf("Repetir?s/n\n");
scanf("%c", resposta[0]);
}
} while ((strcmp(resposta[0], "S") == 0) || (strcmp(resposta[0], "s") == 0));
return 0;
}
when declaring an array in C, the number in the brackets [] is not the index number, it is the total number of elements contained within the array. Setting char resposta[0]; means that resposta contains no elements.
This is different from when accessing elements, which uses a zero-index. So running scanf("%c", resposta[0]); is trying to access the first element, but there are zero elements within resposta, so that's likely why this isn't working.
I also figure I should add that it is entirely possible to just create a char variable that is not an array, since it seems you only need the one character. Just do char resposta; without the brackets.
Some more info on Arrays if you're interested: W3Schools
There are many issues.
You have an array that can contains 0 elements which is rather pointless, you want char resposta[1].
But you don't want an array of chars anyway. You're using the %c specifier you need a single char instead of an array of chars.
You're mixing up strings and chars. Change strcmp(resposta[0], "S")==0 to resposta == 'S', provided you have declared char resposta;.
You're asking the user if he wants to start over again in the for loop which is probably not what you want.
Basically you want this (untested code):
int main() {
char resposta;
resposta = 's';
int num = 0;
do {
num = 0;
for (int i = 0; i < 100; i++) {
num += 1;
printf("%i\n", num);
}
printf("Repetir?s/n\n");
scanf("%c", &resposta); // not the useage of the & operator here
} while (resposta == 'S' || resposta == 's');
}
One of the reasons it is not working is that the program stops to ask the user to continue inside the loop. Another failure is that the counter being used is not reset to zero for the second and subsequent runs. Another failure is the scanf format specifier that does not clear the newline out of the input buffer. The next 'get input' will not find an S/s to continue, but will pickup a LF and terminate the loop.
Things can be simplified if they are separated. The interaction with the user is very clearly defined and should be factored-out into its own function:
#include <stdio.h>
int again(void) {
puts( "Repetir? " );
char ch;
return scanf( " %c", &ch ) == 1 && (ch == 'S' || ch == 's' );
}
// Edit: Just noticed the OP title says "0 to 100"
void out(void) {
int i = 0;
while( i <= 100 ) printf( "%i\n", i++ );
}
int main() {
do out(); while( again() );
return 0;
}
I have been trying to reverse(and print it that way) a given string only using for loops and nothing more. I think I have built up the basic logic, but it has some defects. When run, it only reverses the first two characters and then stops. Please help me find the defect in my logic.
#include<stdio.h>
#include<stdlib.h>
int main()
{
char a[20];
int i;
printf("Enter any String\n");
gets(a);
for(i=0;a[i]!=NULL;i++)
{}
for(i=1;i>=0;i--)
{
printf("%c",a[i]);
}
}
For starters the function gets is not a standard C function any more. it is unsafe. Instead use the standard C function fgets. The function can append the new line character '\n' to the entered string that should be excluded from the string.
It is unclear from your question whether you are allowed to use standard string functions.
Nevertheless here is a demonstrative program that does the task without using standard C string functions and that uses only for loops (neither while loop nor do-while loop).
#include <stdio.h>
int main(void)
{
enum { N = 20 };
char s[N];
printf( "Enter any String less than %d symbols: ", N );
fgets( s, N, stdin );
// remove the new line character and calculate the length of the string
size_t n = 0;
for ( ; s[n] != '\0' && s[n] != '\n'; ) ++n;
s[n] = '\0';
// reverse the string
for ( size_t i = 0; i < n / 2; i++ )
{
char c = s[i];
s[i] = s[n-i-1];
s[n-i-1] = c;
}
puts( s );
return 0;
}
Its output might look the following way
Enter any String less than 20 symbols: Hello dev.aniruddha
ahddurina.ved olleH
If you want just to output the original string in the reverse order then the program can look like
#include <stdio.h>
int main(void)
{
enum { N = 20 };
char s[N];
printf( "Enter any String less than %d symbols: ", N );
fgets( s, N, stdin );
// remove the new line character and calculate the length of the string
size_t n = 0;
for ( ; s[n] != '\0' && s[n] != '\n'; ) ++n;
s[n] = '\0';
// reverse the string
for ( ; n-- != 0; )
{
putchar( s[n] );
}
putchar( '\n' );
return 0;
}
Its output is the same as shown above
Enter any String less than 20 symbols: Hello dev.aniruddha
ahddurina.ved olleH
gets() is a bad idea as you can easily get overflows and it is no longer part of the c standard.
So let's assume that the string entered fits the array and this is just for an excercise with no reallife usage.
Your first loop finds the terminator. That's good.
Your second loop sets the variable that indicates the terminator to 1, destroying the result.
If you remove the assignment i=1, your program compiles with gcc and does what you want.
#include<stdio.h>
#include<stdlib.h>
int main()
{
char a[20];
int i;
printf("Enter any String\n");
gets(a);
for(i=0;a[i]!=NULL;i++)
{}
for(;i>=0;i--) //removed i=1 here
{
printf("%c",a[i]);
}
}
But there are still some issues to be addressed.
You will also reverse the terminator, instead you should start from i-1
I would advise to not use a for loop if you do not have a counter criterion The first loop should rather be a while loop, but as it was part of the assignment you had no choice still I will replace it in my recommendation. As they can easily be swapped.
Then you could use another variable for the second loop for clarity.
Also NULL is the NULL-pointer not the value 0 (also namend NUL apperantly) . So you should replace this either with 0 or with '\0'
Also stdlib.h is not required here
#include<stdio.h>
int main()
{
char a[20];
int i = 0;
printf("Enter any String\n");
gets(a);
while (a[i] != 0)
{
i++;
}
for(int j = i-1; j>=0; j--) // -1 to get the value in front of the terminator
{
printf("%c",a[j]);
}
printf("\n"); //to flush the output.
}
Here is the solution code.
The first for loop is to be used for determining the length of the string and the second for loop is for traversing the string from the last position to the first.
#include<stdio.h>
int main()
{
char a[20];
int i,j,len;
printf("Enter a String\n");
gets(a);
for(i=0;a[i]!=NULL;i++)
{}
len=i;
for(j=len-1;j>=0;j--)
{
printf("%c",a[j]);
}
}
I think why only two chars are been return is because of the condition statement in your second "for loop".
for(i=1;i>=0;i--)
Note:
it repeats from 1~0 (1,0): meaning it will repeat only twice
first iteration: when i == 1
second iteration: when i == 0 ; then it ends .
Please note that you created two "for loops" with the first one having no content.
Bonus:
I tried to fixed your code but realized that my C language skills isnt the best lol . Anyways, i came up with something that you could reference but it only reverse strings of less than 8 elements.
#include<stdio.h>
#include<stdlib.h>
int findlength(char a[]);
int main()
{
char a[20];
int i;
printf("Enter any String\n");
gets(a);
int len = findlength(a);
printf("Lenght of the String is: %d \n",len);
printf("Reversed String is: ");
for(i=len;i>-1;i--){
printf("%c",a[i]);
}
}
int findlength(char a[]){
int result = 0;
int i;
for(i=0;i<sizeof(a) / sizeof(char);i++){ // sizeof(char) is 1
if(a[i] == '\0') //end of string
return result;
result += 1;
}
return result;
}
Reposting because my first post was no good. I have a question that I'm not really sure how to do. I know the process I'm going for, but am not totally sure how to scan a string into an array so that each character/integer is scanned into a independent element of the array. I'll post the question and the code I have so far, and any help would be appreciated.
Question:
Assume that we have a pattern like the following: ([n][letter])+ in which n is an integer number and letter is one of the lowercase letters from a-z. For example, 2a and 3b are valid expressions based on our pattern. Also, “+” at the end of the pattern means that we have at least one expression (string) or more than one expression attached. For instance, 2a4b is another valid expression which is matched with the pattern. In this question, we want to convert these valid expressions to a string in which letters are repeated n times.
o Read an expression (string) from user and print the converted version of the expression in the output.
o Check if input expression is valid. For example, 2ab is not a valid expression. If the expression is not valid, print “Invalid” in the output and ask user to enteranother expression.
o Sample input1 = “2a”, output = aa
o Sample input2 = “2a3b”, output = aabbb
o You will receive extra credit if you briefly explain what concept or theory you can use to check whether an expression is valid or not.
What I have so far:
#include <stdio.h>
int main()
{
int size, i, j;
char pattern[20];
char vowel[20];
int count[20];
printf("Please enter your string: ");
gets(pattern);
size = strlen(pattern);
for(i=0; i<size; i++)
if((i+1)%2 == 0)
vowel[i] = pattern[i];
else if((i+1)%2 != 0)
count[i] = pattern[i];
for(i=0; i<size/2; i++);
for(j=0; j<count[i]; j++)
printf("%s", vowel[i]);
}
I assumed you want to write the "invalid\n" string on stderr. If not just change the file descriptor given to write.
#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <string.h>
#define MAX_INPUT_SIZE 20
int
check_input(char *input)
{
while (*input)
{
if (*input < '0' || *input > '9')
{
write(2, "invalid\n", 8);
return 1;
}
while (*input >= '0' && *input <= '9')
input++;
if (*input < 'a' || *input > 'z')
{
write(2, "invalid\n", 8);
return 1;
}
input++;
}
return 0;
}
void
print_output(char *input)
{
int i;
while (*input)
{
i = atoi(input);
while (*input >= '0' && *input <= '9')
input++;
for (; i > 0; i--)
write(1, input, 1);
input++;
}
write(1, "\n", 1);
}
int
main()
{
char input[MAX_INPUT_SIZE];
do
{
printf("Please enter your string: ");
fgets(input, MAX_INPUT_SIZE, stdin);
input[strlen(input) - 1] = '\0';
}
while (check_input(input));
print_output(input);
return 0;
}
The steps are:
Read pattern
Check if pattern is valid
Generate output
Since the input length is not specified you have to assume a maximum length.
Another assumption is n is a single digit number.
Now you may read the whole expression with fgets() or read it char by char.
The latter allows you to check for validity as you read.
Lets use fgets() for convenience and in case the expression needs to be stored for later use.
char exp[100]; // assuming at most 50 instances of ([n][letter])
int len;
printf("Input: ");
fgets(exp, 100, stdin);
len = strlen(exp) - 1; // Discard newline at end
An empty input is invalid. Also a valid expression length should be even.
if (len == 0 || len%2 != 0) {
printf("Invalid-len\n");
return 1;
}
Now parse the expression and separately store numbers and letters in two arrays.
char nums[50], letters[50];
invalid = 0;
for (i = 0, j = 0; i < len; i += 2, j++) {
if (exp[i] >= '1' && exp[i] <= '9') {
nums[j] = exp[i] - '0';
} else {
invalid = 1;
break;
}
if (exp[i+1] >= 'a' && exp[i+1] <= 'z') {
letters[j] = exp[i+1];
} else {
invalid = 1;
break;
}
}
Notice that in each iteration if first char is not a number or second char is not a letter, then the expression is considered to be invalid.
If the expression is found to be invalid, nothing to do.
if (invalid) {
printf("Invalid\n");
return 1;
}
For a valid expression run nested loops to print the output.
The outer loop iterates for each ([n][letter]) pattern.
The inner loop prints n times the letter.
printf("Output: ");
for (i = 0; i < len/2; i++) {
for (j = 0; j < nums[i]; j++)
printf("%c", letters[i]);
}
This is a rather naive way to solve problems of this type. It is better to use regular expressions.
C standard library doesn't have regex support. However on Unix-like systems you can use POSIX regular expressions.
like this
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <stdbool.h>
#define prompt "Please enter your string: "
void occurs_error(const char *src, const char *curr){
printf("\nInvalid\n");
printf("%s\n", src);
while(src++ != curr){
putchar(' ');
}
printf("^\n");
}
bool invalid(char *pattern){
char *p = pattern;
while(*p){
if(!isdigit((unsigned char)*p)){//no number
occurs_error(pattern, p);
break;
}
strtoul(p, &p, 10);
if(!*p || !islower((unsigned char)*p)){//no character or not lowercase
occurs_error(pattern, p);
break;
}
++p;
}
return *p;
}
int main(void){
char pattern[20];
while(fputs(prompt, stdout), fflush(stdout), fgets(pattern, sizeof pattern, stdin)){
pattern[strcspn(pattern, "\n")] = 0;//chomp newline
char *p = pattern;
if(invalid(p)){
continue;
}
while(*p){
int n = strtoul(p, &p, 10);
while(n--)
putchar(*p);
++p;
}
puts("");
}
}
I'm trying to create a program that checks if a given array/string is a palindrome or not and its not working. The program just prints "0" on every given array, even on palindromes.
int main()
{
char string[100]= {0};
char stringReverse[100]= {0};
int temp = 0;
int firstLetter = 0;
int lastLetter = 0;
printf("Please enter a word or a sentence: ");
fgets(string, 100, stdin);
strcpy(stringReverse , string); // This function copies the scanned array to a new array called "stringReverse"
firstLetter = 0;
lastLetter = strlen(string) - 1; //because in array, the last cell is NULL
// This while reverses the array and insert it to a new array called "stringReverse"
while(firstLetter < lastLetter)
{
temp = stringReverse[firstLetter];
stringReverse[firstLetter] = stringReverse[lastLetter];
stringReverse[lastLetter] = temp;
firstLetter++;
lastLetter--;
}
printf("%s %s", stringReverse, string);
if ( strcmp(stringReverse , string) == 0)
{
printf("1");
}
else
{
printf("0");
}
}
Lets say we implement a simple fun to do that
int check_palindrome (const char *s) {
int i,j;
for (i=0,j=strlen(s)-1 ; i<j ; ++i, --j) {
if (s[i] != s[j]) return 0; // Not palindrome
}
return 1; //Palindrome
}
I think this is far more simpler ;)
For the code posted in question:
Be aware of fgets(). It stops in the first '\n' or EOF and keeps the '\n' character.
So if you give radar for ex, the result string will be "radar\n", which doesn't match with "\nradar"
The Problem:
Let's say you enter the string RACECAR as input for your program and press enter, this puts a newline character or a '\n' in your buffer stream and this is also read as part of your string by fgets, and so your program effectively ends up checking if RACECAR\n is a palindrome, which it is not.
The Solution:
After you initialize lastLetter to strlen(string) - 1 check if the last character in your string (or the character at the lastLetter index is the newline character (\n) and if so, decrease lastLetter by one so that your program checks if the rest of your string (RACECAR) is a palindrome.
lastLetter = strlen(string) - 1; //because in array, the last cell is NULL
// Add these 2 lines to your code
// Checks if the last character of the string read by fgets is newline
if (string[lastLetter] == '\n')
lastLetter--;
fgets adds a '\n' at the end.
So if the user entered "aba", string contains "aba\n".
reverseString contains "\naba".
So it doesn't match.
After the fgets, add this code
int l = strlen(string) - 1;
string[l] = 0;
This will strip out the '\n' at the end before copying it to reverseString.
That aside, you can do this whole program inplace without the need of a second buffer or strcpy or strlen calls.
You have several issues in your code:
first you forgot the last closing brace };
then you forgot to remove the trailing \n (or maybe also \r under Windows) in string;
you don't need to revert the string into a new string; a one-pass check is enough:
Here is a working code:
#include <stdio.h>
#include <string.h>
int main()
{
char string[100]= {0};
int temp = 0;
int firstLetter = 0;
int lastLetter = 0;
printf("Please enter a word or a sentence: ");
fgets(string, 100, stdin);
firstLetter = 0;
lastLetter = strlen(string) - 1; //because in array, the last cell is NULL
while ((string[lastLetter]=='\n')||(string[lastLetter]=='\r')) {
lastLetter--;
}
// This while reverses the array and insert it to a new array called "stringReverse"
temp = 1;
while(firstLetter < lastLetter)
{
if (string[firstLetter] != string[lastLetter]) {
temp = 0;
break;
}
firstLetter++;
lastLetter--;
}
if ( temp )
{
printf("1");
}
else
{
printf("0");
}
}
You can do it by this simpleway also.
#include <stdio.h>
#include <string.h>
int main()
{
char string[10], revString[10];
printf("Enter string for reversing it...\n");
scanf("%s", string);
int stringLength = strlen(string);
for(int i = 0; string[i] != '\0'; i++, stringLength--)
{
revString[i] = string[stringLength - 1];
}
if(strcmp(string, revString) == 0)
printf("Given string is pelindrom\n");
else
printf("Given string is not pelindrom\n");
}
#include<stdio.h>
#include<string.h>`enter code here`
void fun(char *a);
int main ()
{
char p[100];
char *s=p;
printf("enter the string");
scanf("%[^\n]",s);
fun(s);
}
void fun(char *a)
{
if(*a && *a!='\n')
{
fun(a+1);
putchar(*a);
}
}
// use this approach better time complexity and easier work hope this helps
How would you be able to count word lengths and output their occurrences from a string using gets() or fgets()? For example, here is code doing so but using getchar()below. I think writing it in gets() would make it easier to incorporate all of the delimiters in the program rather than having to manually set if statements for each one of those would it not?
#include <string.h>
#include <ctype.h>
const char delim[] = ", . - !*()&^%$##<> ? []{}\\ / \"";
#define SIZE 100
int main(void){
int length[SIZE] = { 0 };
int name[SIZE];
int i = 0, ch, word_len = 0;
int count = 0;
printf("enter sentence: ");
while (1){
ch = getchar();
if (isalpha(ch)){
++word_len;
}
else if (ch == ' ' || ch == '.'){
if (word_len)
length[word_len - 1]++;//-1: to 0 origin
if (ch == '.')
break;
word_len = 0;
}
}
printf("Word Length \tCount \n");
for (i = 0; i<sizeof(length) / sizeof(*length); ++i){
if (length[i])
printf(" %d \t\t%d\n", i + 1, length[i]);
}
return 0;
}
You can build your custom delimiter detection function.
// globals
const char *delim = " .,;:!?\n\0";
const int n_delim = 9;
int is_delim(int c)
{
register int i;
for (i = 0; i < n_delim; i++)
if (c == delim[i]) return 1;
return 0;
}
This function will return 1 every time it can match c with delim. So you can use it like this:
fgets(buffer, 200, stdin);
for (i = 0; i < strlen(buffer); i++) {
if (is_delim(buffer[i])) {
wl[words++] = length;
length = 0;
continue;
}
length++;
}
I'm assuming you're familiar with the fgets function.
You basically will loop through your buffer, making comparisons with each character. Every loop iteration you check if the current character is a word delimiter, if it is, you save the current length and set length=0 for a new word, and at every iteration you increment the length.
You'll need to come up with a way of either not inserting the zero length values due to double delimiters or just ignore them when you're printing the results.
Basically you want to split a string into words, based on some delimiters, and compute their length. The C standard library provides the strtok function, which does exactly what you need: it splits the given string into multiple tokens.