#include <stdio.h>
#include <string.h>
#include <conio.h>
void Pig_Latin_Converter(char *a);
int main() {
char password[50];
printf("Please input password: ");
int passw = 0;
do {
password[passw] = getch();
if(password[passw] != '\r') {
printf("*");
}
passw++;
}
while(password[passw - 1] != '\r');
password[passw - 1] = '\0';
Pig_Latin_Converter(password);
return 0;
}
void Pig_Latin_Converter (char *a) {
int v = 0, c = 0, pos = 0,sv = 0, ch;
strlwr(a);
if(a[0]=='a'|| a[0]=='e'|| a[0]=='i'|| a[0]=='o'|| a[0]=='u'){
++sv;
printf("\n\nPig Latin Equivalent: %syay", *a);
}
else
{
++c;
for (int i = 0; a[i] != '\0'; ++i)
{
if (a[i] == 'a' || a[i] == 'e' || a[i] == 'i' ||
a[i] == 'o' || a[i] == 'u') {
++v;
pos=i;
break;}
}
}
if(c==1 && v==1)
{
printf("\n\n");
printf("Pig Latin Equivalent: ");
for(int i=pos; i<strlen(*a); i++) {
printf("%c", a[i]);
}
for(int i=0;i<pos;i++)
{
printf("%cay", a[i]);
}
}
else if(c==1 && sv==0)
{
printf("\n\n");
printf("Pig Latin Equivalent: %syay", *a);
}
}
Hello, I'm a beginner in using user-defined functions used in C programming. This program is tasked to input masked password then translate it to Pig Latin. But, I kept on coming up with errors on this. Is there a way to solve this? I have a huge gut feel that the problem is only inside the function Pig Latin.
Related
I wrote code that replaces integers from 0 to 3 with strings. I was only allowed to use getchar() and putchar(). If the input is 1, the output will become "one".
#include <stdio.h>
int main()
{
int c;
char* arr[4] = {"zero", "one", "two","three"};
int i;
while ((c = getchar ()) != EOF)
{
if(c==0+'0') {
char* str = arr[0];
for (i = 0; str[i] != '\0'; i++) {
putchar(str[i]);
}
}
else if(c==1+'0') {
char* str = arr[1];
for (i= 0; str[i] != '\0';i++) {
putchar(str[i]);
}
}
else if(c==2+'0') {
char* str = arr[2];
for (i = 0; str[i] != '\0'; i++) {
putchar(str[i]);
}
}
else if(c==3+'0') {
char* str = arr[3];
for (i = 0; str[i] != '\0'; i++) {
putchar(str[i]);
}
}
else
putchar(c);
}
return 0;
}
The code is pretty long. Is there a shorter way to write it?
If I type in 33 the output will be "threethree". Could anyone give me suggestions how can i modify my code not to do that?
note: I am also not allowed to use functions.
You can use a variable to remember last input and compare, so that you will not print continuous char.
#include <stdio.h>
int main()
{
int c;
char* arr[4] = {"zero", "one", "two","three"};
int i;
char last_input = '9';
while ((c = getchar ()) != EOF)
{
if(c != last_input && '0' <= c && c <= '3') {
last_input = c;
int index = c - '0';
char* str = arr[index];
for (i = 0; str[i] != '\0'; i++) {
putchar(str[i]);
}
}
else{
putchar(c);
}
}
return 0;
}
You can compress your if statements using one if condition like this :
#include <stdio.h>
int main()
{
int c;
char* arr[4] = {"zero", "one", "two","three"};
int i;
while ((c = getchar ()) != EOF) {
int k = c-'0';
if(k>=0 && k<=3) {
char* str = arr[k];
for (i= 0; str[i] != '\0';i++) {
putchar(str[i]);
}
}
else {
putchar(c);
}
}
return 0;
}
Here is the simple approach to the same task. I tried to explain the logic in the comments.
int main(void) {
char *arr[11] = {"zero", "one", "two","three","four","five","six","seven","eight","Nine","Ten"};
int *input = malloc(sizeof(*input))/*1st time 4 byte */ , row = 1;
while( (input[row-1] = getchar())!=EOF ) {
if(input[row-1]==10) /* if ENTER key is presed */
break;
input[row-1] = input[row-1] - 48;/* convert it */
printf("%s ",arr[ input[row-1]%10 ]);/* its simple, just think on it */
row++;
input = realloc(input,row * sizeof(*input));/* reallocate based on number of input */
}
/* free dynamically allocated memory #TODO*/
return 0;
}
I just given hint, make it generic like write the condition if input is less than zero etc. I hope it helps.
Here my code using loop to shorten your code.
#include <stdio.h>
int main()
{
int c;
char* arr[4] = {"zero", "one", "two","three"};
int i, j;
while ((c = getchar ()) != EOF)
{
for(j = 0; j < 4; j++)
{
if(c == j + '0')
{
char* str = arr[j];
for (i = 0; str[i] != '\0'; i++)
{
putchar(str[i]);
}
j = 10; // just to detect processed character
break;
}
}
if(j != 10)
{
putchar(c);
}
}
return 0;
}
I have school task. To reverse each word in sentence, so example :
Input: Fried chicken, fried duck.
Output: deirF nekcihc, deirf kcud.
So except dot and comma it's not reversed.
The first code
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main() {
int i, n, titik = 0, coma = 0;
char s[5001];
char c[5001];
char *tok;
scanf("%[^\n]s", s);
if (s[0] == ' ')
printf(" ");
tok = strtok(s, " ");
while (tok != NULL) {
strcpy(c, tok);
n = strlen(c);
for (i = n; i >= 0; i--) {
if (c[i] == ',') {
coma = 1;
} else
if (c[i] == '.') {
titik = 1;
} else
printf("%c", c[i]);
}
if (coma) {
printf(",");
coma = 0;
} else
if (titik){
printf(".");
titik = 0;
}
tok = strtok(NULL," ");
if (tok == NULL)
printf("\n");
else
printf(" ");
}
}
Second code is
#include <stdio.h>
#include <ctype.h>
#include <string.h>
int main() {
int i, j, n, prana = 0, titik = 0, coma = 0, end = 0;
char s[5001];
scanf("%[^\n]s", s);
n = strlen(s);
for (i = 0; i <= n; i++) {
if (isspace(s[i]) || iscntrl(s[i])) {
if (iscntrl(s[i]))
end = 1;
for (j = i - 1; j >= prana; j--) {
if (s[j] == '.') {
titik = 1;
} else
if (s[j] == ',') {
coma = 1;
} else
printf("%c", s[j]);
}
prana = i + 1;
if (titik) {
titik = 0;
if (end)
printf(".");
else
printf(". ");
} else
if (coma) {
coma = 0;
if (end)
printf(",");
else
printf(", ");
} else {
if (end)
printf("");
else
printf(" ");
}
}
}
printf("\n");
return 0;
}
Why the second code is accepted in test case?, but first code is not.
I tested the result it's same. Really identical in md5 hash.
The output of the two codes id different, because you print the terminating null character for each token in the first code. This loop:
for (i = n; i >=0 ; i--) ...
will have i == n in its first iteration. For a C string of length n, s[n] is the terminating null. This character may not show in the console, but it is part of the output.
To fix the loop, you could start with i = n - 1, but C uses inclusive lower bounds and exclusive upper bounds, and a more idomatic loop syntax is:
i = n;
while (i-- > 0) ...
Not related to your question at hand, but your codes are rather complicated, because they rely on many assumptions: words separated by spaces; only punctuation is comma or stop; repeated punctuation marks are ignored, special case for last word.
Here's a solution that treats all chunks of alphabetic characters plus the apostrophe as words and reverses them in place:
#include <stdlib.h>
#include <stdio.h>
#include <ctype.h>
void reverse(char *str, int i, int j)
{
while (i < j) {
int c = str[--j];
str[j] = str[i];
str[i++] = c;
}
}
int main()
{
char str[512];
int begin = -1;
int i;
if (fgets(str, sizeof(str), stdin) == NULL) return -1;
for (i = 0; str[i]; i++) {
if (isalpha((unsigned char) str[i]) || str[i] == '\'') {
if (begin == -1) begin = i;
} else {
if (begin != -1) {
reverse(str, begin, i);
begin = -1;
}
}
}
printf("%s", str);
return 0;
}
I'm writing a program to count spaces and vowels and it didn´t work, I think I did an infinite loop.I'll show you my code:
int contar_p(char a[100]) {
int i = 0, spaces = 1;
while (a[i] != '\0' && i < 100) {
if (a[i] == ' ') {
spaces += 1;
i++;
}
}
return spaces;
}
int contar_v(char b[100]) {
int i = 0, counter = 0;
while (b[i] != '\0' && i < 100) {
if (b[i] == 'a' || b[i] == 'e' || b[i] == 'i' || b[i] == 'o' || b[i] == 'u') {
counter += 1;
}
i++;
}
return counter;
}
int main(void){
char phrase[100];
int words = 0, vowels = 0;
printf("write a phrase ");
gets(phrase);
palabras = contar_p(phrase);
vocales = contar_v(phrase);
printf("%d\n", words);
printf("%d", vowels);
return 0;
}
The loop
while (a[i]!='\0'&&i<100){
if(a[i]==' '){
spaces+=1;
i++;
}
}
is an infinite loop. Place i++ outside the if. Change it to
while (a[i]!='\0'){ // No need of condition i < 100
if(a[i]==' '){
spaces+=1;
}
i++;
}
Maybe another approach will help you to understand things easier, i'm mean you do know that there are A,E,I,O,U also and not only a,e,i,o,u. You should never use gets instead use fgets, anyway take a look at the following program:
#include <stdlib.h>
#include <stdio.h>
void countVowels(char* array){
int i,j,v;
i=0;
int count = 0;
char vowel[]={'a','e','i','o','u','A','E','I','O','U'};
while(array[i]!='\0'){
for(v=0;v<10;v++){
if (array[i]==vowel[v]){
j=i;
while(array[j]!='\0'){
array[j]=array[j+1];
j++;
}
count++;
i--;
break;
}
}
i++;
}
printf("Found %d Vowels\n",count);
}
void contar_p(char a[100]) {
int i = 0, spaces = 0;
for(i=0;a[i]!='\0';i++){
if(a[i]==' ')
spaces++;
}
printf("Found %d Spaces\n",spaces);
}
int main(void){
char a[]="aa bb EOU cc ii";
countVowels(a);
contar_p(a);
return 0;
}
Output:
Found 7 Vowels
Found 4 Spaces
Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 8 years ago.
Improve this question
I am trying to take a string input from user and then reverse only the words starting with a vowel.Then reprint the changed string.I have used the strtok() function to separate the words from the string.But reversing the words,seems to be a problem....I have written a code for this program,but it is having runtime error.So,it will be really useful,if anyone could help me correct my code or provide me with a solution.
Here is my code:
#include <stdio.h>
#include <string.h>
void reverse(char *tok);
int length(char *t);
int main()
{
char sen[50];
const char s[2] = " ";
int i;
printf("Enter a Sentence: ");
gets(sen);
char *token;
token = strtok(sen, s);
printf("Output: ");
while (token != 0)
{
char z[20] =
{ *token };
for (i = 0;; i++)
{
if (z[i] == ' ')
{
z[i] = '\0';
break;
}
}
if (z[0] == 'a' || z[0] == 'A' || z[0] == 'e' || z[0] == 'E' || z[0] == 'i'
|| z[0] == 'I' || z[0] == 'o' || z[0] == 'O' || z[0] == 'u'
|| z[0] == 'U')
{
reverse(token);
}
else
printf("%s ", *token);
token = strtok(NULL, s);
}
printf("\n");
return 0;
}
//function for reversing the particular parts of string
void reverse(char *tok)
{
char x[20] =
{ *tok };
int i, j, len;
char temp;
for (i = 0;; i++)
{
if (x[i] == ' ')
{
x[i] = '\0';
break;
}
}
len = length(tok);
j = len - 1;
for (i = 0; x[i] != len / 2; i++)
{
temp = x[i];
x[i] = x[j];
x[j] = temp;
j--;
}
printf("%s", x);
printf(" ");
}
//function for determining the length of the token string
int length(char *t)
{
int i = 0;
char y[20] =
{ *t };
for (;; i++)
{
if (y[i] == ' ')
{
y[i] = '\0';
break;
}
}
while (y[i] == '\0')
{
i++;
}
return i;
}
This line
printf("%s ", *token);
passes a char where a 0-terminated char[] is expected.
The lesson learned is: Always compile with all warnings on! (-Wall -Wextra -pedantic for gcc)
Problem lies in this line:
if(z[i]==' ')
You've split with the token ' ', so that means there is no ' ' in z[]. So the loop never ends.
#include <stdio.h>
#include <string.h>
void reverse(char *tok);
int length(char *t);
int main(){
char sen[50];
const char s[2] = " ";
printf("Enter a Sentence: ");
scanf("%49[^\n]", sen);//gets(sen); //"gets" : Obsolete !
char *token = strtok(sen, s);
printf("Output: ");
while (token != NULL){
char z = *token;
if (z == 'A' || z == 'E' || z == 'I' || z == 'O' || z == 'U' ||
z == 'a' || z == 'e' || z == 'i' || z == 'o' || z == 'u'){
reverse(token);//print by this function
} else {
printf("%s ", token);
}
token = strtok(NULL, s);
}
printf("\n");
return 0;
}
void reverse(char *tok){//original not change
int i, len = length(tok);
char x[len + 1];
for (i = 0; i<len; ++i){
x[i] = tok[len-i-1];
}
x[i]='\0';
printf("%s ", x);
}
int length(char *t){//use strlen
int i;
for(i=0;*t;++i, ++t)
;
return i;
}
This is a question from K&R:-
Write a program to print a histogram of the lengths of words in its input.It is easy to draw the histogram with bars horizontal; but a vertical orientation is more challenging.
I am not supposed to use any library functions because this is only a tutorial introduction!
I have written the following program to do so but it have some bugs:-
1)If there is more than one white-space character between words, the program doesn't work as expected.
2)How do I know the maximum value of 'k' I mean how to know how many words are there in the input?
Here is the code:-
#include<stdio.h>
#include<ctype.h>
#include<string.h>
#include<stdlib.h>
#define MAX_WORDS 100
int main(void)
{
int c, i=0, k=1, ch[MAX_WORDS] = {0};
printf("enter the words:-\n");
do
{
while((c=getchar())!=EOF)
{
if(c=='\n' || c==' ' || c=='\t')
break;
else
ch[i]++;
}
i++;
}
while(i<MAX_WORDS);
do
{
printf("%3d|",k);
for(int j=1;j<=ch[k];j++)
printf("%c",'*');
printf("\n");
k++;
}
while(k<10);
}
This program will work fine even if there are more than one newline characters in between the two words and numWords will give you the numbers of words.
#include <stdio.h>
#include <stdbool.h>
int main(void)
{
int ch, cha[100] = {0}, k = 1;
int numWords = 0;
int numLetters = 0;
bool prevWasANewline = true; //Newlines at beginning are ignored
printf("Enter the words:-\n");
while ((ch = getchar()) != EOF && ch != '\n')
{
if (ch == ' ' || ch == '\t')
prevWasANewline = true; //Newlines at the end are ignored
else
{
if (prevWasANewline) //Extra nelines between two words ignored
{
numWords++;
numLetters = 0;
}
prevWasANewline = false;
cha[numWords] = ++numLetters;
}
}
do
{
printf("%3d|",k);
for(int j=0;j<cha[k];j++)
printf("%c",'*');
printf("\n");
k++;
} while(k <= numWords);
return 0;
}
Its kind of an old thread, but i had the same problem. The code above works like charm, but its kind of an overkill for the knowledge we have with K&R in the first 20 pages.
I had no idea what they meant by histogram printing and that code helped me with that.
Anyway i wrote a code myself with the knowledge i gained through the book, i hope it will be helpful to someone.
Forgive me if its a bit messy, i am just a beginner myself :D
#include <stdio.h>
#define YES 1
#define NO 0
int main(void)
{
int wnumb [100];
int i, inword, c, n, k;
n = (-1);
for (i = 0; i <=100; ++i) {
wnumb [i] = 0;
}
inword = NO;
while ((c = getchar()) != EOF) {
if ( c == ' ' || c == '\n' || c == '\t' ) {
inword = NO;
}
else if (inword == NO) {
++n;
++wnumb [n];
inword = YES;
}
else {
++wnumb [n];
}
}
for (i = 0; i <= 100; ++i) {
if (wnumb [i] > 0) {
printf ("\n%3d. | ", (i+1));
for (k = 1; k <= wnumb[i]; ++k) {
printf("*");
}
}
}
printf("\n");
}
Here is example of simple Histogram(Vertically)
#include <stdio.h>
int main()
{
int c, i, j, max;
int ndigit[10];
for (i = 0; i < 10; i++)
ndigit[i] = 0;
while ((c = getchar()) != EOF)
if (c >= '0' && c <= '9')
++ndigit[c-'0'];
max = ndigit[0];
for (i = 1; i < 10; ++i) /* for Y-axis */
if (max < ndigit[i])
max = ndigit[i];
printf("--------------------------------------------------\n");
for (i = max; i > 0; --i) {
printf("%.3d|", i);
for (j = 0; j < 10; ++j)
(ndigit[j] >= i) ? printf(" X ") : printf(" ");
printf("\n");
}
printf(" ");
for (i = 0; i < 10; ++i) /* for X-axis */
printf("%3d", i);
printf("\n--------------------------------------------------\n");
return 0;
}