I am trying to take t (test cases) strings as inputs in C. Every string ends with a # and it can have spaces.
I am using getchar and some while loops but I am not able to achieve what I want.
int t;
scanf("%d",&t);
while(t>0){
char expression[100];
char c;
int i=0;
while(1){
c = getchar();
if(c == '#'){
break;
}
else{
expression[i]=c;
i++;
}
}
t--;
printf("%d\n",i);
for(int j=0;j<i;j++){
printf("%c",expression[j]);
}
}
What I am trying to achieve is if # is encountered I want to ignore whatever is after It.
My input:
1
test#wtf
Output:
5
test
So in case of 1 test case it is working fine.
But when I give try this.
2
test#wtf
test#
The corresponding outs are:
5
test
and
8
wtf
test
As you can see the second output should be test and length should be 5 but it is not so.
In order to ignore text after #, you'll have to consume those characters using getchar() and ignore them till you encounter a new line character '\n'.
So, after you exit the inner while loop, you need to add the following code
while(1) {
c = getchar();
if(c == '\n'){
break;
}
}
You can do it this way, you process all characters, once you get a '#' you keep reading characters without adding them to the array and you get out of the outer loop once you get a new line character '\n'.
while (--t > 0)
{
int i,s,c;
char exp[100];
s = 0;
while (s < 100 && (c = getchar()) != '#')
exp[s++] = c;
while ((c = getchar()) != '\n')
;
printf("%d\n", s);
i = -1;
while (++i < s)
printf("%c", exp[i]);
printf("\n");
}
Related
I am reading the C programming language book Dennis M. Ritchie and
trying to solve this question:
Write a program to print a histogram of
the lengths of words in
its input. It is easy to draw the histogram with the bars horizontal; a vertical
orientation is more challenging.
I think my solution works, but the problem is that if I don't press EOF, the terminal won't show the
result. I know that the condition specifies that exactly, but I am
wondering whether there is any way to make the program terminate after
reading a single line? (Sorry if my explanation of the problem is a bit shallow. Feel free to ask more.)
#include <stdio.h>
int main ()
{
int digits[10];
int nc=0;
int c, i, j;
for (i = 0; i <= 10; i++)
digits[i] = 0;
//take input;
while ((c = getchar ()) != EOF) {
++nc;
if (c == ' ' || c=='\n') {
++digits[nc-1];
//is it also counting the space in nc? i think it is,so we should do nc-1
nc = 0;
}
}
for (i = 1; i <= 5; i++) {
printf("%d :", i);
for (j = 1; j <= digits[i]; j++) {
printf ("*");
}
printf ("\n");
}
// I think this is a problem with getchar()
//the program doesn't exit automatically
//need to find a way to do it
}
You could try to make something like
while ((c = getchar ()) != EOF && c != '\n') {
and then adding a line after the while loop to account for the last word:
if (c == '\n') {
++digits[nc-1];
nc = 0;
There is also another problem inside your program. ++digits[nc-1]; is correct, however, for the wrong reason. You should make it because an array starts at zero, i.e. if you have an array of length 10, it will go from 0 to 9, so you should count the length of the words and then add one to the position of the array length - 1 (as there are no words of length zero). The problem is that you are still counting the blank spaces or the newline characters inside the length of a word, so if you have two blank spaces after a word of length 4, the program will add to the array a word of length 5 + a word of length 1. To avoid this, you should do something like this:
while ((c = getchar ()) != EOF) {
if ((c == ' ' || c == '\n' || c == '\t') && nc > 0) {
++digits[nc-1]; // arrays start at zero
nc = 0;
}
else {
++nc;
}
}
In the following code example from K&R's book, if I replace putchar(c) with printf("%c", c) the code works the same. But if I replace it with printf("%d", c) it gives gibberish output.
#include <stdio.h>
int main() {
int c;
c = getchar();
while (c != EOF) {
putchar(c);
c = getchar();
}
}
From here, I learned that the getchar() converts the stdin to an 8-bit character whose value ranges from 0 to 255.
Now I want to print the value of c using putchar(c) in one line and printf("%d", c) in another line. So I wrote the following code:
#include <stdio.h>
int main() {
int c, b;
c = getchar();
b = c;
while (c != EOF && c != 10) {
printf("%c",c);
c = getchar();
}
printf("\n");
while (b != EOF && b != 10) {
printf("%d\t",b);
b = getchar();
}
}
I used the condition c != 10 as the newline character is read as 10 by getchar(). I expected the code to work as
$ ./a.out
783
783
55 56 51
but the program terminates as
$ ./a.out
783
783
55
I understand that getchar() takes input from stdin and the variable b is not stdin. So how should I copy the variable c to b so that my program works as I expect it to?
The problem is that your code does not (and cannot, as it stands) 'remember' the inputs you gave in the first loop. So, after you have finished that loop, your second loop is wanting to read in the characters for b (after it has output the first value, which is remembered from the earlier b = c line).
So, after outputting 55 (the integer value of the character 7), it is waiting for further input.
Probably the easiest way to get the output that you're looking for is to have an array of input characters. Then, you can output the %c values as you read them (as before), then re-run the outputs using the %d format in a subsequent for loop.
Here is a demonstration that does what I think you're after:
#include <stdio.h>
#define MAXINS 20 // Set to the maximum number of input characters you want to allow
int main()
{
int c[MAXINS];
int i = 0, n = 0;
c[0] = getchar();
while (i < MAXINS && c[i] != EOF && c[i] != 10) {
printf("%c", c[i]);
c[++i] = getchar();
++n;
}
printf("\n");
for (i = 0; i < n; ++i) {
printf("%d\t", (int)(c[i]));
}
return 0;
}
Feel free to ask for further clarification and/or explanation.
EDIT: On the point in the your first paragraph, "But if I replace it with printf("%d", c) it gives gibberish output." Well, when I try the following code and give 783 and then hit return (which generates a newline) I get the expected 55565110 as the output:
int main()
{
int c;
c = getchar();
while (c != EOF) {
printf("%d", c);
c = getchar();
}
return 0;
}
This may look like gibberish, but it's just the same output as you 'expect' in your later code, but without the spaces and with the addition of the 10 for the newline.
You need to have every character stored, because once you read a char from stdin, it is not present in stdin anymore.
Since you want the newline character in the end as a part of the input, you should use fgets to take the input.
Say you are taking an input that could have a maximum of 100 characters.
#include <stdio.h>
int main(void) {
char c[100]; // A char array
fgets(c,100,stdin);
int x=0;
while (c[x] != 10 && c[x] != EOF)
printf("%c",c[x++]);
printf("\n");
x = 0;
while (c[x] != 10 && c[x] != EOF) // You can simply compare it with the newline character too.
printf("%d ",c[x++]);
printf("\n");
return 0;
}
There are many ways to do this. You can also read stdin character-by-character ans store it in an array. However, since you need to display the ASCII values of the characters in another line after displaying the characters themselves, you will have to store them in an array.
You are copying only the first input, to copy the whole string you need to store each input in a buffer and check if the string doesn't overflow that buffer on each iteration:
int main(void)
{
enum {size = 256};
char buffer[size];
size_t count = 0;
int c;
while ((c = getchar()) && (c != '\n') && (c != EOF))
{
printf("%c", c);
if (count < size)
{
buffer[count++] = (char)c;
}
}
printf("\n");
for (size_t iter = 0; iter < count; iter++)
{
printf("%d\t", buffer[iter]);
}
printf("\n");
}
If you don't want to limit the buffer to an arbitrary size then you need to change your approach to use dynamic memory (realloc or a linked list)
#include<stdio.h>
int main() {
char text[1000];
int ch;
int index = 0;
while ((ch = getchar()) != EOF) {
text[index] = ch;
index++;
}
text[index] = '\0';
int i =0;
int num_Count=0;
int lower_Count=0;
int upper_Count =0;
while(i < index) {
if((text[i]>='0') && (text[i]<='9')){
num_Count ++;
i++;
}
else if((text[i]>='A') && (text[i]<='Z')){
upper_Count++;
i++;
}
else if((text[i]>='a') && (text[i] <='z')){
lower_Count++;
i++;
}
else
i++;
}
printf("%d %d %d", num_Count, lower_Count, upper_Count);
return 0;
}
It is a program that outputs the number of lower case, upper case, and number when the sentence is inputted.
For example,
Hi
Name
100
Would output 3 4 2
I keep seeing a runtime error.
The (while) part seems to be wrong.. I do not know what's wrong.
I ran your code in my system and checked for the input: Hi Name 100. The output I got is 3 4 2 which is the expected output. I feel the only place where the code can run in an infinite loop is while reading the inputs. Try to use ctrl+ d for EOF or ctrl+ z for windows.
Rest every thing is fine.
EOF means End Of File. It is used when you read data from a file. I suggest put a character like newline ('\n').
As part of my course, I have to learn C using Turbo C (unfortunately).
Our teacher asked us to make a piece of code that counts the number of characters, words and sentences in a paragraph (only using printf, getch() and a while loop.. he doesn't want us to use any other commands yet). Here is the code I wrote:
#include <stdio.h>
#include <conio.h>
void main(void)
{
clrscr();
int count = 0;
int words = 0;
int sentences = 0;
char ch;
while ((ch = getch()) != '\n')
{
printf("%c", ch);
while ((ch = getch()) != '.')
{
printf("%c", ch);
while ((ch = getch()) != ' ')
{
printf("%c", ch);
count++;
}
printf("%c", ch);
words++;
}
sentences++;
}
printf("The number of characters are %d", count);
printf("\nThe number of words are %d", words);
printf("\nThe number of sentences are %d", sentences);
getch();
}
It does work (counts the number of characters and words at least). However when I compile the code and check it out on the console window I can't get the program to stop running. It is supposed to end as soon as I input the enter key. Why is that?
Here you have the solution to your problem:
#include <stdio.h>
#include <conio.h>
void main(void)
{
clrscr();
int count = 0;
int words = 0;
int sentences = 0;
char ch;
ch = getch();
while (ch != '\n')
{
while (ch != '.' && ch != '\n')
{
while (ch != ' ' && ch != '\n' && ch != '.')
{
count++;
ch = getch();
printf("%c", ch);
}
words++;
while(ch == ' ') {
ch = getch();
printf("%c", ch);
}
}
sentences++;
while(ch == '.' && ch == ' ') {
ch = getch();
printf("%c", ch);
}
}
printf("The number of characters are %d", count);
printf("\nThe number of words are %d", words);
printf("\nThe number of sentences are %d", sentences);
getch();
}
The problem with your code is that the innermost while loop was consuming all the characters. Whenever you enter there and you type a dot or a newline it stays inside that loop because ch is different from a blank. However, when you exit from the innermost loop you risk to remain stuck at the second loop because ch will be a blank and so always different from '.' and '\n'. Since in my solution you only acquire a character in the innermost loop, in the other loops you need to "eat" the blank and the dot in order to go on with the other characters.
Checking these conditions in the two inner loops makes the code work.
Notice that I removed some of your prints.
Hope it helps.
Edit: I added the instructions to print what you type and a last check in the while loop after sentences++ to check the blank, otherwise it will count one word more.
int ch;
int flag;
while ((ch = getch()) != '\r'){
++count;
flag = 1;
while(flag && (ch == ' ' || ch == '.')){
++words;//no good E.g Contiguous space, Space at the beginning of the sentence
flag = 0;;
}
flag = 1;
while(flag && ch == '.'){
++sentences;
flag=0;
}
printf("%c", ch);
}
printf("\n");
I think the problem is because of your outer while loop's condition. It checks for a newline character '\n', as soon as it finds one the loop terminates. You can try to include your code in a while loop with the following condition
while((c=getchar())!=EOF)
this will stop taking input when the user presses Ctrl+z
Hope this helps..
You can implement with ease an if statement using while statement:
bool flag = true;
while(IF_COND && flag)
{
//DO SOMETHING
flag = false;
}
just plug it in a simple solution that uses if statements.
For example:
#include <stdio.h>
#include <conio.h>
void main(void)
{
int count = 0;
int words = 1;
int sentences = 1;
char ch;
bool if_flag;
while ((ch = getch()) != '\n')
{
count++;
if_flag = true;
while (ch==' ' && if_flag)
{
words++;
if_flag = false;
}
if_flag = true;
while (ch=='.' && if_flag)
{
sentences++;
if_flag = false;
}
}
printf("The number of characters are %d", count);
printf("\nThe number of words are %d", words);
printf("\nThe number of sentences are %d", sentences);
getch();
}
#include <stdio.h>
#include <ctype.h>
int main(void){
int sentence=0,characters =0,words =0,c=0,inside_word = 0,temp =0;
// while ((c = getchar()) != EOF)
while ((c = getchar()) != '\n') {
//a word is complete when we arrive at a space after we
// are inside a word or when we reach a full stop
while(c == '.'){
sentence++;
temp = c;
c = 0;
}
while (isalnum(c)) {
inside_word = 1;
characters++;
c =0;
}
while ((isspace(c) || temp == '.') && inside_word == 1){
words++;
inside_word = 0;
temp = 0;
c =0;
}
}
printf(" %d %d %d",characters,words,sentence);
return 0;
}
this should do it,
isalnum checks if the letter is alphanumeric, if its an alphabetical letter or a number, I dont expect random ascii characters in my sentences in this program.
isspace as the name says check for space
you need the ctype.h header for this. or you could add in
while(c == ' ') and whie((c>='a' && c<='z') || (c >= 'A' && c<='Z')
if you don't want to use isalpace and isalnum, your choice, but it will be less elegant :)
The trouble with your code is that you consume the characters in each of your loops.
a '\n' will be consumed either by the loop that scans for words of for sentences, so the outer loop will never see it.
Here is a possible solution to your problem:
int sentences = 0;
int words = 0;
int characters = 0;
int in_word = 0; // state of our parser
int ch;
do
{
int end_word = 1; // consider a word wil end by default
ch = getch();
characters++; // count characters
switch (ch)
{
case '.':
sentences++; // any dot is considered end of a sentence and a word
break;
case ' ': // a space is the end of a word
break;
default:
in_word = 1; // any non-space non-dot char is considered part of a word
end_word = 0; // cancel word ending
}
// handle word termination
if (in_word and end_word)
{
in_word = 0;
words++;
}
} while (ch != '\n');
A general approach to these parsing problems is to write a finite-state machine that will read one character at a time and react to all the possible transitions this character can trigger.
In this example, the machine has to remember if it is currently parsing a word, so that one new word is counted only the first time a terminating space or dot is encountered.
This piece of code uses a switch for concision. You can replace it with an if...else if sequence to please your teacher :).
If your teacher forced you to use only while loops, then your teacher has done a stupid thing. The equivalent code without other conditional expressions will be heavier, less understandable and redundant.
Since some people seem to think it's important, here is one possible solution:
int sentences = 0;
int words = 0;
int characters = 0;
int in_word = 0; // state of our parser
int ch;
// read initial character
ch = getch();
// do it with only while loops
while (ch != '\n')
{
// count characters
characters++;
// count words
while (in_word)
{
in_word = 0;
words++;
}
// skip spaces
while (ch == ' ')
{
ch = -1;
}
// detect sentences
while (ch == '.')
{
sentences++;
ch = -1;
}
// detect words
while ((ch != '\n')
{
word_detected = 1;
ch = -1;
}
// read next character
ch = getch();
}
Basically you can replace if (c== xxx) ... with while (c== xxx) { c = -1; ... }, which is an artifical, contrieved way of programming.
An exercise should not promote stupid ways of doing things, IMHO.
That's why I suspect you misunderstood what the teacher asked.
Obviously if you can use while loops you can also use if statements.
Trying to do this exercise with only while loops is futile and results in something that as little or nothing to do with real parser code.
All these solutions are incorrect. The only way you can solve this is by creating an AI program that uses Natural Language Processing which is not very easy to do.
Input:
"This is a paragraph about the Turing machine. Dr. Allan Turing invented the Turing Machine. It solved a problem that has a .1% change of being solved."
Checkout OpenNLP
https://sourceforge.net/projects/opennlp/
http://opennlp.apache.org/
My task is:
Write a program that reads input up to # and reports the number of times that the sequence ei occurs.
I wrote something that in most of the times works, but there are inputs when it dosent...
Like this input:(suppose to return 1)
sdlksldksdlskd
sdlsklsdks
sldklsdkeisldksdlk
#
number of combination is: 0
This is the code:
int main(void)
{
int index = 0;
int combinationTimes = 0;
int total = 0;
char userInput;
char wordChar[index];
printf("please enter your input:\n");
while ((userInput = getchar()) != '#')
{
if (userInput == '\n')
continue;
wordChar[index] = userInput;
index++;
total++;
}
for (index = 1; index < total; index++)
{
if (wordChar[index] == 'i')
{
if (wordChar[--index] == 'e')
{
combinationTimes++;
++index;
}
}
}
printf("number of combination is: %d", combinationTimes);
return 0;
}
Can you please tell me what am I not getting 1 using this input?
in the book he said to test it with "Receive your eieio award" and it worked...but after i played with it a little i see that not always.
It really doesn't seem necessary to read the file into an array. You just need to keep track of how many times ei is found before you read a # or reach EOF:
#include <stdio.h>
int main(void)
{
int c;
int ei_count = 0;
while ((c = getchar()) != EOF && c != '#')
{
if (c == 'e')
{
int c1 = getchar();
if (c1 == 'i')
ei_count++;
else if (c1 != EOF)
ungetc(c1, stdin);
}
}
printf("ei appeared %d times\n", ei_count);
return(0);
}
Testing (the program is called ei and is built from ei.c):
$ ei < ei.c
ei appeared 0 times
$ sed 1d ei.c | ei
ei appeared 1 times
$ sed 's/#/#/' ei.c | ei
ei appeared 4 times
$
The first one stops at the #include line, the second stops at the # in the comparison, and the third reads the entire file. It also gives the correct output for the sample data.
Analysing the code
Your primary problem is that you do not allocate any space for the array. Change the dimension of the array from index to, say, 4096. That'll be big enough for your testing purposes (but really the program should pay attention to the array and not overflowing it — but then I don't think the array is necessary at all; see the code above).
The next primary problem is that despite its name, getchar() returns an int, not a char. It can return any valid character plus a distinct value, EOF. So it must return a value that's bigger than a char. (One of two things happens if you use char. If char is a signed type, some valid character — often ÿ, y-umlaut, U+00FF, LATIN SMALL LETTER Y WITH DIAERESIS — is also treated as EOF even though it is just a character. If char is an unsigned type, then no input matches EOF. Neither is correct behaviour.)
Fixing that is easy, but your code does not detect EOF. Always handle EOF; the data may be malformatted. That's a simple fix in the code.
A tertiary problem is that the printf() statement does not end with a newline; it should.
Your test condition here is odd:
if (wordChar[--index] == 'e')
{
combinationTimes++;
++index;
}
It's odd to use one pre-increment and one post-increment, but that's just a consistency issue.
Worse, though, is what happens when the character i appears in the input and is not preceded by e. Consider the line #include <stdio.h>: you start with index as 1; that is an i, so you decrement index, but wordChar[0] is not an e, so you don't increment it again, but the end of the loop does, so the loop checks index 1 again, and keeps on going around the loop testing that the i is i and # is not e for a long time.
There's no reason to decrement and then increment index; just use:
if (wordChar[index-1] == 'e')
combinationTimes++;
With those fixed, your code behaves. You trouble was largely that you were using an array that was not big enough (being size 0), and you were overwriting quasi-random memory with the data you were reading.
#include <stdio.h>
int main(void)
{
int index = 0;
int combinationTimes = 0;
int total = 0;
int userInput;
char wordChar[4096];
printf("please enter your input:\n");
while ((userInput = getchar()) != '#' && userInput != EOF)
{
if (userInput == '\n')
continue;
wordChar[index] = userInput;
index++;
total++;
}
printf("total: %d\n", total);
for (index = 1; index < total; index++)
{
if (wordChar[index] == 'i')
{
if (wordChar[index-1] == 'e')
combinationTimes++;
}
}
printf("number of combination is: %d\n", combinationTimes);
return 0;
}
Note that you could reasonably write the nested if as:
if (wordChar[index] == 'i' && wordChar[index-1] == 'e')
combinationTimes++;
change your wordChar array value.
int main(void)
{
int index = 0;
int combinationTimes = 0;
int total = 0;
char userInput;
//char wordChar[index]; // index = 0
char wordChar[255]; // should change the value of array.
printf("please enter your input:\n");
while ((userInput = getchar()) != '#')
{
if (userInput == '\n')
continue;
wordChar[index] = userInput;
index++;
total++;
}
for (index = 1; index < total; index++)
{
if (wordChar[index] == 'i')
{
if (wordChar[--index] == 'e')
{
combinationTimes++;
++index;
}
}
}
printf("number of combination is: %d", combinationTimes);
return 0;
}
or maybe you can use pointer and then use malloc and realloc.