So I was working on an assignment for school, and had written up a variation of this code:
#include <stdio.h>
#include <ctype.h>
#include <string.h>
#define MAX 100
// This program takes an input of strings and prints them out with a new line separating each one.
int main() {
char *WordArray[MAX]; //initializing variables
int i = 0;
int count = 0;
printf("enter up to 100 words, that are 20 characters maximum \n");
for (i = 0; i <100; i++){ //runs while there's less than 100 inputs
char Array[1];
scanf("%s",Array); //stores string in the array
if (strcmp(Array, "STOP") == 0) { //compares the string with stop, and if it is, it breaks out of the loop
break;
}
WordArray[i]=Array; //stores the string in the pointer array
}
printf("The output is\n");
for (count = 0; count<i; count++){ //counts up to the amount of words stored
printf("%s\n",WordArray[count]); //outputs each pointer string
}
}
and I noticed that the output was printing "STOP" instead of the values stored. Anyone have any answers to why and/or how to fix it? I know one of the methods is to switch to a 2D array instead of using pointers, but I'm still baffled as to why a program like this wouldn't work.
Your char Array[1]; isn't large enough to store any but an empty string. Also, when it works, every pointer will point to the same string, which will be the last entry you made. This makes some corrections where commented.
#include <stdio.h>
#include <stdlib.h> // instead of ctype.h
#include <string.h>
#define MAX 100
// This program takes an input of strings and prints them out with a new line separating each one.
int main() {
char *WordArray[MAX];
int i = 0;
int count = 0;
printf("enter up to 100 words, that are 20 characters maximum \n");
for (i = 0; i <100; i++){
char Array[21]; // increase size of array
scanf("%20s",Array); // limit entry length
if (strcmp(Array, "STOP") == 0) {
break;
}
WordArray[i] = strdup(Array); // allocate memory for and copy string
}
printf("The output is\n");
for (count = 0; count<i; count++){
printf("%s\n",WordArray[count]);
}
// free each string's memory
for (count = 0; count<i; count++){
free(WordArray[count]);
}
}
Program output:
enter up to 100 words, that are 20 characters maximum
one two three STOP
The output is
one
two
three
Edit: note that your code contains another undefined behaviour besides the too-short string char Array[1] which is that you dereference the pointer you stored in char *WordArray[MAX];. The scope of Array is inside the for loop, and theoretically ceases to exist after the loop completes, so the pointer you store is invalid. Here, the word entered is duplicated with strdup so that doesn't apply.
Related
I'm writing a code a that prompts the user to enter a string
&
create a function that is a type void that prints out the character that was used the most
(As in where it appeared more than any other ones)
&
also shows the number of how many times it was in that string.
Therefore here is what I have so far...
#include <stdio.h>
#include <string.h>
/* frequent character in the string along with the length of the string (use strlen from string.h – this will require you to #include <string.h> at the top of your program).*/
/* Use array syntax (e.g. array[5]) to access the elements of your array.
* Write a program that prompts a user to input a string,
* accepts the string as input, and outputs the most
* You should implement a function called mostfrequent.
* The function prototype for mostfrequent is: void mostfrequent(int *counts, char *most_freq, int *qty_most_freq, int num_counts);
* Hint: Consider the integer value of the ASCII characters and how the offsets can be translated to ints.
* Assume the user inputs only the characters a through z (all lowercase, no spaces).
*/
void mostfrequent(int *counts, char *most_freq, int *qty_most_freq, int num_counts_)
{
int array[255] = {0}; // initialize all elements to 0
int i, index;
for(i = 0; most_freq[i] != 0; i++)
{
++array[most_freq[i]];
}
// Find the letter that was used the most
qty_most_freq = array[0];
for(i = 0; most_freq[i] != 0; i++)
{
if(array[most_freq[i]] > qty_most_freq)
{
qty_most_freq = array[most_freq[i]];
counts = i;
}
num_counts_++;
}
printf("The most frequent character was: '%c' with %d occurances \n", most_freq[index], counts);
printf("%d characters were used \n", num_counts_);
}
int main()
{
char array[5];
printf("Enter a string ");
scanf("%s", array);
int count = sizeof(array);
mostfrequent(count , array, 0, 0);
return 0;
}
I'm getting the wrong output too.
output:
Enter a string hello
The most frequent character was: 'h' with 2 occurances
5 characters were used
should be
The most frequent character was: 'l' with 2 occurances
5 characters were used
let's do it short (others will correct me if I write something wrong ^_^ )
you declare a int like this:
int var;
use it like this :
var = 3;
you declare a pointer like this :
int* pvar;
and use the pointed value like this:
*pvar = 3;
if you declared a variable and need to pass a pointer to it as function parameters, use the & operator like this :
functionA(&var);
or simply save its address in a pointer var :
pvar = &var;
that's the basics. I hope it will help...
The function prototype you are supposed to use seems to include at least one superfluous parameter. (you have the total character count available in main()). In order to find the most frequently appearing character (at least the 1st of the characters that occur that number of times), all you need to provide your function is:
the character string to be evaluated;
an array sized so that each element represents on in the range of values you want to find the most frequent (for ASCII characters 128 is fine, for all in the range of unsigned char, 256 will do); and finally
a pointer to return the index in your frequency array that holds the index to the most frequently used character (or the 1st character of a set if more than one are used that same number of times).
In your function, your goal is to loop over each character in your string. In the frequency array (that you have initialized all zero), you will map each character to an element in the frequency array and increment the value at that element each time the character is encountered. For example for "hello", you would increment:
frequency['h']++;
frequency['e']++;
frequency['l']++;
frequency['l']++;
frequency['o']++;
Above you can see when you are done, the element frequency['l']; will hold the value of 2. So when you are done you just loop over all elements in frequency and find the index for the element that holds the largest value.
if (frequency[i] > frequency[most])
most = i;
(which is also why you will get the first of all characters that appear that number of times. If you change to >= you will get the last of that set of characters. Also, in your character count you ignore the 6th character, the '\n', which is fine for single-line input, but for multi-line input you need to consider how you want to handle that)
In your case, putting it altogether, you could do something similar to:
#include <stdio.h>
#include <ctype.h>
enum { CHARS = 255, MAXC = 1024 }; /* constants used below */
void mostfrequent (const char *s, int *c, int *most)
{
for (; *s; s++) /* loop over each char, fill c, set most index */
if (isalpha (*s) && ++c[(int)*s] > c[*most])
*most = *s;
}
int main (void) {
char buf[MAXC];
int c[CHARS] = {0}, n = 0, ndx;
/* read all chars into buf up to MAXC-1 chars */
while (n < MAXC-1 && (buf[n] = getchar()) != '\n' && buf[n] != EOF)
n++;
buf[n] = 0; /* nul-terminate buf */
mostfrequent (buf, c, &ndx); /* fill c with most freq, set index */
printf ("most frequent char: %c (occurs %d times, %d chars used)\n",
ndx, c[ndx], n);
}
(note: by using isalpha() in the comparison it will handle both upper/lower case characters, you can adjust as desired by simply checking upper/lower case or just converting all characters to one case or another)
Example Use/Output
$ echo "hello" | ./bin/mostfreqchar3
most frequent char: l (occurs 2 times, 5 chars used)
(note: if you use "heello", you will still receive "most frequent char: e (occurs 2 times, 6 chars used)" due to 'e' being the first of two character that are seen the same number of times)
There are many ways to handle frequency problems, but in essence they all work in the same manner. With ASCII characters, you can capture both the most frequent character and the number of times it occurs in a single array of int and an int holding the index to where the max occurs. (you don't really need the index either -- it just save looping to find it each time it is needed).
For more complex types, you will generally use a simple struct to hold the count and the object. For example if you were looking for the most frequent word, you would generally use a struct such as:
struct wfreq {
char *word;
int count;
}
Then you simply use an array of struct wfreq in the same way you are using your array of int here. Look things over and let me know if you have further questions.
Here is what I came up with. I messed up with the pointers.
void mostfrequent(int *counts, char *most_freq, int *qty_most_freq, int num_counts_)
{
*qty_most_freq = counts[0];
*most_freq = 'a';
int i;
for(i = 0; i < num_counts_; i++)
{
if(counts[i] > *qty_most_freq)
{
*qty_most_freq = counts[i];
*most_freq = 'a' + i;
}
}
}
/* char string[80]
* read in string
* int counts[26]; // histogram
* zero counts (zero the array)
* look at each character in string and update the histogram
*/
int main()
{
int i;
int num_chars = 26;
int counts[num_chars];
char string[100];
/*zero out the counts array */
for(i = 0; i < num_chars; i++)
{
counts[i] = 0;
}
printf("Enter a string ");
scanf("%s", string);
for(i = 0; i < strlen(string); i++)
{
counts[(string[i] - 'a')]++;
}
int qty_most_freq;
char most_freq;
mostfrequent(counts , &most_freq, &qty_most_freq, num_chars);
printf("The most frequent character was: '%c' with %d occurances \n", most_freq, qty_most_freq);
printf("%d characters were used \n", strlen(string));
return 0;
}
I am trying to make an array of character Arrays. The program will read in sentences and store the sentences in a character array, and then that character array will be stored in another array. After reading numerous websites and Stack Over Flow pages I think it can be done like this. The program breaks when trying to store my character array into another array, so i'm not sure how to correct my code.
#include <stdio.h>
#include <math.h>
#include <time.h>
int main(int ac, char *av[])
{
int size; //number of sentences
char strings[100];// character array to hold the sentences
char temp;
printf("Number of Strings: ");
scanf_s("%d", &size); // read in the number of sentences to type
char **c = malloc(size); //array of character arrays
int i = 0;
int j = 0;
while (i < size) //loop for number of sentences
{
printf("Enter string %i ",(i+1));
scanf_s("%c", &temp); // temp statement to clear buffer
fgets(strings, 100, stdin);
// **** this next line breaks the program
c[i][j] = strings; // store sentence into array of character arrays
j++;
i++;
}
printf("The first character in element 0 is: %d\n", c[0][0]);
system("PAUSE");
return 0;
}
Al you need to do is allocate the memory for the string just read and copy the string:
c[i][j] = strings; // replace this with:
c[i]= malloc(strlen(strings)+1);
strcpy(c[i],strings);
Sadly char **c is not array of characters arrays. But this will properly allocate a 2d array dynamically if you follow
char (*c)[SIZE];
And then doing this
c = malloc(sizeof(char[LEN][SIZE]));
Then you do what you are trying to do.
for(size_t i = 0; i < LEN; i++){
if(fgets(c[i],SIZE,stdin)){
...
}
}
Or you can do it like this
char **c = malloc(LEN);
..
for(size_t i = 0; i < LEN; i++){
c[i] = malloc(SIZE);
...
}
But again c is nothing but jagged array of characters.
Check the return value of malloc and free the dynamically allocated memory when you are done working with it.
Listen dude,
Do you want to store the words in the character array and then store this in another array.
It is what you want ?
If yes then you can do the following.
Use stdlib by using # include <stdlib.h> .
This lets you use string functions directly.
Now read every word as a string and make array of strings.
So now if you number of strings is n, define an array for that as string my_array[n] and scan each word using scanf("%s",&my_array [i]).
In this way you will get an array of strings.
the objective of my question is very simple. The first input that I get from the user is n (number of test cases). For each test case, the program will scan a string input from the user. And each of these strings I will process separately.
The question here is how can I get string inputs and process them separately in C language??? The idea is similar to the dictionary concept where we can have many words which are individual arrays inside one big array.
The program I have written so far:
#include <stdio.h>
#define max 100
int main (){
int n; // number of testcases
char str [100];
scanf ("%d\n",&n);
for (int i =0;i <n;i++){
scanf ("%s",&str [i]);
}
getchar ();
return 0;
}
Can someone suggest what should be done?
The input should be something like this:
Input 1:
3
Shoe
Horse
House
Input 2:
2
Flower
Bee
here 3 and 2 are the values of n, the number of test cases.
First of all, Don't be confused between "string" in C++ , and "Character Array" in C.
Since your question is based on C language, I will be answering according to that...
#include <stdio.h>
int main (){
int n; // number of testcases
char str [100][100] ; // many words , as individual arrays inside one big array
scanf ("%d\n",&n);
for (int i =0;i <n;i++){
scanf ("%s",str[i]); // since you are taking string , not character
}
// Now if you want to access i'th word you can do like
for(int i = 0 ; i < n; i++)
printf("%s\n" , str[i]);
getchar ();
return 0;
}
Now here instead of using a two-dimensional array, you can also use a one-dimensional array and separate two words by spaces, and store each word's starting position in some another array. (which is lot of implementation).
First of all yours is not C program, as you can't declare variable inside FOR loop in C, secondly have created a prototype using Pointer to Pointer, storing character array in matrix style datastructure, here is the code :-
#include <stdio.h>
#include <stdlib.h>
#define max 100
int main (){
int n,i; // number of testcases
char str [100];
char **strArray;
scanf ("%d",&n);
strArray = (char **) malloc(n);
for (i =0;i <n;i++){
(strArray)[i] = (char *) malloc(sizeof(char)*100);
scanf ("%s",(strArray)[i]);
}
for (i =0;i <n;i++){
printf("%s\n",(strArray)[i]);
free((strArray)[i]);
}
getchar ();
return 0;
}
#include <stdio.h>
#define MAX 100 // poorly named
int n=0; // number of testcases
char** strs=0;
void releaseMemory() // don't forget to release memory when done
{
int counter; // a better name
if (strs != 0)
{
for (counter=0; counter<n; counter++)
{
if (strs[counter] != 0)
free(strs[counter]);
}
free(strs);
}
}
int main ()
{
int counter; // a better name
scanf("%d\n",&n);
strs = (char**) calloc(n,sizeof(char*));
if (strs == 0)
{
printf("outer allocation failed!")
return -1;
}
for (counter=0; counter<n; counter++)
{
strs[counter] = (char*) malloc(MAX*sizeof(char));
if (strs[counter] == 0)
{
printf("allocate buffer %d failed!",counter)
releaseMemory();
return -1;
}
scanf("%s",&strs[counter]); // better hope the input is less than MAX!!
// N.B. - this doesn't limit input to one word, use validation to handle that
}
getchar();
// do whatever you need to with the data
releaseMemory();
return 0;
}
I would like to count the number of elements in an integer array(sized, like: array[1000]) after input without counting manually while inputting(using scanf() which is = number of arguments passed to scanf).Though int arrays are not null terminated as well as scanf() cannot be used like getchar() or gets() and no availale function like strlen() for int arrays,is it possible to write a C program to prompt the user to input as many numbers as he wishes and the program will count them(total arguments passed to scanf()) and print the maximum using arrays or pointers?
Without having a termination value, you will have to count the inputs as they are made. You could do this by defining a struct to hold the array. Your program does not know how many integers you will enter, and this code allocates more memory when the array is full, keeping track of the array size and elements used.
#include <stdio.h>
#include <stdlib.h>
#define ARRAY_STEP 10 // increment in array size
typedef struct {
int maxlen; // capacity of array
int length; // elements used
int *array; // the array pointer
} istruct;
int main(void) {
istruct intarr = {0}; // empty array
int i;
printf ("Enter some integers:\n");
while (scanf("%d", &i) == 1) {
if (intarr.length >= intarr.maxlen) {
intarr.maxlen += ARRAY_STEP; // increase array size
intarr.array = realloc(intarr.array, intarr.maxlen * sizeof(int));
if (intarr.array == NULL) {
printf("Memory allocation error\n");
exit (1);
}
}
intarr.array[intarr.length++] = i;
}
printf ("You entered:\n");
for (i=0; i<intarr.length; i++)
printf ("%-10d", intarr.array[i]);
printf ("\n");
if (intarr.array)
free(intarr.array);
return 0;
}
strlen() iterates on the character array until the C string termination character (ASCII \0) is encountered. So it is COUNTING the elements. If you want to do that for an int array I guess you could also have a reserved 'terminator' value for your array, make room for it (allocate one more int in your array for the terminator) and implement your own int_array_len() similar to strlen. However, in your case counting the elements while inputting them seems like a much better way to go.
Smart vertion:
#include <stdio.h>
int main()
{
double a[100000],mx=0;
int i,j,c=0;
printf("Enter as many numbers as you wish . Press Ctrl+D or any character to stop inputting :\n");
/*for(i=0;i<100000;i++)
a[i]=0;*/
for(i=0;i<100000;i++)
{
if((scanf("%lf",&a[i]))==1)
c++;
//else break;
}
for(j=0;j<c;j++)
{
if(a[j]>mx) mx=a[j];
}
printf("You have inserted %d values and the maximum is:%g",c,mx);
return 0;
}
It's pretty simple.
just declare of your desired "initial size". And keep checking if input is a valid integer or not.
if((scanf("%d",&a[i]))==1) this would be true if and only if the input is a valid integer. hence the moment the input is not an int, it exits the loop. You can count the number of input values within the same loop and also the max values.
Here is the code:
#include <stdio.h>
#include<limits.h>
int main()
{
int a[100],max = INT_MIN;
int i,j,count=0;
printf("Start Entering the integers... Give any non-integer input to stop:\n");
for(i=0;i<100;i++)
{
if((scanf("%d",&a[i]))==1) {
count++;
if(a[i]>max) {
max = a[i];
}
}
else
break;
}
printf("number of input values: %d;\nThe maximum input value: %d",count,max);
return 0;
}
I will have to take the input from user and find the longest input from those 10 strings..
#include<stdio.h>
#include<conio.h>
void main() {
char str[10][10]
printf("Enter strings:")
scanf("%s", str)
}
If I take the user input like this, would it store the strings in str two dimensional array? To find out the longest string first I would find the length of each strings and use max_length function to determine the longest string.
You do not need to store all of the strings, just the longest one entered so far.
Note that you do need to define a maximum length of string to avoid buffer overrun.
For example:
#define MAX_STRING_SIZE 1024
char last_entered_string[MAX_STRING_SIZE];
char longest_entered_string[MAX_STRING_SIZE] = ""; /* Must be initialized. */
scanf("%1023s", last_entered_string); /* Read one less to allow for
null terminator. */
Use a loop to accept the ten inputs and compare with the longest string. If the last entered string is longer then copy it into the longest string. As this is homework I'll not provide any further code.
No, it won't. You have to loop through and read all strings.
for(i=0;i<10;i++)
scanf("%s", str[i]);
Also, you are missing some semi-colons!
You can find the longest string put and save it for all the string received.
int main()
{
char *str = NULL;
char *compare;
printf("Enter strings:");
scanf("%s", compare);
if (strlen(str) < strlen(compare))
str = strdup(compare);
return(0);
}
And if you want to store all of users input (considering you can have just 10 string from the user)you can do this :
int main()
{
char **array;
char *str;
int x = 0;
int shortest;
array = malloc(sizeof(char*) * 10);
while (x < 10)
{
scanf("%s", str)
array[x] = strdup(str);
x++;
}
x = 0;
shortest = x;
while (x < 10)
{
if (strlen(array[x]) > strlen(shortest))
shortest = x;
x++;
}
return (0);
}
shortest will be the index of the longest string in your array.
I hope this will help you.
Store all input in an array, then do qsort() it on the array entries length and then take the first (or the last, depending on how you sorted) entry.
Ok, ok ... - this might be over-engineered ... ;-)
The program presented will take 10 input strings from the user and then finally print out the longest string and its length. It will not store any other input strings than the biggest one.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_STR_LEN 1024
int main(int argc, char **argv){
char str_final[MAX_STR_LEN];
char str_temp[MAX_STR_LEN];
unsigned int i, j, len_str;
unsigned int num_string = 10;
unsigned int len_max = 0;
for (i=0; i<num_string; i++){
printf("Enter string number: %d\n", i);
gets(str_temp);
for (j=0; str_temp[j]; j++);
len_str = j;
if(len_str > len_max){
len_max = len_str;
memset(str_final, 0, MAX_STR_LEN);
memcpy(str_final, str_temp, len_str);
}else{
memset(str_temp, 0, MAX_STR_LEN);
}
}
printf("The biggest string is: %s\n", str_final);
printf("It's size is: %d\n", len_max);
exit(EXIT_SUCCESS);
}
I think what you can do is take a nested loop and search for a '\0' character in the row and run a counter simultaneously. as soon as you find a '\0' stop the counter and store the value of counter in a separate array . so now you will have a array of 10 integers.
Now search for smallest integer in the array and... Bingo!
The corresponding row will have the shortest string.
I know this approach is very raw but I think it will be helpful for people with only basic knowledge of C.