Error: value computed is not used [-Werror=unused-value] - arrays

Ive written this code to try and read if there are any duplicate letters in a word, but I keep coming across this error:
Error: value computed is not used [-Werror=unused-value]
The line in question is:
arr[(int)(str[i])++];
The whole code is:
#include<stdio.h>
#include<string.h>
int main()
{
char str[30];
printf("Enter your String:");
scanf("%[^\n]",str);
int i;
int arr[256]={0};
for(i=0;i<strlen(str);i++)
{
if(str[i]==' '){
continue;
arr[(int)(str[i])++];
}
}
printf("Repeated character in a string are:\n");
for(i=0;i<256;i++)
{
if(arr[i]>1)
{
printf("%c occurs %d times\n",(char)(i),arr[i]);
}}
return 0;
}
Here is the error message from the console:
https://i.stack.imgur.com/GpMOW.png
Any help is appreciated :)

I've made some changes to the code, shown in comment.
#include <stdio.h>
#include <string.h>
int main()
{
char str[30];
printf("Enter your String:");
scanf("%29[^\n]", str); // limit the input length
int i;
int arr[256] = { 0 };
for(i = 0; i < strlen(str); i++)
{
if(str[i] == ' '){
continue;
}
// this line was repositioned so that it can execute
// char can be signed, so don't index by a negative value
// the post-increment should apply to the `arr[]` array element
arr[ (unsigned char)str[i] ]++;
}
printf("Repeated character in a string are:\n");
for(i = 0; i < 256; i++)
{
if(arr[i] > 0) // inform *any* usage
{
// (char)i gets promoted to int anyway
printf("%c occurs %d times\n", i, arr[i]);
}
}
return 0;
}
Session:
Enter your String:one two
Repeated character in a string are:
e occurs 1 times
n occurs 1 times
o occurs 2 times
t occurs 1 times
w occurs 1 times

Related

Need some help how to enter a number and count it using array

I have to code in an array that can count an element. For example, if the user enters a 2, 2, 2, 1,1 then the user wants to count the number 2 then the result will be ELEMENT is 2 and FREQUENCY is 3. but I have a problem with the parts of " ENTER THE NUMBER YOU WANT TO BE COUNTED". I use scanf but when I run it I cannot enter any number.
Here's my code:
void frequency()
{
system("cls");
int num;
int count=0;
printf("Enter a number you want to be count: \n ");
scanf("i%", &num);
printf(" ELEMENT | FREQUENCY \n ");
for (i = 0; i<=n; i++)
{
if (a[i]==a[num])
count++;
}
printf(" \n %i ", num);
printf(" \t\t");
printf("%i \n ", count);
getch();
}
Your program requires understanding on two parts:
Get input and split input by delimiter, which can be done by using strtok.
Algorithm for finding the duplicated elements in an array.
#include <stdio.h>
#include <string.h>
int main() {
frequency();
return 0;
}
void frequency() {
char str[100];
printf("Enter a number you want to be count: \n ");
gets(str);
int init_size = strlen(str);
char delim[] = " ";
char *ptr = strtok(str, delim);
char *pch;
int arr[20];
int count = 0;
int ncount, i, j;
int a[count], Freq[count];
while(ptr != NULL) {
/*printf("'%s'\n", ptr);*/
/*Converts the string argument str to an integer (type int)*/
arr[count] = atoi(ptr);
/*strtok accepts two strings - the first one is the string to split, the second one is a string containing all delimiters*/
ptr = strtok(NULL, delim);
/*Initialize frequency value to -1*/
Freq[count] = -1;
count += 1;
}
/*Count the frequency of each element*/
for (i = 0; i < count; i++) {
ncount = 1;
for(j = i + 1; j < count; j++) {
/*Part to perform checking for duplicate elements*/
if(arr[i] == arr[j]) {
ncount++;
/*Make sure not to count frequency of same element again*/
Freq[j] = 0;
}
}
/*If frequency of current element is not counted*/
if(Freq[i] != 0) {
Freq[i] = ncount;
}
}
printf(" ELEMENT | FREQUENCY \n");
printf("-------------------------\n");
for (i = 0; i < count; i++) {
if(Freq[i] != 0) {
printf("\t%d\t\t\t%d\n", arr[i], Freq[i]);
}
}
}
Also, from your code:
You did not define i and n, which is required by your for loop. Also, since your for loop is for (i = 0; i<=n; i++), you have to define the value of n, which is the length of elements inputted by the user, in order to loop through the number of elements you expected.
int i, n, num;
...
...
for (i = 0; i<=num; i++)
Your scanf("i%", &num); should be scanf("%i", &num); instead.
You did not initialize your array a. You should have this line of code before assigning values to your array a. The value 20 can be adjusted by yourself depending on how many inputs are expected. Also, it can be coded in a flexible way instead of hardcoded as 20.
...
int i, num;
int count=0;
int a[20];
...
...
Lastly, it is a good practice to include the function's library before using it. In your case, you should include #include <conio.h> to use the getch() function.

How can I pass a stdin as an argument for my function?

My first program. I would like it if the user enters a word made of letters and then it uses my loop function to output mixed up even and odd characters. Currently I cannot get it to compile. Bonus points if someone can show me how to loop the users input so after it asks the size to make the array, it prompts the user that many times for an "element" or word so that the function can scramble it and output it.
#include <stdio.h>
char transform(char str[]);
int main()
{ //Declare an array and size variable
int size = 0;
char str[size];
printf("How many elements?");
scanf("%d", &size);
printf("Please type an element: ");
//Get input from user
str[0] = scanf("%s", str);
transform(str);
printf("Please type another element: ");
//Get another input from user
str[1] = scanf("%s", str);
transform(str);
//This is the loop function that I programmed
char transform(char str[]);
{
//Loop that prints even characters
for (int i = 0; str[i] != '\0'; i++)
{
if(i % 2 == 0)
{
printf("%c", str[i]);
}
} //Space between even/odd characters
printf(" ");
//Loop that prints odd characters
for (int i = 0; str[i] != '\0'; i++)
{
if(i % 2 != 0)
{
printf("%c", str[i]);
}
}
printf("\n");
return 0;
}
}
#include <stdlib.h>
#include <stdio.h>
char transform(char str[]);
int main()
{ //Declare an array and size variable
int size = 0;
printf("How many elements?");
scanf("%d", &size);
for (int i = 0; i < size; ++i)
{
printf("Please type an element: ");
char str[2048]; //declare a wide buffer to be able to store lots of chars
scanf("%s", str);
transform(str);
}
return 0;
} //end your main here, by putting closing brace
char transform(char str[]) //define transform without semicolon, and outside of main
{ //This is the loop function that I programmed
//Loop that prints even characters
for (int i = 0; str[i] != '\0'; i++)
{
if (i % 2 == 0)
printf("%c", str[i]);
} //Space between even/odd characters
printf(" ");
//Loop that prints odd characters
for (int i = 0; str[i] != '\0'; i++)
{
if (i % 2 != 0)
printf("%c", str[i]);
}
printf("\n");
return 0;
}

Keep characters in an array

I want to do a program that ask to the user to give one character, then enter... until he wants to stop by pressing enter and no caracters.
Then, the program will say: "you gave the caracters ...."
for example:
give the caracter 1: k + enter
give the caracter 2: l + enter
give the caracter 3: just enter ('\n')
result: You gave the caracters: kl
My code doesnet work because when i just press enter, nothing happen. Here is the code:
#include <stdio.h>
#define N 1000
int main() {
int i = 0;
int j = 0;
char str[N];
while (str[i] != '\n') {
printf("element number str[%d] : ", i);
scanf("%s", &str[i]);
i++;
}
printf("The string is: ");
while (j < i) {
printf("%s", str[j]);
j += 1;
}
return 0;
}
You can do it with c = getchar(); or c = fgetc(stdin) function:
#include <stdio.h>
#define N 1000
int
main ()
{
int i = 0;
int j = 0;
int c;
char str[N];
while (1)
{
c = fgetc(stdin); // or c = getchar();
if ( (c != EOF) && (c != 0x0A ) ) // 0x0A = 'nl' character
{
str[i] = (char) c;
printf ("element number str[%d]=%c \n", i, str[i++] );
}
else
{
str[i] = 0;
break;
}
}
printf ("The string is: %s", str);
return 0;
}
OUTPUT:
This is my string!
element number str[1]=T
element number str[2]=h
element number str[3]=i
element number str[4]=s
element number str[5]=
element number str[6]=i
element number str[7]=s
element number str[8]=
element number str[9]=m
element number str[10]=y
element number str[11]=
element number str[12]=s
element number str[13]=t
element number str[14]=r
element number str[15]=i
element number str[16]=n
element number str[17]=g
element number str[18]=!
The string is: This is my string!
Or you can use your original scanf("%s", &str1);
#include <stdio.h>
#define N 1000
int main ()
{
int i = 0;
int k = 0;
int c;
int len;
char str[N];
char str1[N];
scanf("%s", &str1);
len = strlen(str1);
for(k = 0; k < len; k++)
{
c = str1[k];
if ( (c != EOF) && c != '\n') // EOF will work for ^D on UNIX
{
str[i] = (char) c;
printf ("element number str[%d]=%c \n", i, str[i++] );
}
else
{
str[i] = 0;
break;
}
}
printf ("The string is: %s", str);
return 0;
}
OUTPUT:
12345
element number str[1]=1
element number str[2]=2
element number str[3]=3
element number str[4]=4
element number str[5]=5
The string is: 12345
As stated in this answer scanf will not return until you give it a string, i.e. it skips whitespace.
As suggested in the answer and in general, using fgets is the better option.
Edit: A way to accomplish what you want would look like this:
#include <stdio.h>
#define N 1000
int main() {
int i = 0;
int j = 0;
char str[N];
do {
printf("element number str[%d] : ", i);
fgets(&str[i], 3, stdin);
i++;
} while (str[i - 1] != '\n');
printf("The string is: ");
while (i > j) {
printf("%c", str[j]);
j++;
}
return 0;
}
In the fgets you use the number 3 because pressing enter gives both a newline character [/n] and a return carriage [/r].

making a word search puzzle?

I've made a program that allows you to choose the size of the grid and it allows you to enter up to 20 words. Now I have to insert the entered words horizontally into the original array using a function. The function must return a value for success and a value for failure to enter the word into the puzzle board. I need help getting started with what the actual function should look like along with the function prototype. Pseudocode would be helpful. I'm a fairly new programmer so any help is great. Thank you
#include<stdio.h>
#include<string.h>
void printmatrix(char matrix[][20],int);
void inserthor(char matrix[][20],int);
int main(void)
{
//declare variables
char matrix[20][20];
char words[20][100];
int x;
int a,b;
int i=0;
int n=0;
for (a=0;a<20;a++)
{
for (b=0;b<20;b++)
{
matrix[a][b] = '+';
}
}
while (x<10 || x>20)
{
printf("How large would you like the puzzle to be (between 10 and 20):\n");
scanf("%d",&x);
}
printmatrix(matrix,x);
//part 3
printf("Enter up to 20 words to hide in the puzzle.\n");
printf("Enter the word 'done' after your last word if entering less than 20 words.\n");
for (i = 0; i < 20; i++)
{
printf("Enter word %2d:\n", i+1);
if (scanf("%99s", words[i]) != 1 || strcmp(words[i], "done") == 0)
break;
}
n = i;
printf("%d words entered\n", n);
for (i = 0; i < n; i++)
printf("Word %2d = [%s]\n", i+1, words[i]);
return 0;
}
void printmatrix(char matrix[][20],int x)
{
int i,j;
printf("Empty Puzzle:\n");
for (i=0;i<x;i++)
{
for (j=0;j<x;j++)
{
printf(" %c ", matrix[i][j]);
}
printf("\n");
}
}
Your function prototype
void inserthor(char matrix[][20],int);
lacks the parameter with the word to be entered and the value to be returned. You could use
char *inserthor(char matrix[][20], int order, char *word)
{
int i, j, l = strlen(word);
for (i = 0; i < order; ++i)
for (j = 0; j <= order-l; ++j)
if (matrix[i][j] == '+') return memcpy(&matrix[i][j], word, l);
return NULL;
}
which returns the address of the inserted word for success and NULL for failure.

Count all character occurrences in a text file

The following code snippet is intended to count all the symbols met in a file after text is entered, next step is counting the occurrences of all characters (For instance 'a' met 3 times, 'b' 0 times etc.). However when I compile the loop goes infinite and the counting is always 0. My question is if it could be fixed or rewritten in another way.
char type, c, text[100]; counts[100];
int count=0, i;
while((type=getchar())!=EOF) {
fputc(type, f); count++;
}
printf("Symbols found: %d", count-1);
rewind(f);
while(fscanf(f, "%s", &text)) {
for (i = 0; i < strlen(text); i++) {
counts[(text[i])]++;
printf("The %d. character has %d occurrences.\n", i, counts[i]);
}
}
You can build your histogram as you read the input. The return value from getchar() is an int, not a char, since it has to represent EOF in addition to the 256 char values. Once the histogram has been built, you can iterate over the buckets and print them. Here, I have assumed that all 256 char values are possible, and included code to display unprintable characters in hex notation.
#include <stdio.h>
#include <ctype.h>
#include <string.h>
int main(int argc, char **argv)
{
int c;
int i;
int histogram[256];
int total;
memset(histogram, 0, sizeof(histogram));
total = 0;
while ((c = getchar()) != EOF) {
histogram[c]++;
total++;
}
printf("Symbols found: %d\n", total);
for (i = 0; i < 256; i++) {
if (histogram[i]) {
char repr[5];
sprintf(repr, isprint(i) ? "%c" : "\\x%02x", i);
printf("The '%s'. character has %d occurrences.\n", repr, histogram[i]);
}
}
return 0;
}
Your for loop scans the string with variable i being an index to the character tested, but your printf says i is a symbol accounted.
You should separate counting and printing results:
char * ptr;
while(fscanf(f, "%s", text))
for (ptr = text; * ptr != 0; ptr++)
counts[ (unsigned char)*ptr ]++;
for( i = 0; i < 256; i++)
printf("The %d. character has %d occurrences.\n", i, counts[i]);
Don't forget to declare count[ 256] and note that scanf gets text, not `&text~as a destination.

Resources