Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 3 years ago.
Improve this question
So i have this problem where i have to input a string of unknown size with only lowercase letters then output the number of distinct letters.this is the main program
#include <stdio.h>
#include <ctype.h>
int test(char *T);
int main(){
char T[100];int i=-1,j=0,nd=0;
do{
gets(T);
}while((test(T))==1);
do {
i++;
j=i;
do{j++;
}while ((T[i]!=T[j])||((T[j])!=""));
if (T[j]=="")
nd++;
}while (T[i+1]!="");
and this is my function test
int test(char *T){
int i=-1,s;
do {
i++;
}while (((islower(T[i])==1))||(T[i]==""));
if ((T[i]=="")&&(i!=0))
s=0;
else s=1;
return s;
}
the problem is that i get a lot of warnings "comparison between integer and pointer" everytime i compare a char of the array T and i don't knowhow to fix that.your help would be much appreciated.
Update:So i tried fixing the program following your advices and this is the new main program
#include <stdio.h>
#include <ctype.h>
int test(char *T);
int main(){
char T[100];int i=-1,j=0,nd=0;
do{
gets(T);
}while((test(T))==1);
do {
i++;
j=i;
do{j++;
}while ((T[i]!=T[j])||((T[j])!='\0'));
if (T[j]=='\0')
nd++;
}while (T[i+1]!='\0');
printf("%d",nd);}
and this is function test
int test(char *T){
int i=-1,s;
do {
i++;
}while (((islower(T[i])==1))||(T[i]=='\0'));
if ((T[i]=='\0')&&(i!=0))
s=0;
else s=1;
return s;
}
I don't get anymore warnings and the program gets compiled with no problems but after i input the string in the execution nothing happens.
You function test should return 1 if the string contains only lowercase letters, and 0 otherwise. Unfortunately, it is not doing that.
You should first test if the character is a letter and then if it's a lowercase letter. Or more efficiently, you test if the character is in the range 'a' to 'z'.
Another problem of your code is the use of do while loops which makes the code difficult to understand and executes the loop once.
Here is a better implementation of the test function:
int test(char *T){
// reject empty strings
if(T[0] == '\0')
return 0;
// reject strings containing non lowercase letter
for(int i = 0; T[i] != '\0'; i++)
if((T[i] < 'a') || (T[i] > 'z'))
return 0;
// string is not empty and contains only lowercase letters
return 1;
Counting the different letters can be made more readable by using a for loop instead of a go while loop.
int nd = 0;
for(int i = 0; T[i] != '\0'; i++) {
for(int j = 0; j < i; j++) {
if(T[j] == T[i])
break; // quit inner loop
nd++;
}
}
This code examine each letter and see if it has been seen before. It is thus different from yours.
A problem in your code is the test (T[i]!=T[j])||((T[j])!='\0'). It should be && instead of ||, and testing if the end of string is reached should be performed first. The test should be (T[j]!='\0')&&(T[i]!=T[j]).
So my code is finally working here's the final main program
#include <stdio.h>
#include <ctype.h>
int test(char *T);
int main(){
char T[100];int i,j,nd=0;
do{
gets(T);
}while((test(T))==0);
for(i = 0; T[i] != '\0'; i++) {
j=i;
do{
j++;
}while ((T[j]!='\0')&&(T[j]!=T[i]));
if (T[j]=='\0')
nd++;
}
(#chmike i used the code you posted with a little adjustment on the loop)
and for the function test i used the code that posted #chmike as well.
Huge thanks to all of you guys for the help you provided :)
Related
Closed. This question needs to be more focused. It is not currently accepting answers.
Want to improve this question? Update the question so it focuses on one problem only by editing this post.
Closed 1 year ago.
Improve this question
I am trying to make a program which prints all the possible permutations of the string "A?B?AB?" via replacing question marks with A or B. I can't understand why my code works the way it does and need help improving it.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char* rec(char s[8]){
int i=0;
for(;i<strlen(s);i++){
if(s[i]=='?'){
s[i]='A';
rec(s);
s[i]='B';
rec(s);
s[i]='?';
}
}
for(int k=0;k<strlen(s);k++){
if(s[k]=='?')
i=-1;
}
if(i!=-1)
printf("%s\n",s);
return s;
}
int main(){
char s[8]="A?B?AB?";
rec(s);
}
This should do the trick:
#include <stdio.h>
#include <string.h>
void rec(char *str, int size, int depth){
for (int i = depth; i < size; i++){
if (str[i] == '?'){
str[i] = 'A';
rec(str, size, i + 1);
str[i] = 'B';
rec(str, size, i + 1);
str[i] = '?';
return;
}
}
printf("%s\n", str);
}
int main(){
char s[8] = "A?B?AB?";
rec(s, strlen(s), 0);
}
It's much like August's solution, but I did decide to do some looping until it found the next question mark. That should avoid having too big of a callstack, which could lead to stack overflow, with really big strings. (Note: I didn't test it, so there could still be some minor problems)
You only need to look at one character at a time. Try this:
#include <stdio.h>
void PrintPermutationsFrom(int i, char s[])
{
if (s[i] == '?') {
s[i] = 'A';
PrintPermutationsFrom(i + 1, s);
s[i] = 'B';
PrintPermutationsFrom(i + 1, s);
s[i] = '?';
} else if (s[i] != '\0') {
PrintPermutationsFrom(i + 1, s);
} else {
puts(s);
}
}
void PrintPermutations(char s[])
{
PrintPermutationsFrom(0, s);
}
int main(void)
{
char s[] = "A?B?AB?";
PrintPermutations(s);
return 0;
}
I can't understand why my code works the way it does and need help improving it.
The rec code does simply too much in that after having replaced a ? with both A and B and called itself, it continues iterating over s and generates further output. That is too much because after the first found ?, the recursive invocations already have handled all following ? and generated all arrangements. To correct this, just insert a break; after s[i]='?';.
I am new to C programming and I want to display how many times did "ma" showed up for example "mamama" there is 3 "ma" but in my code when I write "mamam" it displays 3 "ma". Sorry for my bad explanation... I recently started studying C programming.
I tried with do-while but I end up screwing even more..
#include <string.h>
#include <stdio.h>
int main()
{
char a[81];
int i, j;
int zbroj=0;
i=0;
fgets(a,81,stdin);
for(i=0; i<'m';i++)
{
for(j=0; j<'a';j++)
{
while(a[i] !='\0')
{
if(a[i] == 'm' || a[j]=='a')
{
zbroj++;
}
i++;
}
}
}
printf("%d\n", zbroj);
return 0;
}
Well, the end goal is when I type "mamam" program should write down there is 2 "ma".
This is what you need:
int main()
{
char a[81];
int i, j;
int zbroj=0;
i=0;
fgets(a,81,stdin);
while(a[i] !='\0') {
if(a[i] == 'm' && a[i+1]=='a')
{
zbroj++;
i++;
}
i++;
};
printf("%d\n", zbroj);
return 0;
}
Your first iteration are not needed. You need to iterate though the string and check if you have found the letter you want and then check the next one. Then increase the counter. Try to understand what is that each line of code does though. You will need it.
hint I have also changed || to &&
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 5 years ago.
Improve this question
I wrote this program to find the total number of digits from a line of text entered by user. I am having error on using getchar(). I can't seem to figure out what am I doing wrong?
#include <stdio.h>
#define MAX_SIZE 100
void main() {
char c[MAX_SIZE];
int digit, sum, i;
digit, i = 0;
printf("Enter a line of characters>");
c = getchar();
while (c[i] != '\n') {
digit = 0;
if (c [i] >= '0' && c[i] <= '9') {
digit++;
}
}
printf("%d\n", digit);
}
I will be adding all the digits I found using sum variable. but I am getting error on getchar() line. HELP??
You can enter a "line of text" without using an array.
#include <stdio.h>
#include <ctype.h>
int main(void) { // notice this signature
int c, digits = 0, sum = 0;
while((c = getchar()) != '\n' && c != EOF) {
if(isdigit(c)) {
digits++;
sum += c - '0';
}
}
printf("%d digits with sum %d\n", digits, sum);
return 0;
}
Note that c is of type int. Most of the library's character functions do not use char type.
Edit: added the sum of the digits.
Weather Vane's answer is the best and simplest answer. However, for future reference, if you want to iterate (loop through) an array, it would be easier to use a for loop. Also, your main function should return an int and should look like this: int main(). You will need to put a return 0; at the end of your main function. Here is a modified version of your program that uses a for loop to loop through the character array. I used the gets function to read a line of characters from the console. It does wait for the user to enter the string.
#include <stdio.h>
#define MAX_SIZE 100
int main()
{
char c[MAX_SIZE];
int digit = 0;
printf("Enter a line of characters>");
gets(c);
for (int i = 0; i < MAX_SIZE; i++)
{
if (c[i] == '\n') break; // this line checks to see if we have reached the end of the line. If so, exit the for loop (thats what the "break" statment does.)
//if (isdigit(c[i])) // uncomment this line and comment or delete the one below to use a much easier method to check if a character is a digit.
if (c [i]>= '0' && c[i] <= '9')
{
digit++;
}
}
printf("%d\n", digit);
return 0;
}
An easy of getting the number of digits in an integer is the use of log10() function which is defined in math.h header. Consider this program
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main (int argc, const char *argv[]) {
system("clear");
unsigned int i, sum = 0, numberOfDigits;
puts("Enter a number");
scanf("%u", &i);
numberOfDigits = (int) (log10(x) + 1);
system("clear");
while(i != 0) {
sum += (i % 10);
i /= 10;
}
fprintf(stdout, "The sum is %i\n", sum);
fflush(stdin);
return 0;
}
Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 2 years ago.
Improve this question
I am very new to programming and i wonder if there is a way to print out the first word of a string with gets() in C?
void printFirstWord(char string[])
{
int i;
for(i = 0; i < (string[i] != '\0'); i++)
{
if(isalpha(string[i]))
printf("%c", string[i]);
}
}
int main()
{
char string[MAX];
printf("Type in a scentence");
gets(string);
printFirstWord(string);
return 0;
}
This is the function that i have written and called in main right now. Is it because i have isalpha in the function?
In your implementation, you might add the following line in the loop:
if (string[i] == ' ')
break;
also, fix your loop parameters e.g. like this:
for (i = 0; i < strlen(string); i++)
Overall implementation in you way will be as below.
Consider choosing another design according to comments you got, e.g. not using gets.
void printFirstWord(char string[])
{
int i;
for (i = 0; i < strlen(string); i++)
{
if (isalpha(string[i]))
printf("%c", string[i]);
if (string[i] == ' ')
break;
}
}
int main()
{
#define MAX 100
char string[MAX];
printf("Type in a scentence\n");
gets_s(string, MAX);
printFirstWord(string);
getchar();
return 0;
}
I just found a way with isblank(); function, hope it helps to anybody :)
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main (){
int length, number, counter, position;
char name[50];
printf("Please type your complete name:\n");
gets(name);
//strlen();
//Returns the length of the given null-terminated byte string, that is, the number of characters in a character array
length=strlen(name);
//Counts each position until it finds a space
for(counter=0;counter<length;counter++)
{
if(isblank(name[counter]))
position=counter;
}
//Prints each character until the counter reaches the position number given by the counter variable
printf("\nThe first word you typed is: ");
for(number=0; number<=position; number++){
printf("%c", name[number]);
}
}
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 7 years ago.
Improve this question
I am writing a program of string pallindrome, code is compiling successfully but on running it accepting the string but nothing after that, the output window stays on hold with cursor blinking, help me what is wrong with this code.
I am using dev-c++
gets(ch); // the program stops here
p=ch;
while(ch!='\0')
{ p++;
size++;
}
for(i=0;i<20;i++)
{
if(c[i]==c[j])
printf("string is pallindrome");
else printf("string is not pallindrome");
}
getch();
return 0;
}
Here is the problem:
while(ch!='\0')
ch is a char array, and you are comparing it with a single char.
Also, size is not initialised.
I would suggest something like this:
size=0;
while(ch[size]!='\0')
{ p++;
size++;
}
or, using the pointer method:
while(*p!=0)
{
p++;
size++;
}
Also, instead of printing inside the for loop (which would make it print several times), use a flag variable.
You only need one loop, for example while (i < i).
Look at this example that will do the job:
#include <stdio.h>
#include <string.h>
/* in c99 use <stdbool.h> instead*/
typedef int bool;
#define true 1;
#define false 0;
int main(void)
{
char ch[20];
puts("enter the string: ");
gets(ch);
size_t size = strlen(ch);
bool pallindrome = true;
int j = size-1;
int i = 0;
while (i < j)
{
if(ch[i] != ch[j]) {
pallindrome = false;
break;
}
++i;
--j;
}
if (pallindrome)
printf("\"%s\" is pallindrome\n", ch);
else
printf("\"%s\" is not pallindrome\n", ch);
getch();
return 0;
}