#include <stdio.h>
int count(char str[100])
{
int i=0;
while(str[i]!=0)
{
i++;
}
return i;
}
int main() {
char palv[100] = "vinirj";
int i = (count(palv) - 2);
int m = count(palv);
scanf("%d", &n);
if(m<=n && m>=1)
{
printf("%s", palv);
}
else
{
printf("%c", palv[0]);
printf("%d", i);
printf("%c", palv[m-1]);
}
return 0;
}
i was trying to solve the problem "Way Too Long Words" exercise 71A and managed to solve it at least in my personal compiler poi when I submit it on the codeforces page directly in the compiler of the wrong answer page it itself, I experienced the sample inputs given by the exercise and give exactly the same output request. Where did I go wrong?
Related
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 &&
I'm trying to write a program that gets a string, and a number, and calculates the length of it and shifting all the elents right.
I have 2 errors:
1.assignment makes pointer from integer without a cast.
2.assignment makes integer from pointer without a cast.
#include <stdio.h>
#include <string.h>
#define N 10
int myStrlen(char*);
void shiftRight(char*, int);
int main() {
char str[N] = {0};
int num = 0;
int len;
/* input of the string */
scanf("%s",str);
scanf("%d",&num);
len=myStrlen(str);
if(num>=0) {
shiftRight(str, num);
printf("%s\n",str);
}
else
{
printf("%s\n", str);
}
return 0;
}
int myStrlen(char*str)
{
int my_len=0;
while (str[my_len] != '\0')
{
my_len++;
}
return my_len;
}
void shiftRight(char* str, int num)
{
int i;
char* j;
int count;
j=(str[N-1]);
for(count=0;count<num;count++)
{
for(i=N-1;i>0;--i)
{
str[i]=str[i-1];
}
str[0]=j;
}
}
Your answers are welcome,anf if you anything wrong with this code,please mention it.
As your compiler will have told you, pointer from integer without a cast is at
j=(str[N-1]);
And integer from pointer is at
str[0]=j;
You should have declared j as char j;
But now when i run it, and typing lets say ball as a string and 1 to
be a number, i get nothing from the program instead of getting "lbal"
You have all the correct elements but that's not enough. Writing a program is telling a story, you need to set the scene, describe what happens along the way and conclude your narrative. A story with elements out of order is nonsense, as is a program.
Specific issues with your code: you're saving of the last character (to restore it to the beginning of the string) is in the wrong place; you're using the allocation of the string when you should be using it's length (and conveniently, you have a function for that!); this is really more of a rotation than a shift; use the most descriptive variable names you can, not the shortest you can get away with; pick one indentation style and stick with it -- it can change between programs you write but shouldn't change within an individual program.
Below is a rework of your code addressing some of the issues above:
#include <stdio.h>
#define STRING_SIZE 10
int myStrlen(char *string)
{
int length = 0;
while (string[length] != '\0')
{
length++;
}
return length;
}
void rotateRight(char *string, int number)
{
int length = myStrlen(string);
for (int count = 0; count < number; count++)
{
char j = string[length - 1];
for (int i = length - 1; i > 0; i--)
{
string[i] = string[i - 1];
}
string[0] = j;
}
}
int main()
{
char string[STRING_SIZE] = {0};
int number = 0;
/* input of the string */
scanf("%s", string);
scanf("%d", &number);
if (number > 0)
{
rotateRight(string, number);
printf("%s\n", string);
}
else
{
printf("%s\n", string);
}
return 0;
}
OUTPUT
% ./a.out
elephant
3
anteleph
%
This is a code that has to take an input array from the user and input the same after removing the duplicates. However, I am unsure on how to incorporate an input array in this, and right now it has the elements hardcoded. This is my first week of programming so I apologize if this is a silly question. This is the code:
#include <stdio.h>
#include <stdbool.h>
#define nelems 8
int main()
{
int l[nelems] = {1,2,3,1,4,4,5,6};
for(int m=0;m<nelems;m++)
{
bool wase = 0;
for(int n=0;n<nelems && m>n;n++)
{
if (l[m] == l[n] && m != n)
wase = 1;
}
if (wase == 0){
printf("%d\n", l[m]);
}
}
return 0;
}
Try using a for loop and scanf.
int i;
for(i=0;i<nelems;i++){
scanf("%d",&l[i]);
}
This is what you need.
#include <stdio.h>
#include <stdbool.h>
#define nelems 8
int main()
{
int i;
int l[nelems] ;
for(i=0;i<nelems;i++)
{
printf("enter %d number :",i);
scanf("%d",&l[i]);
}
for(int m=0;m<nelems;m++)
{
bool wase = 0;
for(int n=0;n<nelems && m>n;n++)
{
if (l[m] == l[n] && m != n)
wase = 1;
}
if (wase == 0){
printf("%d\n", l[m]);
}
}
return 0;
}
If you like int-type array, you can just declare another one:
int input[nelems];
and follow the user968000 advice, remembering that when you are typing the sequence in your console you have to put a white space between each number.
To avoid that, I'd rather use char-type arrays, declared as follows:
char l[nelems] = {'1', '2', '3' /*etc.*/};
char input[nelems];
Then you make a for loop, as user968000 suggested:
int i;
for(i=0;i<nelems;i++)
scanf("%c", &input[i]);
In this case you won't need the white spaces between the digits. Notice the '&' character in the scanf function: just put it as I showed, you'll surely learn what it is in next lessons.
So you have an input array and you can handle it as you want.
No matter what the 5th input the output was clover symbol, program purpose was to align right the inputs:
EDIT
im not using scanf(%[\^n],a[i]), the output was horrible, using gets instead
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
char a[6][50];
int i;
for (i=1;i<=5;i++)
{
printf("insert name%d :\n",i);
gets(a[i]);
}
for (i=1;i<=5;i++)
{
printf("%d%25s\n",i,a[i]);
}
return 0;
}
Example output:
1 far cry
2 iron man
3 new super mario
4 program
5 "clover"
Using your code, I got strange chars on output as you mentioned.
So, I made some fixes on your code and now I think it works properly.
#include <unistd.h>
int main(void)
{
char a[6][50];
int i, r;
for (i = 1; i <= 5; i++) {
printf("insert name%d :\n", i);
r = read(STDIN_FILENO, a[i], 49);
a[i][r] = '\0';
}
for (i = 1; i <= 5; i++) {
printf("%d%25s\n", i, a[i]);
}
return 0;
}
Avoid to use deprecated functions in your code.
You need to ensure that after reading from stdin, your buffer is going to get the null terminator right after the last position written in buffer.
I am trying to make a program that converts a hex string into decimal. However I am having an issue assigning a returned integer value from the findLength function. Going by the printf statements I can tell that findLength(theString) will yield the correct value however length is showing a value of 0 despite the fact that I have length = findlength(theString).
This isn't a homework problem, I'm just absolutely stumped as to why this simple assignment isn't working. I've already declared length so I know that's not the issue. I'm also getting no compiler messages. Any help would be greatly appreciated.
Edit: I know convert doesn't do anything useful and the for loop needs to be fixed however that shouldn't be effecting the findLength return right?
Second Edit:
I've always submitted a string of '324' to be tested.
#include <stdio.h>
int convert(char s[], int theLength);
int findLength(char s[]);
int main(){
char theString[100];
int result;
int i;
int length;
printf("%s","Hello, please enter a string below. Press enter when finished.");
scanf("%s",theString); //Apparently scanf is bad but we'll learn better input methods later.
//For my tests I submitted a string of '324'.
length = (findLength(theString)); //length = findLength('324')
printf("%d",findLength(theString)); //yields 3
printf("%d",length); //yields value of 0 always.
result = convert(theString, length);
printf("%d\n result is",result);
return 0;
} //End of main
int convert(char s[], int theLength){ //This function will eventually converts a string of hex into ints. As of now it does nothing useful.
int i;
int sum;
for(i = theLength; i=0; i--){
sum = sum + s[i];
printf("%d\n",sum);
}
return sum;
} //End of convert
int findLength(char s[]){
int i;
for(i = 0; s[i]!='\0'; ++i){
}
return(i);
} //End of findLength
The variable length is storing the correct value. I think what has you confused is how you've laid out your printf statements. If you were to try something like the below it would be much easier to see that your code works.
#include <stdio.h>
int findLength(char s[]);
int main(){
char theString[100];
int result;
int i;
int length;
printf("Hello, please enter a string below. Press enter when finished.\n");
scanf("%s",theString);
length = (findLength(theString));
printf("findLength(theString) = %d\n",findLength(theString));
printf("length = %d\n",length);
return 0;
}
int findLength(char s[]){
int i;
for(i = 0; s[i]!='\0'; ++i){
}
return(i);
}
Just to clarify in your post you have
printf("%d",findLength(theString));
printf("%d",length);
printf("%d\n result is",result);
Note the \n before the %d in the last printf statement. This is 0 because your convert function needs to be fixed and this is the value of result NOT length.