I am writing a program to count the occurrence of '2' followed by '1' in a sting.
I dynamically allocated string
Code is:
#include <stdio.h>
#include<stdlib.h>
int penalty_shoot(char* s){
int count=0,i=0;
while(s[i]!='\0'){
if(s[i]=='2')
if(s[i+1]=='1')
count++;
i++;
}
return count;
}
int main() {
int t;
int i=0;
scanf("%d",&t); //t is for number of test cases.
while(t--){
char *str, c;
str = (char*)malloc(1*sizeof(char));
while(c = getc(stdin),c!='\n')
{
str[i] = c;
i++;
str=realloc(str,i*sizeof(char));
}
str[i] ='\0';
printf("%s\n",str);
printf("%d\n",penalty_shoot(str));
free(str);
str=NULL;
i=0;
}
return 0;
}
Input is :
3
101201212110
10101
2120
I am facing 2 problems:
1) I feel the dynamic allocation is not working fine.I wrote the code for dynamic allocation seeing various codes on stackoverflow . (Can anyone suggest some changes.)
2) The code isn't reading '2120' as the 3rd input.
(why is it so ?)
Three errors:
Not checking for EOF:
Change while(c = getc(stdin),c!='\n') to while(c=getc(stdin),c!='\n'&&c!=EOF)
Reallocating with the wrong number of bytes:
Change str=realloc(str,i*sizeof(char)); to str=realloc(str,(i+1)*sizeof(char));
After taking one character input we increment i (i++), so the next character will be stored at the ith position. Now, in order to store the character at ith position, the length of the character array must be i+1. So, we realloc with i+1.
Just for the sake of brevity, as suggested by Basile, you
might as well do this:
Change str=realloc(str,(i+1)*sizeof(char)); to str=realloc(str,i+1);
Why? Because sizeof char is 1 byte
Not consuming the \n after inputting t:
Change scanf("%d",&t); to scanf("%d ",&t); or scanf("%d\n",&t);
scanf("%d ",&t); or scanf("%d\n",&t);.
Either of them works. Why, you ask? Read this explanation taken from another SO answer here:
An \n - or any whitespace character - in the format string consumes
an entire (possibly empty) sequence of whitespace characters in the
input. So the scanf only returns when it encounters the next
non-whitespace character, or the end of the input stream.
Tested here.
you can use scanf("%d ", &t); when user input to test
then just before second while loop, which condition should be c != '\n' write c = getchar();
and then make sure you create a char variable, i called mine clear, that receives 0 so when you loop after initiating your string you write c = clear; and under it c = getchar() again. and when you use realloc make sure you make it bigger by (i+1) since char is only the size of 1 byte.
we create the clear variable in order to clear the buffer.
it worked for me. make sure you insert the string all at once.
Related
The assignment asks to print out the number of times a chosen character appears in an input (no length limit) string. I wanted to solve it by only using do or do-while loops, and after a bit of googling I found this code (source: https://www.tutorialgateway.org/c-program-to-count-all-occurrence-of-a-character-in-a-string/.).
I get the gist of it, but there are many things I still haven't covered, such as the meaning of str[i], the meaning of the variable ch, and kind of how the structure works. How can I interpret it piece by piece? Or if there's any simpler way, how can I work on it? I'm a beginner and I fear this is much easier than expected but I don't have the base to move on, thank you
#include <stdio.h>
#include <string.h>
int main() {
char str[10], ch;
int i, Count;
i = Count = 0;
printf("\n Please Enter any String : ");
gets(str);
printf("\n Please Enter the Character that you want to Search for : ");
scanf("%c", &ch);
while (str[i] != '\0') {
if (str[i] == ch) {
Count++;
}
i++;
}
printf("\n The Total Number of times '%c' has Occurred = %d ", ch, Count);
return 0;
}
Well i am giving an easy example regarding that problem with a proper explanation. Hope you might understand.
char is a datatype which will accepts character type of variable. Here str[100] will be an array of length 100, where we will store our search example. ch is a character type variable where we will store the character for which we will find the concurrence.
i and count are integer variables where i will be loop variable and the count will keep count of the concurrence.
after taking the text string using puts function we are storing it in the str[100] array.
then we are taking the search letter and stored it in ch.
we are now running for loop from 0 to the length of the string we have given.
strlen() function returning us the length.
now str[i] will search from i=0 to the length size of the string. each time loop will go forward one by one letter and compare the letter with the letter inside ch.
if match found then we will increase the count value.
after the ending of the loop count will be the result of the concurrency.
reference: Concurrency of a letter in a string
#include <stdio.h>
#include <string.h>
int main()
{
char str[100], ch;
int i, Count;
Count = 0;
printf("\n Please Enter any String : ");
gets(str);
printf("\n Please Enter the Character that you want to Search for : ");
scanf("%c", &ch);
for(i = 0; i <= strlen(str); i++)
{
if(str[i] == ch)
{
Count++;
}
}
printf("\n The Total Number of times '%c' has Occured = %d ", ch, Count);
return 0;
}
The code has the following parts:
1)The header files: These contain predefined functions like scanf() which you use in your program
2)The main function: Here you are performing your character count. Usually, this function contains the driver code for the program
ch is a variable for the character you want to count in your string. You are taking this as input from the user using scanf()
str is the string you are performing your count operation in. It is also taken as input.
str[i] is used to denote the index i in the string. For example in "test", the index of 's' will be 2 as it is the 3rd character and your index starts from 0.
Final note: I recommend going through the basic syntax and usage of arrays if you do not know indexing. Also, as someone commented, do not use gets(). It causes security issues in programs as user can access stack values by giving malicious inputs containing format specifiers.
Instead, use a scanf with proper format specifiers or use fgets() when accessing content from files.
I have an assignment to make which has been giving me a headache. I have an array of characters
called "word[]". It's at most 64 chars long, which is why 'i' is counting up to that number.
Now I want to count all the vowels in that array, but for some reason it's telling me that
there is one more vowel than there really is, up to a number of 9 vowels. From 9 vowels onward, it seems to be giving me the correct number of vowels.
#include<stdio.h>
#include<ctype.h>
int vowelcounter(char word[]) {
int count = 0;
for (int i=0; i<64; i++) {
word[i] = toupper(word[i]);
if (word[i]=='A' || word[i]=='E' || word[i]=='I' || word[i]=='O' || word[i] == 'U') {
count++;
}
}
return count;
}
int main() {
char word[64];
scanf(" %s", word);
printf("Number of vowels is %i\n", vowelcounter(word));
return 0;
}
I tried to have a switch case instead of a if case, but to no avail.
OP's Problem:
The user may enter a string shorter than the max length, you should iterate through the array until you find the \0 character. Or calculate it's length with strlen and use it as a stopping point.
As is, you haven't initialised the array to 0, and iterate through the whole array regardless of the length of the string. Hence undefined behaviour.
Buffer Overflow Vulnerability:
The program is vulnerable to a buffer overflow. What's stopping one to enter more than 64 characters at a time? Consider using fgets to read a whole line instead, and then parse it accordingly. It reads at most n - 1 characters and nul-terminates the string.
as a homework assignment for my computing 1 college course, my professor has given me the task of having the user input a string of characters into the terminal, taking that string, adding it into an array, then printing the array and printing the array backwards. I think that I know of a way to print the array backwards, however, I cannot come up with a way to read from the terminal and add the characters from the terminal to an array. I have tried doing the following:
char ch;
for (int i = 0; i <= 80 || str[i] == '\n'; ++i) {
scanf_s("%c", &str[i]);
}
I am wondering if someone could explain to me why this section of code does not operate as expected, and if someone could give me some other ideas to try. Thank you.
You are using scanf_s with %c specifier incorrectly.
Please take notice of compiler warnings, there is a size argument missing.
Microsoft's scanf_s is not a direct replacement for scanf.
Unlike scanf ... scanf_s ... requires the buffer size to be specified for all input parameters of type c, C, s, S, or string control sets that are enclosed in []. The buffer size in characters is passed as an additional parameter immediately following the pointer to the buffer or variable.
scanf_s("%c", &str[i], 1);
You might also want to filter out any newline which may have been left in the buffer, with
scanf_s(" %c", &str[i], 1);
notice the added space.
Why your code is showing this type of behaviour...
use scanf instead of scanf_s
the conditions you have provided in the for loop are wrong
#include <stdio.h>
int main()
{
char ch;
char str[1000];
int i;
for (i = 0; i <= 80 ; i++)
{
scanf("%c", &str[i]);
if(str[i]=='\n')
{
str[i]='\0';
break;
}
}
printf(str);
}
I could show you the same task in simple manner. I have tried to answer your question in your way. That's why it may seem complicated.
#include <stdio.h>
#define MAX 25
int main()
{
char buf[MAX];
fgets(buf, MAX, stdin);
printf("%s\n", buf);
return 0;
}
fgets- Reads until new line character encountered or maximum limit of character array.
I'm trying to set up a code that counts the whole string and doesn't stop after the first space that it finds. How do I do that?
I tried this kind of code but it just counts the first word then goes to display the number of letters in that first word.
So far this is what I have tried.
int main(){
char get[100];
int i, space=0, len=0, tot;
scanf("%s", get);
for (i=0; get[i]!='\0'; i++)
{
if (get[i] == ' ')
space++;
else
len++;
}
tot = space + len;
printf("%i", tot);
}
And
int main(){
char get[100];
int len;
scanf("%s", &get);
len = strlen(get);
printf("%i", len);
}
But would still get the same answer as the first one.
I expected that if the
input: The fox is gorgeous.
output: 19
But all I get is
input: The fox is gorgeous.
output: 3
strlen already includes spaces, since it counts the length of the string up to the terminating NUL character (zero, '\0').
Your problem is that that the %s conversion of scanf stops reading when it encounters whitespace, so your string never included it in the first place (you can verify this easily by printing out the string). (You could fix it by using different scanf conversions, but in general it's easier to get things right by reading with fgets – it also forces you to specify the buffer size, fixing the potential buffer overflow in your current code.)
The Answer by Arkku is correct in its diagnose.
However, if you wish to use scanf, you could do this:
scanf("%99[^\n]", get);
The 99 tells scanf not to read more than 99 characters, so your get buffer won't overflow. The [^\n] tells scanf to read any character until it encounters the newline character (when you hit enter).
As Chux pointed out, the code still has 2 issues.
When using scanf, it is always a good idea to check its return value, which is the number of items it could read. Also, indeed the \n remains in the input buffer when using the above syntax. So, you could do this:
if(scanf("%99[^\n]", get) == 0){
get[0] = 0; //Put in a NUL terminator if scanf read nothing
}
getchar(); //Remove the newline character from the input buffer
I will take one example to explain the concept.
main()
{
char s[20], i;
scanf("%[^\n]", &s);
while(s[i] != '\0') {
i++;
}
printf("%d", i);
return 0;
}
i have used c language and u can loop through the ending the part of the string and you will get the length. here i have used "EDIT SET CONVESRION METHOD" to read string, you can also gets to read.
Trying to take strings as input and place it in 2d array. Why is this given code showing different behavior. The last for loop "arr[i][j]" is not printing the string.It is not even printing a character also.
Why this code does not work.only this code.Not a new way to write it
This code takes input just fine(or at least the way needed.each row a single string no white space)And when a short string is stored remaining are filled with null after carriage return. When the arr[] is passed in last for loop everything seems fine only when arr[][] is passed ,the problem arises.But again arr[][] is initialized as arr[1][0] then arr[2][0] so should not it work!
#include <stdio.h>
#include <stdbool.h>
int main(void){
int i,j,m;
scanf("%d",&m);
char arr[m][50];
for(i=0;i<m;i++){
for(j=0;j<50;j++){
printf("please enter a string");
scanf("%s",&arr[i][j]);
/*j is always 0. arr[i] takes string without space and store ending with null*/
break;
}
}
//Everything fine upto this,including storing a small continuous string in arr[i](where i<50) and null terminating*/
for(i=0;i<m;i++){
for(j=0;j<50;j++){
printf("%s\n",arr[i][j]);
break;
}
}
}
You program has several issues, like using wrong format specifier:
scanf("%s",&arr[i][j]);
arr[i][j] is a character and you are using %s format specifier. If you want your program should take string as input, you just need to do:
scanf("%s",arr[i]);
Since, you have given the size 50 characters, put a restriction in scanf() to not to read more than 49 characters (the remain one character space is for null terminating character) like this:
scanf("%49s",arr[i]);
^^
Beware with this, it does not discard the remaining input from input stream when the input characters are more than 49 and the remaining characters will be consumed by consecutive scanf() call.
If you want to drop the extra input which wasn't consumed by scanf(), one way of doing it is to read and discard the extra input using a loop, like this:
int c;
while((c = getchar()) != '\n' && c != EOF)
/* discard the character */;
In case if you have any doubt on how this will discard the extra input, I would suggest first go through getchar().
Putting all these together, you can do:
#include <stdio.h>
int main(void){
int i,m;
scanf("%d",&m);
char arr[m][50];
for(i=0;i<m;i++){
printf("please enter a string");
scanf("%49s",arr[i]);
int c;
while((c = getchar()) != '\n' && c != EOF) // <=== This loop read the extra input characters and discard them
/* discard the character */;
}
for(i=0;i<m;i++){
printf("%s\n",arr[i]);
}
return 0;
}
EDIT
The below edit is because OP updated question and added - Why this code does not work.only this code.Not a new way to write it
Above in my answer, I have already stated that you are using wrong format specifier in the scanf(). In this part of your code:
for(i=0;i<m;i++){
for(j=0;j<50;j++){ // <====== Nested for loop
printf("please enter a string");
scanf("%s",&arr[i][j]);
// since the nested loop is supposed to run 50 times, assuming you are trying to read character by character and using %s format specifier
break;
// the nested loop will break in the first iteration itself unconditionally, do you really need nested loop here!
}
}
Check the inline comments. Hope this might give an idea of the mistakes you are doing.
Seems that you want to read string character by character using scanf(). If this is the case than make sure to take care of null terminating character because, in C, strings are actually one-dimensional array of characters terminated by a null character '\0'.
You can do:
#include <stdio.h>
void discard_extra_input() {
int c;
while((c = getchar()) != '\n' && c != EOF)
/* discard the character */;
}
int main(void){
int i,j,m;
printf ("Enter number of strings: ");
scanf("%d",&m);
discard_extra_input();
char arr[m][50];
for(i=0;i<m;i++){
printf("please enter string number %d: ", i+1);
for(j=0;j<49;j++){
scanf("%c",&arr[i][j]);
if (arr[i][j] == '\n') {
//need to add null terminating character manually
arr[i][j] = '\0';
break;
}
}
if (j==49) {
// In case where the input from user is more than 50 characters,
// need to add null terminating character manually.
arr[i][j] = '\0';
// discard the extra input when input from user is more than 50 characters.
discard_extra_input();
}
}
for(i=0;i<m;i++){
for(j=0;j<50 && arr[i][j]!='\0';j++){
printf("%c",arr[i][j]);
}
printf ("\n");
}
return 0;
}
The code is self explanatory except one thing - call to discard_extra_input() function after first input from user scanf("%d",&m);. Reason -
Look at the statement:
scanf("%c",&arr[i][j]);
the %c format specifier will consume the leftover newline character '\n' from the input stream due to the ENTER key pressed after first input by the user (number of strings input from user). Hence, in order to discard it, calling discard_extra_input() function. In the other place it has been used to discard the characters when user entered string of size more than 49.
Hope this helps.
I know the code. But looking for specific ans. Where the problem lies with the code
The problem is here:
scanf("%s",&arr[i][j]);
and here:
printf("%s", arr[i][j]);
This is the specific answer you are looking for.
%s won't do any bound checking. It adds the characters starting from the memory location arr + i * m + j to arr + i * m + j + (length of input) + 1 (one extra char for the additional null character that scanf appends). Take a sample input. Assume an arbitrary starting address for arr and do the maths.
Also consider any writes beyond the allocated space for arr leads to undefined behavior.
Similarly printf("%s", arr[i][j]); will try to start reading from the address arr[i][j] till it finds a null character. It would usually lead to crash of the code because if your string has ascii characters, the address would be too low to point to any valid user-mapped memory.
If your code is working, its mostly because you already have a UB in your scanf.
Get a pen and paper and do some dry runs
This is a pretty simple problem buddy. You've got the idea right actually, that you need to use 2d array to store strings. Just that the usage is slightly wrong.
First of all let me tell you how 2d arrays need to be used to store in c. In your 2-D array, you've got rows and columns. Say, row represented by i and columns by j, i.e, each row arr[i] contains j elements. So in your context, each row arr[i] contains each string of upto 50 chars. So scanf should be just for arr[i]. And you need to loop with for, m times to accept m strings.
Same applies to printing as well.
Here is the working code:
#include <stdio.h>
#include <stdbool.h>
int main(void){
int i,j,m;
printf("\nenter m value:");
scanf("%d",&m);
char arr[m][50];
for(i=0;i<m;i++){
printf("\nplease enter a string no %d: ", (i+1));
scanf("%s",arr[i]);
}
printf("\nthe strings are: \n");
for(i=0;i<m;i++){
printf("\n%s\n",arr[i]);
}
}
And the output in case you want to cross check:
OUTPUT:
enter m value: 3
please enter a string no 1: qwerty
please enter a string no 2: asdfgh
please enter a string no 3: zxcvbn
the strings are:
qwerty
asdfgh
zxcvbn