#include <stdio.h>
#define MAX 1000
void any(char s1[], char s2[], char s3[]);
int main()
{
char string1[MAX], string2[MAX], string3[MAX];
printf("Jepni stringen 1\n");
scanf("%s", &string1); //saving string 1
printf("Jepni stringen 2\n");
scanf("%s", &string2); //saving string 2
any(string1, string2, string3); /*comparing characters from string 2 to string 1 and saving the places where they are equal on third string */
printf("%d", string3[0]); //printing the first character of the third string
return 0;
}
void any(char s1[], char s2[], char s3[])
{
int i, j, k;
k = 0;
for (j = 0; j != '\0'; j++) {
for (i = 0; i != '\0'; i++) {
if (s1[i] == s2[j]) {
s3[k] = i;
j++;
k++;
}
}
}
}
I am trying to create a c program that scans 2 strings (saves them on string 1 and 2) than the program using function any will see character by character if the string 2 characters are equal with the string 1,If they are it will give the first position where they are found.In case nothing is found it displays a -1.The program asks for the first character that is equal,thats why i am printing always the first character from string 3.The program isnt working cuz it always prints -1.
Example if i put on string 1 dad
and on string 2 the character d
dhe program should display the 0 position
if i put dad on string 1
and on string 2 i put a
it should display 1.
First of all the for loop doesn't begin because the condition is that j shall be different from zero.In ASCII '\0' is zero (maybe not on all machines), so you rather want to check that s2[j] is different from zero.Same for i.
Another thing is that s3 is an array of chars, so putting s3[k]=i doesn't make it equal to '1' or '2', but to 1 or 2 (ASCII values), so you should add 48 to i (good till '9', then you'll have two digits), or print the string character per character, with the %d format specifier:
void any(char s1[], char s2[], char s3[])
{
int i, j, k;
k = 0;
for (j = 0; s2[j] != '\0'; j++) {
for (i = 0; s1[i] != '\0'; i++) {
if (s1[i] == s2[j]) {
s3[k] = i;
j++;
k++;
}
}
}
}
Maybe I'm missing some other error, try the code and run it to see if it's right (also remember to use the %d specifier to print s3).
Related
I'm having trouble with trying to manipulate 2d dynamic arrays in C. What I want to do is to store a char string in every row of the the 2d array then perform a check to see if the string contains a certain character, if so remove all occurrences then shift over the empty positions. What's actually happening is I get an exit status 1.
More about the problem, for example if I have
Enter string 1: testing
Enter string 2: apple
Enter string 3: banana
I would want the output to become
What letter? a // ask what character to search for and remove all occurences
testing
pple
bnn
Here is my full code:
#include <stdio.h>
#include <stdlib.h>
void removeOccurences2(char** letters, int strs, int size, char letter){
// Get size of array
// Shift amount says how many of the letter that we have removed so far.
int shiftAmt = 0;
// Shift array says how much we should shift each element at the end
int shiftArray[strs][size];
// The first loop to remove letters and put things the shift amount in the array
int i,j;
for(i=0;i < strs; i++){
for(j = 0; j < size - 1; j++) {
if (letters[i][j] == '\0'){
break;
}
else {
// If the letter matches
if(letter == letters[i][j]){
// Set to null terminator
letters[i][j] = '\0';
// Increase Shift amount
shiftAmt++;
// Set shift amount for this position to be 0
shiftArray[i][j] = 0;
}else{
// Set the shift amount for this letter to be equal to the current shift amount
shiftArray[i][j] = shiftAmt;
}
}
}
}
// Loop back through and shift each index the required amount
for(i = 0; i < strs; i++){
for(j = 0; j < size - 1; j++) {
// If the shift amount for this index is 0 don't do anything
if(shiftArray[i][j] == 0) continue;
// Otherwise swap
letters[i][j - shiftArray[i][j]] = letters[i][j];
letters[i][j] = '\0';
}
//now print the new string
printf("%s", letters[i]);
}
return;
}
int main() {
int strs;
char** array2;
int size;
int cnt;
int c;
char letter;
printf("How many strings do you want to enter?\n");
scanf("%d", &strs);
printf("What is the max size of the strings?\n");
scanf("%d", &size);
array2 = malloc(sizeof(char*)*strs);
cnt = 0;
while (cnt < strs) {
c = 0;
printf("Enter string %d:\n", cnt + 1);
array2[cnt] = malloc(sizeof(char)*size);
scanf("%s", array2[cnt]);
cnt += 1;
}
printf("What letter?\n");
scanf(" %c", &letter);
removeOccurences2(array2,strs,size,letter);
}
Thanks in advance!
You can remove letters from a string in place, because you can only shorten the string.
The code could simply be:
void removeOccurences2(char** letters, int strs, int size, char letter){
int i,j,k;
// loop over the array of strings
for(i=0;i < strs; i++){
// loop per string
for(j = 0, k=0; j < size; j++) {
// stop on the first null character
if (letters[i][j] == '\0'){
letters[i][k] = 0;
break;
}
// If the letter does not match, keep the letter
if(letter != letters[i][j]){
letters[i][k++] = letters[i][j];
}
}
//now print the new string
printf("%s\n", letters[i]);
}
return;
}
But you should free all the allocated arrays before returning to environment, and explicitely return 0 at the end of main.
Well, there are several issues on your program, basically you are getting segmentation fault error because you are accessing invalid memory which isn't allocated by your program. Here are some issues I found:
shiftAmt isn't reset after processing/checking each string which lead to incorrect value of shiftArray.
Values of shiftArray only set as expected for length of string but after that (values from from length of each string to size) are random numbers.
The logic to delete occurrence character is incorrect - you need to shift the whole string after the occurrence character to the left not just manipulating a single character like what you are doing.
1 & 2 cause the segmentation fault error (crash the program) because it causes this line letters[i][j - shiftArray[i][j]] = letters[i][j]; access to unexpected memory. You can take a look at my edited version of your removeOccurences2 method for reference:
int removeOccurences2(char* string, char letter) {
if(!string) return -1;
int i = 0;
while (*(string+i) != '\0') {
if (*(string+i) == letter) {
memmove(string + i, string + i + 1, strlen(string + i + 1));
string[strlen(string) - 1] = '\0'; // delete last character
}
i++;
}
return 0;
}
It's just an example and there is still some flaw in its logics waiting for you to complete. Hint: try the case: "bananaaaa123"
Happy coding!
"...if the string contains a certain character, if so remove all occurrences then shift over the empty positions."
The original string can be edited in place by incrementing two pointers initially containing the same content. The following illustrates.:
void remove_all_chars(char* str, char c)
{
char *pr = str://pointer read
char *pw = str;//pointer write
while(*pr)
{
*pw = *pr++;
pw += (*pw != c);//increment pw only if current position == c
}
*pw = '\0';//terminate to mark last position of modified string
}
This is the cleanest, simplest form I have seen for doing this task. Credit goes to this answer.
I did this program to reverse the order of the words in the give string. (And it works)
i.e. Output: sentence first the is This
However I am stuck when it comes to adding another sentence to the array.
For example I need to have an array {"This is the first sentence", "And this is the second"} producing as output: sentence first the is This , second the is this And
int main() {
char str[] = {"This is the first sentence"};
int length = strlen(str);
// Traverse string from end
int i;
for (i = length - 1; i >= 0; i--) {
if (str[i] == ' ') {
// putting the NULL character at the position of space characters for
next iteration.
str[i] = '\0';
// Start from next character
printf("%s ", &(str[i]) + 1);
}
}
// printing the last word
printf("%s", str);
return 0;
}
I am new to C so its not surprising that I got stuck even if the solution is quite easy. Any help would be appreciated! Thanks!
Since you already have the code to print the words of one string in reverse order, I would suggest making that a function which takes a single string as an argument, i.e.:
void print_words_reverse(char * const str) {
// your current code here
}
Then you can call it separately for each string:
char strings[][30] = {
"This is the first sentence",
"And this is the second"
};
for (int i = 0; i < sizeof(strings) / sizeof(*strings); ++i) {
print_words_reverse(strings[i]);
}
Note that since you are modifying the string (by replacing spaces with NUL bytes), the argument needs to be modifiable, which means you are not allowed to call it (in standard C) with a pointer to a string literal, which means you can't simply use const char *strings[] = { "first", "second" }. You could get rid of the ugly constant length (here 30) reserved for every string by making your code not modify the argument string. Or you could have a separate char array for each sentence and then use pointers to those (modifiable) strings.
First, you can try with a two-dimensional array or use an array of pointers.
Secondly, in your approach, you lose the initial value of your string, I don't know how important it is.
This is my fast approach using arrray of pointers.
#include <stdio.h>
#include <string.h>
static void print_word(const char *str)
{
for (int i = 0; str[i] && str[i] != ' '; i++)
printf("%c", str[i]);
putchar(' ');
}
int main(void)
{
int len;
const char *str[] = {"This is the first sentence",
"And this is second", NULL};
for (int i = 0; str[i]; i++) {
for (len = strlen(str[i]); len >= 0; len--) {
if (len == 0)
print_word(&str[i][len]);
else if (str[i][len] == ' ')
print_word(&str[i][len + 1]);
}
putchar('\n');
}
printf("Initial value of array of strings [%s | %s] \n", str[0], str[1]);
return 0;
}
output is:
sentence first the is This
second is this And
Initial value of array of strings [This is the first sentence | And this is second]
I suggest you using memcpy but without altering too much your code this seems to work
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define MAX_STRING_LENGTH 100
int main()
{
char *str[] = {"This is the first", "And this is the second sentence"};
const size_t NUM_STRING = sizeof(str)/sizeof(char*);
/*%z used to print size_t variables*/
printf("%zd strings found\n", NUM_STRING);
int length[2];
int i;
for (i=0; i<NUM_STRING; i++)
{
length[i] = strlen(str[i]);
}
printf("length initialized %d %d\n", length[0], length[1]);
// Traverse string from end
int j = 0;
char temp[MAX_STRING_LENGTH];
printf("\n\n");
for (j=0; j<NUM_STRING; j++)
{
/*Make sure the string respect the MAX_STRING_LENGTH limit*/
if (strlen(str[j])>MAX_STRING_LENGTH)
{
printf("ERROR: string %d exceding max string length %d defined in constant "
"MAX_STRING_LENGTH. Exiting from program.\n", j, MAX_STRING_LENGTH);
exit(1);
}
//reset temporary string
memset(temp, '\0', sizeof(temp));
//printf("temp variable reinitialized\n");
for (i = length[j] - 1; i >= 0; i--)
{
temp[i] = str[j][i];
if (str[j][i] == ' ')
{
// putting the NULL character at the position of space characters for next iteration.
temp[i] = '\0';
// Start from next character
printf("%s ", &(temp[i]) + 1);
}
}
// printing the last word
printf("%s ", temp);
}
printf("\n");
return 0;
}
I have noticed that when you try to print a null character in C, nothing will get printed.
printf("trying to print null\n");
printf("%c", '\0');
However, I am trying to print the characters in the following array one by one, up to the sixth character which is the null character.
char s[] = "Hello\0Bye";
int i;
for(i = 0; i < 7; i++) {
printf("%c", s[i]);
}
printf("\n");
I was expecting "Hello" to be printed, as since the sixth character is null nothing will be printed. However my output was: "HelloB".
It seems like printf skipped the null character and just went to the next character. I am not sure why the output is "HelloB" instead of "Hello".
Any insights would be really appreciated.
The construction '\0' is commonly used to represent the null character. Here
printf("%c", '\0');
it prints nothing.
And in the decalaration of s
char s[] = "Hello\0Bye";
when you print like
for(i = 0; i < 7; i++) {
printf("%c", s[i]);
}
printf() prints upto 0<7(h), 1<7(e)..5<7(nothing on console),6<7(B) iterations only and 6th charactar is B hence its prints HelloB.
I was expecting "Hello" to be printed ? For that you should rotate loop until \0 encountered. For e.g
for(i = 0; s[i] != '\0'; i++) { /* rotate upto \0 not 7 or random no of times */
printf("%c", s[i]);
}
Or even you no need to check s[i] != '\0'
for(i = 0; s[i]; i++) { /* loop terminates automatically when \0 encounters */
printf("%c", s[i]);
}
You can use below two options
1. size_t fwrite(const void *ptr, size_t size, size_t nmemb, FILE *stream);
2.Iterate through each character and print it.
int i; for(i=0; i<7; i++) generates 7, not 6 iterations.
Printing \0 generates no pixels on the console (character 6) and then the 7th character (B follows).
This is for Homework
I have to write a program that asks the user to enter a string, then my program would separate the even and odd values from the entered string. Here is my program.
#include <stdio.h>
#include <string.h>
int main(void) {
char *str[41];
char odd[21];
char even[21];
int i = 0;
int j = 0;
int k = 0;
printf("Enter a string (40 characters maximum): ");
scanf("%s", &str);
while (&str[i] < 41) {
if (i % 2 == 0) {
odd[j++] = *str[i];
} else {
even[k++] = *str[i];
}
i++;
}
printf("The even string is:%s\n ", even);
printf("The odd string is:%s\n ", odd);
return 0;
}
When I try and compile my program I get two warnings:
For my scanf I get "format '%s' expects arguments of type char but argument has 'char * (*)[41]". I'm not sure what this means but I assume it's because of the array initialization.
On the while loop it gives me the warning that says comparison between pointer and integer. I'm not sure what that means either and I thought it was legal in C to make that comparison.
When I compile the program, I get random characters for both the even and odd string.
Any help would be appreciated!
this declaration is wrong:
char *str[41];
you're declaring 41 uninitialized strings. You want:
char str[41];
then, scanf("%40s" , str);, no & and limit the input size (safety)
then the loop (where your while (str[i]<41) is wrong, it probably ends at once since letters start at 65 (ascii code for "A"). You wanted to test i against 41 but test str[i] against \0 instead, else you get all the garbage after nul-termination char in one of odd or even strings if the string is not exactly 40 bytes long)
while (str[i]) {
if (i % 2 == 0) {
odd[j++] = str[i];
} else {
even[k++] = str[i];
}
i++;
}
if you want to use a pointer (assignement requirement), just define str as before:
char str[41];
scan the input value on it as indicated above, then point on it:
char *p = str;
And now that you defined a pointer on a buffer, if you're required to use deference instead of index access you can do:
while (*p) { // test end of string termination
if (i % 2 == 0) { // if ((p-str) % 2 == 0) { would allow to get rid of i
odd[j++] = *p;
} else {
even[k++] = *p;
}
p++;
i++;
}
(we have to increase i for the even/odd test, or we would have to test p-str evenness)
aaaand last classical mistake (thanks to last-minute comments), even & odd aren't null terminated so the risk of getting garbage at the end when printing them, you need:
even[k] = odd[j] = '\0';
(as another answer states, check the concept of even & odd, the expected result may be the other way round)
There are multiple problems in your code:
You define an array of pointers char *str[41], not an array of char.
You should pass the array to scanf instead of its address: When passed to a function, an array decays into a pointer to its first element.
You should limit the number of characters read by scanf.
You should iterate until the end of the string, not on all elements of the array, especially with (&str[i] < 41) that compares the address of the ith element with the value 41, which is meaningless. The end of the string is the null terminator which can be tested with (str[i] != '\0').
You should read the characters from str with str[i].
You should null terminate the even and odd arrays.
Here is a modified version:
#include <stdio.h>
int main(void) {
char str[41];
char odd[21];
char even[21];
int i = 0;
int j = 0;
int k = 0;
printf("Enter a string (40 characters maximum): ");
if (scanf("%40s", str) != 1)
return 1;
while (str[i] != '\0') {
if (i % 2 == 0) {
odd[j++] = str[i];
} else {
even[k++] = str[i];
}
i++;
}
odd[j] = even[k] = '\0';
printf("The even string is: %s\n", even);
printf("The odd string is: %s\n", odd);
return 0;
}
Note that your interpretation of even and odd characters assumes 1-based offsets, ie: the first character is an odd character. This is not consistent with the C approach where an even characters would be interpreted as having en even offset from the beginning of the string, starting at 0.
Many answers all ready point out the original code`s problems.
Below are some ideas to reduce memory usage as the 2 arrays odd[], even[] are not needed.
As the "even" characters are seen, print them out.
As the "odd" characters are seen, move them to the first part of the array.
Alternative print: If code used "%.*s", the array does not need a null character termination.
#include <stdio.h>
#include <string.h>
int main(void) {
char str[41];
printf("Enter a string (40 characters maximum): ");
fflush(stdout);
if (scanf("%40s", str) == 1) {
int i;
printf("The even string is:");
for (i = 0; str[i]; i++) {
if (i % 2 == 0) {
str[i / 2] = str[i]; // copy character to an earlier part of `str[]`
} else {
putchar(str[i]);
}
}
printf("\n");
printf("The odd string is:%.*s\n ", (i + 1) / 2, str);
}
return 0;
}
or simply
printf("The even string is:");
for (int i = 0; str[i]; i++) {
if (i % 2 != 0) {
putchar(str[i]);
}
}
printf("\n");
printf("The odd string is:");
for (int i = 0; str[i]; i++) {
if (i % 2 == 0) {
putchar(str[i]);
}
}
printf("\n");
here is your solution :)
#include <stdio.h>
#include <string.h>
int main(void)
{
char str[41];
char odd[21];
char even[21];
int i = 0;
int j = 0;
int k = 0;
printf("Enter a string (40 characters maximum): ");
scanf("%s" , str);
while (i < strlen(str))
{
if (i % 2 == 0) {
odd[j++] = str[i];
} else {
even[k++] = str[i];
}
i++;
}
odd[j] = '\0';
even[k] = '\0';
printf("The even string is:%s\n " , even);
printf("The odd string is:%s\n " , odd);
return 0;
}
solved the mistake in the declaration, the scanning string value, condition of the while loop and assignment of element of array. :)
I am trying to write an application in C and I have a problem: I have some unwanted chars in my string2 array in the code part below:
#include <stdio.h>
#include <string.h>
#define max_len 100
int main() {
char string1[max_len], string2[max_len], string3[max_len], temp[max_len];
int m, n, i, j, k, min_str_len;
puts("Enter your first word: ");
gets(string1);
puts("Enter your second word: ");
gets(string2);
m = strlen(string1);
n = strlen(string2);
min_str_len = ((m < n) ? m : n);
for (i = 0; i <= m; i++) {
for (j = 0; j <= n; j++) {
for (k = 0; k <= min_str_len; k++) {
if (string1[i] == string2[j]) {
//printf("%s",string2[j]);
temp[k] = string2[j];
break;
}
...
}
}
}
So you see here a part of my code. I am assigning "placozoa" as string1, and "placement" as string2. Please see the line: //printf("%s",string2[j]); When I make this line active, I see:
I did not understand why [] and ' characters are included in string2, I was just expecting to see "plac" there in the output.. The weird thing is, I am printing string2 array right after getting from the user (under gets(string2); line), and I see it is "placement" there.
What is happening in the middle, where is my mistake?
In your print statement, printf("%s",string2[j]); use %c instead of %s,
Because %s always try to convert the printing item to string. And to do that, your compiler has added a null char at the end of it.
So what actually happening right now,
when j == 0, string2[j] is equal to 'p' as a single charcter, but as string it becomes "p\0" where \0 is the null character which means the end of a string.
So, I think, just printing as a single character using %c (character) will solve the problem.