I'm fairly new to C, I have an encoding rule that is basicaly, if there is Z in the string, we add another Z, if there are no Z, just repeat
Example
input: STZCK -> output: STZZCK
I managed to add another Z but just at the end of string, I have to add after the found one.
I tried
char * encoding (char * str){
int size = strlen(str);
for(int i=0; i<size; i++){
if(str[i] == 'Z'){
char ch = 'Z';
strncat(str, &ch, 1);
}
else if(str[i] != 'Z'){
str[i] = str[i];
}
}
return str;
}
Thanks in advance
I would do it like this
char* encoding(char* str) {
int len = strlen(str);
// count Z to see how much bigger the result will be
int countZ = 0;
for (int i = 0; i < len; i++) {
if (str[i] == 'Z') countZ++;
}
// make return string large enough
char* retStr = malloc(len + countZ + 1);
// copy old to new with added Zs
int retOff = 0;
for (int i = 0; i < len; i++) {
retStr[retOff++] = str[i];
if (str[i] == 'Z') retStr[retOff++] = 'Z';
}
retStr[retOff] = 0;
return retStr;
}
You need to clarify if your function receives a buffer with enough space for the output string or if your function needs to dynamically allocate memory.
In the first case you can do it like this:
#include <stdio.h>
void encode(char *d, const char *s)
{
do {
*d++ = *s;
if (*s == 'Z') {
*d++ = *s;
}
} while (*s++ != 0);
}
int main(void) {
char *src[] = {"STZCK", "HELLO", "ZZ", "", NULL};
char dst[256];
for (char **s = src; *s != NULL; ++s) {
encode(dst, *s);
printf("%s -> %s\n", *s, dst);
}
return 0;
}
Related
I'm learning C now
I need to make a program that remove char that I'll input from string. I've seen an algorithm and I write this code
#define MAX_LEN 200
int main()
{
char str[MAX_LEN];
char rem;
int i = 0;
printf("Enter the setence:");
gets(str);
printf("\nEnter the char to remove");
rem = getchar();
char* pDest = str;
char* pS= str;
printf("sent:\n%s", str);
while (str[i]!='\0'){
if (*pDest != rem) {
*pDest = *pS;
pDest++;
pS++;
}
else if (*pDest == rem) {
pS++;
}
i++;
}
*pDest = '\0';
while (str[i] != '\0') {
printf("number%d", i);
putchar(str[i]);
printf("\n");
i++;
}
}
But it returns nothing, like the value str gets, i think \0 and retuns nothing.
May you help me to find the problem?
Use functions!!
If dest is NULL then this function will modify the string str otherwise, it will place the string with removed ch in dest.
It returns reference to the string with removed character.
char *removeChar(char *dest, char *str, const char ch)
{
char *head = dest ? dest : str, *tail = str;
if(str)
{
while(*tail)
{
if(*tail == ch) tail++;
else *head++ = *tail++;
}
*head = 0;
}
return dest ? dest : str;
}
int main(void)
{
char str[] = "ssHeslsslsos sWossrlssd!ss";
printf("Removal of 's' : `%s`\n", removeChar(NULL, str, 's'));
}
It would be easier to use array style indexing to go through the string. For example use str[i] = str[i + 1] instead of *pstr = *other_pstr. I leave this incomplete method, since this looks like homework.
int main()
{
char str[] = "0123456789";
char ch = '3';
for (int i = 0, len = strlen(str); i < len; i++)
if (str[i] == ch)
{
for (int k = i; k < len; k++)
{
//Todo: shift the characters to left
//Hint, it's one line
}
len--;
}
printf("%s\n", str);
return 0;
}
I just added new char array char dest[MAX_LEN] that store string with deleted symbols:
#define MAX_LEN 200
int main()
{
char str[MAX_LEN];
char rem;
int i = 0;
printf("Enter the setence:");
gets(str);
printf("\nEnter the char to remove");
rem = getchar();
char dest[MAX_LEN] = "\0";
char* pDest = dest;
char* pS = str;
printf("sent:\n%s", str);
while (str[i]!='\0')
{
if (*pS != rem)
{
*pDest = *pS;
pDest++;
pS++;
}
else if (*pS == rem)
{
pS++;
}
i++;
}
i = 0;
printf("\nres:\n %s \n", dest);
while (dest[i] != '\0') {
printf("number%d", i);
putchar(dest[i]);
printf("\n");
i++;
}
}
I have a char array containing a number.
char number[] = "12000000"
I need to have a function to insert a divider in every 3 digits. Like:
char result[] = "12,000,000"
My function accepts the number as a char pointer and it needs to return result as a char pointer too.
char* insert_divider(char* number) {
some magic;
return result;
}
I have no idea of working with pointers. Thanks.
Here you have a function that adds char c every num characters starting from the end. You need to make sure that the string buffer is long enough to accommodate the amended string.
char *addEvery(char *str, char c, unsigned num)
{
char *end = str;
if(str && *str && num)
{
size_t count = 1;
while(*(end)) end++;
while(end != str)
{
end--;
count++;
if(!(count % (num + 1)) && str != end)
{
memmove(end + 1, end, count);
*end = c;
count++;
}
}
}
return str;
}
int main(void)
{
char str[100] = "120000000000";
printf("%s", addEvery(str,',',3));
}
I came up with this piece of code:
char *result;
result = (char*) malloc(15);
int len= strlen(input);
uint8_t cursor= 0;
for(int i = 0; i < len; i++) {
if ((len- i) > 0 && (len- i) % 3 == 0) {
result[i + cursor] = ',';
cursor++;
}
result[i + cursor] = input[i];
}
result[len+ cursor] = '\0';
Thanks everyone for help and advice.
Here is another way to do it:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
char* insert_divider(char* number, size_t length) {
int j = length + length/3; // every 3 digits a ',' will be inserted
char *out = (char*)malloc(j + 1);
out[j--] = '\0';
for (int i = length - 1, k = 1; i >= 0; i--, k++) {
out[j--] = number[i];
if ((k%3) == 0) {
out[j--] = ',';
}
}
return out;
}
int main(){
char number[] = "12000000";
char *outNumber = insert_divider(number, strlen(number));
printf("%s", outNumber);
free(outNumber);
return 0;
}
Please, help with the code.
Requirement:
Write a function my_union that takes two strings and returns, without doubles, the characters that appear in either one of the strings.
Example:
Input: "zpadinton" && "paqefwtdjetyiytjneytjoeyjnejeyj"
Output: "zpadintoqefwjy"
My code:
#include <stdbool.h>
#include <stdio.h>
#include <string.h>
char *my_union(char *a, char *b) {
char *str;
// Algorithm for excluding nonunique characters from string a(given in
// parameters).
str[0] = a[0];
int k = 1;
str[k] = '\0';
for (int i = 1; a[i] != '\0'; i++) {
bool is = true;
for (int j = 0; str[j] != '\0'; j++) {
if (str[j] == a[i]) {
is = false;
break;
}
}
if (is) {
str[k] = a[i];
k++;
str[k] = '\0';
}
} // In this case we are excluding excess character 'n' from "zpadinton", so
// str is equal to "zpadinto".
// Algorithm for adding unique characters from array b(given in parameters)
// into str.
for (int i = 0; b[i] != '\0'; i++) {
bool is = true;
for (int j = 0; str[j] != '\0'; j++) {
if (str[j] == b[i]) {
is = false;
break;
}
}
if (is) {
strncat(str, &b[i], 1);
}
}
return str;
}
The first algorithm is almost identical with second, but it doesn't work(. Mb I messed up with memory, give some advice, pls.
If you mean, get the unique characters from two strings and store them into a new string, try this code ;
First, you must allocate a memory for str. In your code, str is not pointing allocated memory location, so you will probably get segmentation fault.
int contains(const char * str,char c)
{
for (int i = 0; i < strlen(str); ++i)
if(str[i] == c)
return 1;
return 0;
}
char * my_union(char *a, char*b)
{
char * res = (char*)malloc(sizeof(char)*(strlen(a) + strlen(b)));
int pushed = 0;
for (int i = 0; i < strlen(a); ++i)
{
if(!contains(res,a[i])){
res[pushed] = a[i];
pushed++;
}
}
for (int i = 0; i < strlen(b); ++i)
{
if(!contains(res,b[i])){
res[pushed] = b[i];
pushed++;
}
}
return res;
}
int main(int argc, char const *argv[])
{
char string1[9] = "abcdefgh";
char string2[9] = "abegzygj";
char * result = my_union(string1,string2);
printf("%s\n", result);
return 0;
}
Also, do not forget the free the return value of my_union after you done with it.
#include <stdio.h>
#include <string.h>
#define SIZE 40
int main(void)
{
char buffer1[SIZE] = "computer program";
char *ptr;
int ch = 'p', j = 0, i;
for (i = 0; i<strlen(buffer1); i++)
{
ptr = strchr(buffer1[i], ch);
if (ptr != 0) j++;
printf(" %d ", j);
}
}
I want to count how many times a character occurs in a string.
In my program I chose the character 'p'.
I know Pascal, I am learning C now. In pascal is a function called Pos(x,y) which is searching for x in y. Is something familiar to this? I think what I used here is not.
The function signature of strchr is
char *strchr(const char *s, int c);
You need to pass a char* but you have passed a char. This is wrong.
You have used the strlen in loop - making it inefficient. Just calculate the length of the string once and then iterate over it.
char *t = buffer;
while(t!= NULL)
{
t = strchr(t, ch);
if( t ) {
t++;
occurences++;
}
}
And without using standard library functions you can simply loop over the char array.
size_t len = strlen(buffer);
for(size_t i = 0; i < len; i++){
if( ch == buffer[i]) occurences++;
}
Or alternatively without using strlen
char *p = buffer;
while(*p){
if( *p == ch ){
occurences++;
}
p++;
}
Or
for(char *p = buffer; *p; occurences += *p++ == ch);
Try this example :
int main()
{
char buffer1[1000] = "computer program";
char ch = 'p';
int i, frequency = 0;
for(i = 0; buffer1[i] != '\0'; ++i)
{
if(ch == buffer1[i])
++frequency;
}
printf("Frequency of %c = %d", ch, frequency);
return 0;
}
I came across a interview question that asked to remove the repeated char from a given string, in-place.
So if the input was "hi there" the output expected was "hi ter". It was also told to consider only alphabetic repititions and all the
alphabets were lower case. I came up with the following program. I have comments to make my logic clear. But the program does not work as expectd for some inputs. If the input is "hii" it works, but if its "hi there" it fails. Please help.
#include <stdio.h>
int main()
{
char str[] = "programming is really cool"; // original string.
char hash[26] = {0}; // hash table.
int i,j; // loop counter.
// iterate through the input string char by char.
for(i=0,j=0;str[i];)
{
// if the char is not hashed.
if(!hash[str[i] - 'a'])
{
// hash it.
hash[str[i] - 'a'] = 1;
// copy the char at index i to index j.
str[j++] = str[i++];
}
else
{
// move to next char of the original string.
// do not increment j, so that later we can over-write the repeated char.
i++;
}
}
// add a null char.
str[j] = 0;
// print it.
printf("%s\n",str); // "progamin s ely c" expected.
return 0;
}
when str[i] is a non-alphabet, say a space and when you do:
hash[str[i] - 'a']
your program can blow.
ASCII value of space is 32 and that of a is 97 so you are effectively accessing array hash with a negative index.
To solve this you can ignore non-alphabets by doing :
if(! isalpha(str[i]) {
str[j++] = str[i++]; // copy the char.
continue; // ignore rest of the loop.
}
This is going to break on any space characters (or anything else outside the range 'a'..'z') because you are accessing beyond the bounds of your hash array.
void striprepeatedchars(char *str)
{
int seen[UCHAR_MAX + 1];
char *c, *n;
memset(seen, 0, sizeof(seen));
c = n = str;
while (*n != '\0') {
if (!isalpha(*n) || !seen[(unsigned char) *n]) {
*c = *n;
seen[(unsigned char) *n]++;
c++;
}
n++;
}
*c = '\0';
}
This is code golf, right?
d(s){char*i=s,*o=s;for(;*i;++i)!memchr(s,*i,o-s)?*o++=*i:0;*o=0;}
...
// iterate through the input string char by char.
for(i=0,j=0;str[i];)
{
if (str[i] == ' ')
{
str[j++] = str[i++];
continue;
}
// if the char is not hashed.
if(!hash[str[i] - 'a'])
{
...
#include <stdio.h>
#include <string.h>
int hash[26] = {0};
static int in_valid_range (char c);
static int get_hash_code (char c);
static char *
remove_repeated_char (char *s)
{
size_t len = strlen (s);
size_t i, j = 0;
for (i = 0; i < len; ++i)
{
if (in_valid_range (s[i]))
{
int h = get_hash_code (s[i]);
if (!hash[h])
{
s[j++] = s[i];
hash[h] = 1;
}
}
else
{
s[j++] = s[i];
}
}
s[j] = 0;
return s;
}
int
main (int argc, char **argv)
{
printf ("%s\n", remove_repeated_char (argv[1]));
return 0;
}
static int
in_valid_range (char c)
{
return (c >= 'a' && c <= 'z');
}
static int
get_hash_code (char c)
{
return (int) (c - 'a');
}
char *s;
int i = 0;
for (i = 0; s[i]; i++)
{
int j;
int gap = 0;
for (j = i + 1; s[j]; j++)
{
if (gap > 0)
s[j] = s[j + gap];
if (!s[j])
break;
while (s[i] == s[j])
{
s[j] = s[j + gap + 1];
gap++;
}
}
}