#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char first[15];
printf("\n\tWrite here your code");
scanf("%s",first);
if()//there is "n" in the char
//change n with 1
else
//quit
return 0;
}
Replace character with character
This code will replace every occurrence of the character 'n' with the character '1'.
#include <stdio.h>
int main(void)
{
char str[15];
printf("Input: ");
scanf("%14s", str); // prevent buffer overflow
for (int i = 0; str[i]; ++i) { // iterate over str
if (str[i] == 'n')
str[i] = '1';
}
printf("Modified: %s\n", str);
return 0;
}
Replace character with string
This code will replace every occurrence of the character 'n' with the string "1+k" using the function strcat. The modified string is saved to char mod[].
#include <stdio.h>
#include <string.h>
#include <stdio.h>
int main(void)
{
char str[15];
printf("Input: ");
scanf("%14s", str); // prevent buffer overflow
char mod[] = "";
for (int i = 0; str[i]; ++i) { // iterate over str
if (str[i] == 'n') {
char *repl = "1+k";
strcat(mod, repl); // add "1+k" to mod
} else {
strncat(mod, &str[i], 1); // add str[i] to mod
}
}
printf("Modified: %s\n", mod);
return 0;
}
Such a trivial question deserves a non trivial answer
Note that this approach with branching, demonstrated by #Andy Sukowski-Bang, is - when compiled with -O3 - roughly 6.5x time slower than my approach, illustratating the efficiency of bitwise operations over branching instructions (not to mention Meltdown and Spectre vulnerabilities and its on branching if mitigations are enabled).
The following program will convert every occurence of the letter from to the designated letter.
It is able to replace all characters from 'A' to 'z' to '8' in a string of 8,286,208 bytes in only 0.07s:
#include <stdio.h>
#include <unistd.h>
#include <errno.h>
#include <string.h>
#include <time.h>
// gcc -O3 replace.c && ./a.out
void replace(char *s, char from, char to, long n) {
char eq;
while (n--) {
eq = !(*s ^ from);
*s = !eq * *s + eq * to;
s++;
}
}
void replace_branching(char *s, char from, char to, long n) {
while (n--) {
if (*s == from)
*s = to;
s++;
}
}
void read_file_to_buffer(FILE *f, long length) {
char buffer[length + 1];
fseek(f, 0, SEEK_SET);
void (*fptr[2]) (char *s, char from, char to, long n) = {replace_branching, replace};
if (fread (buffer, 1, length, f) ) {
buffer[length] = '\0';
fclose (f);
char subbuff[33];
subbuff[32] = '\0';
char from = 'A';
char to = '8';
clock_t start, end;
double cpu_time_used[2];
for (int j = 0; j < 2; j++){
start = clock();
for (int i = 0; i < 0x20 + 26; i++) {
fptr[j](buffer, from + i, to, length);
//memcpy( subbuff, &buffer[0], 32 );
//printf("Modified: %s\n", subbuff);
}
end = clock();
cpu_time_used[j] = ((double) (end - start)) / CLOCKS_PER_SEC;
printf("Time: %fs\n", cpu_time_used[j]);
}
printf("Without branching it is %fx faster\n", cpu_time_used[0] / cpu_time_used[1]);
}
}
int main(void)
{
FILE * f;
long length;
if ((f = fopen ("test.txt", "rb"))) // NB: test.txt is a file of 8,286,208 bytes
{
fseek (f, 0, SEEK_END);
read_file_to_buffer(f, ftell(f));
}
return 0;
}
/*
Time: 0.475005s
Time: 0.070859s
Without branching it is 6.703524x faster
*/
Some explanations
eq = !(*s ^ from); // eq will equal 0 if letters are same, else it will equal 1
*s = !eq * *s + eq * to; // we assign to the pointer its same old value with !eq if the character 'from' was absent, else we will assign the character 'to'.
PS: No malloc were used, I used VLA and it is bad, very bad, don't do this at home!
I have an array of strings and am trying to reverse each string in the array to see if that string is a palindrome. I am using a for loop to increment an int i (the index). However after the I call the reverse function, the value of i becomes some really large number and I cant figure out why this is happening.
#include <stdio.h>
#include <string.h>
void revString(char *dest, const char *source);
int main() {
const char *strs[] = {
"racecar",
"radar",
"hello",
"world"
};
int i;
char res[] = "";
for (i = 0; i < strlen(*strs); i++) {
printf("i is %d\n", i);
revString(&res[0], strs[i]); //reversing string
printf("i is now %d\n", i);
//comparing string and reversed string
if (strcmp(res, strs[i]) == 0) {
printf("Is a palindrome");
} else {
printf("Not a palindrome");
}
}
return 0;
}
void revString(char *dest, const char *source) {
printf("%s\n", source);
int len = strlen(source);
printf("%d\n", len);
const char *p;
char s;
for (p = (source + (len - 1)); p >= source; p--) {
s = *p;
*(dest) = s;
dest += 1;
}
*dest = '\0';
}
This is the output showing the value of i before and after the revString function is called.
i is 0
i is now 1667588961
Illegal instruction: 4
There are multiple problems in your code:
You pass a destination array char res[] = ""; that is much too small for the strings you want to reverse. It's size is 1. This causes buffer overflow, resulting in undefined behavior.
Use char res[20]; instead.
You enumerate the array of string with an incorrect upper bound. Use this instead:
for (i = 0; i < sizeof(strs) / sizeof(*strs); i++)
The termination test for the loop in revString() is incorrect too: decrementing p when is equal to source has undefined behavior, although it is unlikely to have an consequences. You can simplify this function this way:
void revString(char *dest, const char *source) {
size_t len = strlen(source);
for (size_t i = 0; i < len; i++) {
dest[i] = source[len - i - 1];
}
dest[len] = '\0';
}
Here is the resulting code:
#include <stdio.h>
#include <string.h>
void revString(char *dest, const char *source) {
size_t len = strlen(source);
for (size_t i = 0; i < len; i++) {
dest[i] = source[len - i - 1];
}
dest[len] = '\0';
}
int main(void) {
const char *strs[] = { "racecar", "radar", "hello", "world" };
char res[20];
for (size_t i = 0; i < sizeof(strs) / sizeof(*strs); i++) {
revString(res, strs[i]);
//comparing string and reversed string
if (strcmp(res, strs[i]) == 0) {
printf("Is a palindrome\n");
} else {
printf("Not a palindrome\n");
}
}
return 0;
}
Here is Final Code with some change
#include <stdio.h>
#include <string.h>
void revString(char* dest, const char* source);
int main(){
const char* strs[] = {
"racecar",
"radar",
"hello",
"world"
};
static int i;
char res[] = "";
int length = (int) sizeof(strs)/sizeof(char*);
for(i = 0; i < length; i++)
{
printf("i is %d\n", i);
revString(&res[0], strs[i]); //reversing string
printf("i is now %d\n", i);
//comparing string and reversed string
if(strcmp(res, strs[i]) == 0){
printf("Is a palindrome");
}else{
printf("Not a palindrome");
}
}
return 0;
}
void revString(char* dest, const char* source){
printf("%s\n", source);
int len = (int) strlen(source);
printf("%d\n", len);
const char* p;
char s;
for(p = (source + (len - 1)); p >= source; p--){
s = *p;
*(dest) = s;
dest += 1;
}
*dest = '\0';
}
Change 1 :-
int i; to static int i; (Reason:- i is local variable you are calling
function so when function call the value of i will remove and after
that it will assign garbage value.)
change 2 :-
strlen(*strs) to length of array (because strlen(*strs) will give the
length of first string)
I have made two functions that find a substring index and substitute that substring in the string. I'm glad I jury rigged this at all, given that similar questions previously asked were never answered/marked as closed without any help. Is there a cleaner method?
void destroy_substr(int index, int len)
{
int i;
for (i = index; i < len; i++)
{
string[i] = '~';
}
}
void find_substr_index(char* substr)
{
int i;
int j;
int k;
int count;
int len = strlen(substr);
for (i = 0; i < strlen(string); i++)
{
if (string[i] == substr[0])
{
for(j = i, k = 0; k < len; j++, k++)
{
if (string[j] == substr[k])
{
count++;
}
if (count == len)
destroy_substr((j - len + 1), len);
}
j = 0;
k = 0;
count = 0;
}
}
}
Your code seems like you're trying to re-inventing your own wheel.
By using standard C functions, which is strstr() and memset(), you can achieve the same result as you expected.
#include <stdio.h>
#include <string.h>
char string[] = "foobar foobar foobar";
char substr[] = "foo";
char replace = '~';
int main() {
int substr_size = strlen(substr);
// Make a copy of your `string` pointer.
// This is to ensure we can safely modify this pointer value, without 'touching' the original one.
char *ptr = string;
// while true (infinite loop)
while(1) {
// Find pointer to next substring
ptr = strstr(ptr, substr);
// If no substring found, then break from the loop
if(ptr == NULL) { break; }
// If found, then replace it with your character
memset(ptr, replace, substr_size);
// iIncrement our string pointer, pass replaced substring
ptr += substr_size;
}
printf("%s\n", string);
return 0;
}
How about this:
#include <stdio.h>
#include <string.h>
int main(int argc, char **argv)
{
char string[] = "HELLO hello WORLD world HELLO hello ell";
char substring[] = "ell";
int stringLength = strlen(string);
int substringLength = strlen(substring);
printf("Before: %s\n", string);
if(substringLength <= stringLength)
{
int i;
int j;
for(i = 0, j = stringLength - substringLength + 1; i < j; )
{
if(memcmp(&string[i], substring, substringLength) == 0)
{
memset(&string[i], '~', substringLength);
i += substringLength;
}
else
{
i++;
}
}
}
printf("After: %s\n", string);
return 0;
}
Key ideas are:
You only need to scan the string (stringLength - substringLength) times
You can use functions from string.h to do the comparison and to replace the substring
You can copy the new string in place. If you want to support insertion of longer strings you will need to manage memory with malloc()/realloc(). If you want to support insertion of smaller strings you'll need to advance the pointer to the beginning by the length of the replacement string, copy the rest of the string to that new location, then zero the new end of the string.
#include <stdio.h>
#include <string.h>
#include <err.h>
int main(int argc, char **argv)
{
char *str = strdup("The fox jumps the dog\n");
char *search = "fox";
char *replace = "cat";
size_t replace_len = strlen(replace);
char *begin = strstr(str, search);
if (begin == NULL)
errx(1, "substring not found");
if (strlen(begin) < replace_len)
errx(1, "replacement too long");
printf("%s", str);
memcpy(begin, replace, replace_len);
printf("%s", str);
return 0;
}
for exemple i need to invers "Paris" to "siraP"...
My main:
int main(void)
{
char w1[] = "Paris";
ReverseWord(w1);
printf("The new word is: %s",w1);
return0;
}
and my function:
void ReverseWord(char *Str)
{
int counter=0;
for(int i=0; *(Str+i)!='\0'; i++)
counter++;
int length = counter-1;
char temp[length];
for(int j=0; temp[j]=='\0'; j++)
temp[j]=Str[length-j];
}
Now I have my renverse word in temp[].
I need to put it in my pointer *Str.
How can I do it??
Thanks
If you want use temp must then your function like this
void ReverseWord(char *Str)
{
int i,j;
if(str)
{
int length=strlen(Str);
char temp[length+1];
for( j=0; j<length; j++)
temp[j]=Str[length-1-j];
temp[j]='\0';
strcpy(Str,temp);
}
}
Without using temp as follows
void ReverseWord(char *Str)
{
int end= strlen(Str)-1;
int start = 0;
while( start<end )
{
Str[start] ^= Str[end];
Str[end] ^= Str[start];
Str[start]^= Str[end];
++start;
--end;
}
}
void ReverseWord(char *Str)
{
size_t len;
char temp, *end;
len = strlen(Str);
if (len < 2)
return;
end = Str + len - 1;
while (end > Str)
{
temp = *end;
*end-- = *Str;
*Str++ = temp;
}
}
One more option, this time with dangerous malloc(3).
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
char *rev(char s[]) {
char *buf = (char *)malloc(sizeof(char) * strlen(s));
int i, j;
if(buf != NULL)
for(i = 0, j = strlen(s) - 1; j >= 0; i++, j--)
buf[i] = s[j];
return buf;
}
int main(int argc, char **argv) {
printf("%s\n", rev(argv[1]));
return 0;
}
Run with "foo bar foobar baz" and get zab raboof rab oof back:
~/tmp$ ./a.out "foo bar foobar baz"
zab raboof rab oof
Here I think you can study two algorithms:
C string length calculate: the end of the c string is '\0'
How to reverse a c string in place
And if you need to test the code, you should alloc testing strings in heap or strack. If you write a literal string, you may meet a bus error because of the literal string being saved in text-area which is a read only memory.
And the following is the demo:
#include <stdio.h>
#include <stdlib.h>
void reverse_string(char* str)
{
size_t len;
char tmp, *s;
//Get the length of string, in C the last char of one string is \0
for(s=str;*s;++s) ;
len = s - str;
//Here we use the algorithm for reverse the char inplace.
//We only need a char tmp place for swap each char
s = str + len - 1;
while(s>str){
tmp = *s;
*s = *str;
*str = tmp;
s--;
str++;
}
}
int main()
{
char* a = "abcd";
//Here "abcd" will be saved in READ Only Memory. If you test code, you will get a bus error.
char* b = (char*)calloc(1,10);
strcpy(b,a);
reverse_string(b);
printf("%s\n",b);
a = "abcde";
strcpy(b,a);
reverse_string(b);
printf("%s\n",b);
}
you can do it simply by following code
for(int k=0;k<strlen(temp);k++)
{
Str[k]=temp[k];
}
How do I remove a character from a string?
If I have the string "abcdef" and I want to remove "b" how do I do that?
Removing the first character is easy with this code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char word[] = "abcdef";
char word2[10];
strcpy(word2, &word[1]);
printf("%s\n", word2);
return 0;
}
and
strncpy(word2, word, strlen(word) - 1);
will give me the string without the last character, but I still didn't figure out how to remove a char in the middle of a string.
memmove can handle overlapping areas, I would try something like that (not tested, maybe +-1 issue)
char word[] = "abcdef";
int idxToDel = 2;
memmove(&word[idxToDel], &word[idxToDel + 1], strlen(word) - idxToDel);
Before: "abcdef"
After: "abdef"
Try this :
void removeChar(char *str, char garbage) {
char *src, *dst;
for (src = dst = str; *src != '\0'; src++) {
*dst = *src;
if (*dst != garbage) dst++;
}
*dst = '\0';
}
Test program:
int main(void) {
char* str = malloc(strlen("abcdef")+1);
strcpy(str, "abcdef");
removeChar(str, 'b');
printf("%s", str);
free(str);
return 0;
}
Result:
>>acdef
My way to remove all specified chars:
void RemoveChars(char *s, char c)
{
int writer = 0, reader = 0;
while (s[reader])
{
if (s[reader]!=c)
{
s[writer++] = s[reader];
}
reader++;
}
s[writer]=0;
}
char a[]="string";
int toBeRemoved=2;
memmove(&a[toBeRemoved],&a[toBeRemoved+1],strlen(a)-toBeRemoved);
puts(a);
Try this . memmove will overlap it.
Tested.
Really surprised this hasn't been posted before.
strcpy(&str[idx_to_delete], &str[idx_to_delete + 1]);
Pretty efficient and simple. strcpy uses memmove on most implementations.
int chartoremove = 1;
strncpy(word2, word, chartoremove);
strncpy(((char*)word2)+chartoremove, ((char*)word)+chartoremove+1,
strlen(word)-1-chartoremove);
Ugly as hell
The following will extends the problem a bit by removing from the first string argument any character that occurs in the second string argument.
/*
* delete one character from a string
*/
static void
_strdelchr( char *s, size_t i, size_t *a, size_t *b)
{
size_t j;
if( *a == *b)
*a = i - 1;
else
for( j = *b + 1; j < i; j++)
s[++(*a)] = s[j];
*b = i;
}
/*
* delete all occurrences of characters in search from s
* returns nr. of deleted characters
*/
size_t
strdelstr( char *s, const char *search)
{
size_t l = strlen(s);
size_t n = strlen(search);
size_t i;
size_t a = 0;
size_t b = 0;
for( i = 0; i < l; i++)
if( memchr( search, s[i], n))
_strdelchr( s, i, &a, &b);
_strdelchr( s, l, &a, &b);
s[++a] = '\0';
return l - a;
}
This is an example of removing vowels from a string
#include <stdio.h>
#include <string.h>
void lower_str_and_remove_vowel(int sz, char str[])
{
for(int i = 0; i < sz; i++)
{
str[i] = tolower(str[i]);
if(str[i] == 'a' || str[i] == 'e' || str[i] == 'i' || str[i] == 'o' || str[i] == 'u')
{
for(int j = i; j < sz; j++)
{
str[j] = str[j + 1];
}
sz--;
i--;
}
}
}
int main(void)
{
char str[101];
gets(str);
int sz = strlen(str);// size of string
lower_str_and_remove_vowel(sz, str);
puts(str);
}
Input:
tour
Output:
tr
Use strcat() to concatenate strings.
But strcat() doesn't allow overlapping so you'd need to create a new string to hold the output.
I tried with strncpy() and snprintf().
int ridx = 1;
strncpy(word2,word,ridx);
snprintf(word2+ridx,10-ridx,"%s",&word[ridx+1]);
Another solution, using memmove() along with index() and sizeof():
char buf[100] = "abcdef";
char remove = 'b';
char* c;
if ((c = index(buf, remove)) != NULL) {
size_t len_left = sizeof(buf) - (c+1-buf);
memmove(c, c+1, len_left);
}
buf[] now contains "acdef"
This might be one of the fastest ones, if you pass the index:
void removeChar(char *str, unsigned int index) {
char *src;
for (src = str+index; *src != '\0'; *src = *(src+1),++src) ;
*src = '\0';
}
This code will delete all characters that you enter from string
#include <stdio.h>
#include <string.h>
#define SIZE 1000
char *erase_c(char *p, int ch)
{
char *ptr;
while (ptr = strchr(p, ch))
strcpy(ptr, ptr + 1);
return p;
}
int main()
{
char str[SIZE];
int ch;
printf("Enter a string\n");
gets(str);
printf("Enter the character to delete\n");
ch = getchar();
erase_c(str, ch);
puts(str);
return 0;
}
input
a man, a plan, a canal Panama
output
A mn, pln, cnl, Pnm!
Edit : Updated the code zstring_remove_chr() according to the latest version of the library.
From a BSD licensed string processing library for C, called zString
https://github.com/fnoyanisi/zString
Function to remove a character
int zstring_search_chr(char *token,char s){
if (!token || s=='\0')
return 0;
for (;*token; token++)
if (*token == s)
return 1;
return 0;
}
char *zstring_remove_chr(char *str,const char *bad) {
char *src = str , *dst = str;
/* validate input */
if (!(str && bad))
return NULL;
while(*src)
if(zstring_search_chr(bad,*src))
src++;
else
*dst++ = *src++; /* assign first, then incement */
*dst='\0';
return str;
}
Exmaple Usage
char s[]="this is a trial string to test the function.";
char *d=" .";
printf("%s\n",zstring_remove_chr(s,d));
Example Output
thisisatrialstringtotestthefunction
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX 50
void dele_char(char s[],char ch)
{
int i,j;
for(i=0;s[i]!='\0';i++)
{
if(s[i]==ch)
{
for(j=i;s[j]!='\0';j++)
s[j]=s[j+1];
i--;
}
}
}
int main()
{
char s[MAX],ch;
printf("Enter the string\n");
gets(s);
printf("Enter The char to be deleted\n");
scanf("%c",&ch);
dele_char(s,ch);
printf("After Deletion:= %s\n",s);
return 0;
}
#include <stdio.h>
#include <string.h>
int main(){
char ch[15],ch1[15];
int i;
gets(ch); // the original string
for (i=0;i<strlen(ch);i++){
while (ch[i]==ch[i+1]){
strncpy(ch1,ch,i+1); //ch1 contains all the characters up to and including x
ch1[i]='\0'; //removing x from ch1
strcpy(ch,&ch[i+1]); //(shrinking ch) removing all the characters up to and including x from ch
strcat(ch1,ch); //rejoining both parts
strcpy(ch,ch1); //just wanna stay classy
}
}
puts(ch);
}
Let's suppose that x is the "symbol" of the character you want to remove
,my idea was to divide the string into 2 parts:
1st part will countain all the characters from the index 0 till (and including) the target character x.
2nd part countains all the characters after x (not including x)
Now all you have to do is to rejoin both parts.
This is what you may be looking for while counter is the index.
#include <stdio.h>
int main(){
char str[20];
int i,counter;
gets(str);
scanf("%d", &counter);
for (i= counter+1; str[i]!='\0'; i++){
str[i-1]=str[i];
}
str[i-1]=0;
puts(str);
return 0;
}
I know that the question is very old, but I will leave my implementation here:
char *ft_strdelchr(const char *str,char c)
{
int i;
int j;
char *s;
char *newstr;
i = 0;
j = 0;
// cast to char* to be able to modify, bc the param is const
// you guys can remove this and change the param too
s = (char*)str;
// malloc the new string with the necessary length.
// obs: strcountchr returns int number of c(haracters) inside s(tring)
if (!(newstr = malloc(ft_strlen(s) - ft_strcountchr(s, c) + 1 * sizeof(char))))
return (NULL);
while (s[i])
{
if (s[i] != c)
{
newstr[j] = s[i];
j++;
}
i++;
}
return (newstr);
}
just throw to a new string the characters that are not equal to the character you want to remove.
Following should do it :
#include <stdio.h>
#include <string.h>
int main (int argc, char const* argv[])
{
char word[] = "abcde";
int i;
int len = strlen(word);
int rem = 1;
/* remove rem'th char from word */
for (i = rem; i < len - 1; i++) word[i] = word[i + 1];
if (i < len) word[i] = '\0';
printf("%s\n", word);
return 0;
}
This is a pretty basic way to do it:
void remove_character(char *string, int index) {
for (index; *(string + index) != '\0'; index++) {
*(string + index) = *(string + index + 1);
}
}
I am amazed none of the answers posted in more than 10 years mention this:
copying the string without the last byte with strncpy(word2, word, strlen(word)-1); is incorrect: the null terminator will not be set at word2[strlen(word) - 1]. Furthermore, this code would cause a crash if word is an empty string (which does not have a last character).
The function strncpy is not a good candidate for this problem. As a matter of fact, it is not recommended for any problem because it does not set a null terminator in the destination array if the n argument is less of equal to the source string length.
Here is a simple generic solution to copy a string while removing the character at offset pos, that does not assume pos to be a valid offset inside the string:
#include <stddef.h>
char *removeat_copy(char *dest, const char *src, size_t pos) {
size_t i;
for (i = 0; i < pos && src[i] != '\0'; i++) {
dest[i] = src[i];
}
for (; src[i] != '\0'; i++) {
dest[i] = src[i + 1];
}
dest[i] = '\0';
return dest;
}
This function also works if dest == src, but for removing the character in place in a modifiable string, use this more efficient version:
#include <stddef.h>
char *removeat_in_place(char *str, size_t pos) {
size_t i;
for (i = 0; i < pos && str[i] != '\0'; i++)
continue;
for (; str[i] != '\0'; i++)
str[i] = str[i + 1];
return str;
}
Finally, here are solutions using library functions:
#include <string.h>
char *removeat_copy(char *dest, const char *src, size_t pos) {
size_t len = strlen(src);
if (pos < len) {
memmove(dest, src, pos);
memmove(dest + pos, src + pos + 1, len - pos);
} else {
memmove(dest, src, len + 1);
}
return dest;
}
char *removeat_in_place(char *str, size_t pos) {
size_t len = strlen(str);
if (pos < len) {
memmove(str + pos, str + pos + 1, len - pos);
}
return str;
}
A convenient, simple and fast way to get rid of \0 is to copy the string without the last char (\0) with the help of strncpy instead of strcpy:
strncpy(newStrg,oldStrg,(strlen(oldStrg)-1));