I am having trouble with the very last line in my function, where I am stilly learning the basics of C. I have the signature of this function given and am tasked to write a function to concatenate two strings. The commented line outputs the correct result.
#include <stdio.h>
#include <stdlib.h>
// 1) len = dst-len + max_dst_len
int strlcat(char *dst, const char *src, int max_dst_len) {
int len = 0;
while (dst[len] != '\0') {
len++;
}
int total_len = len + max_dst_len;
char *new_str = malloc(sizeof(char) * total_len);
for (int i = 0; i < len; i++) {
new_str[i] = dst[i];
}
for (int i = len; i < total_len; i++) {
new_str[i] = src[i - len];
}
new_str[total_len] = '\0';
//printf("%s <--\n", new_str);
dst = *new_str;
return total_len;
}
int main() {
char test1[] = "dst";
char test1src[] = "src";
printf("%s\n", test1);
printf("%d\n", strlcat(test1, test1src, 10));
printf("%s\n", test1);
}
You should not be adding max_dst_len to the length of dst. max_dst_len is the amount of memory that's already allocated in dst, you need to ensure that the concatenated string doesn't exceed this length.
So you need to subtract len from max_dst_len, and also subtract 1 to allow room for the null byte. This will tell you the maximum number of bytes you can copy from src to the end of dst.
In your main() code, you need to declare test1 to be at least 10 bytes if you pass 10 as the max_dst_len argument. When you omit the size in the array declaration, it sizes the array just big enough to hold the string you use to initialize it. It's best to use sizeof test1 as this argument, to ensure that it's correct for the string you're concatenating to.
#include <stdio.h>
int strlcat(char *dst, const char *src, int max_dst_len) {
int len = 0;
while (dst[len] != '\0') {
len++;
}
int len_to_copy = max_dst_len - len - 1;
int i;
for (i = 0; i < len_to_copy && src[i] != '\0'; i++) {
dst[len+i] = src[i];
}
dst[i] = '\0';
//printf("%s <--\n", new_str);
return i + len;
}
int main() {
char test1[6] = "dst";
char test1src[] = "src";
printf("%s\n", test1);
printf("%d\n", strlcat(test1, test1src, sizeof test1));
printf("%s\n", test1);
}
I have this function which receives a pointer to char array and initializes it to be len repetitions of "a":
void test(char ** s, int len) {
*s = (char *)malloc(sizeof(char) * len);
int i;
for(i = 0; i < len; i++) {
*(s[i]) = 'a';
}
printf("%s\n", *s);
}
in the main() I have this code:
char * s;
test(&s, 3);
but I get EXC_BAD_ACCESS (code=1, address=0x0) error when I run main(). The error occurs on the second iteration of the for loop in this line: *(s[i]) = 'a';
As far as I understand I'm not accessing the elements correctly, what is the correct way?
s is declared as a pointer to a pointer. In reality, it's a pointer to a pointer to the start of an array, but that cannot be inferred from the type system alone. It could just as well be a pointer to a start of an array of pointers, which is how s[i] treats it. You need to first derefence s (to get the pointer to the array's start), and then index on it:
(*s)[i] = 'a';
Also, as #MFisherKDX correctly pointed out in comments, if you're going to pass *s to printf or any other standard string-manipulation function, you have to turn into a proper C string by terminating it with a 0 character.
This more clearly shows what you should be doing, while staying close to your original code:
void test(char ** s, int len) {
char *p = malloc(len);
int i;
for(i = 0; i < len; i++) {
p[i] = 'a';
}
printf("%s\n", p);
*s = p;
}
Note that sizeof(char) is always one by definition, and in C there's no need to cast the result of malloc().
That code also has all your original problems in that it doesn't actually create a string that you can send to printf( "%s... ). This fixes that problem:
void test(char ** s, int len) {
char *p = malloc(len+1);
int i;
for(i = 0; i < len; i++) {
p[i] = 'a';
}
p[i]='\0';
printf("%s\n", p);
*s = p;
}
And this is even easier, with no need to use a double-* pointer:
char *test(int len) {
char *p = malloc(len+1);
int i;
for(i = 0; i < len; i++) {
p[i] = 'a';
}
p[i]='\0';
printf("%s\n", p);
return(p);
}
Instead of assignment
*(s[i]) = 'a';
use this:
(*s)[i] = 'a';
But do not forget that C/C++ strings are null terminated
Or you can use this approach to get more readable code:
void test(char** s, int len) {
// 1 char extra for zero
char *str = (char*)malloc(sizeof(char) * (len+1));
int i;
for(i = 0; i < len; i++)
str[i] = 'a';
// zero terminated string
str[len] = 0;
printf("%s\n", str);
*s = str;
}
Im trying to rewrite strcpy function but im getting stuck in this, my function does not run when the first arg is the static array of chars, only run if it is an allocated pointer. I think the problem is the realloc statement in my function. Please help me for my problem...
#include <stdio.h>
#include <stdlib.h>
int myStrlen(const char *);
char *myStrcpy(char *, const char *);
int main() {
char str1[40];
char str2[] = "one piece";
char *str;
str = (char*)malloc(1 * sizeof(char));
myStrcpy(str, str2);
printf("%s", str);
myStrcpy(str1, str2);
printf("%s", str1);
return 0;
}
int myStrlen(const char *str) {
int i = 0, len = 0;
while (str[i++] != '\0') {
len++;
}
return len;
}
char *myStrcpy(char *str1, const char *str2) {
int len1, len2;
len1 = myStrlen(str1);
len2 = myStrlen(str2);
str1 = (char*)realloc(str1, len2 * sizeof(char));
for (int i = 0; i < len2; i++) {
str1[i] = str2[i];
}
str1[len2] = '\0';
return str1;
}
Please help me to make my code using pointer work...
Btw, here is my function without using pointer, any ideas?
void myStrcpy(char str1[], char str2[]) {
int i = 0;
while (str2[i] != '\0') {
str1[i] = str2[i];
i++;
}
str1[i] = '\0';
}
void myStrncpy(char str1[], char str2[], int n) {
int i = 0;
while (str2[i] != '\0' || i <= n) {
str1[i] = str2[i];
i++;
}
str1[i] = '\0';
}
Your first implementation uses realloc() for no valid purpose. For standard function strcpy(), str1 is assumed to point to an array with enough space to store a copy of str2.
If your intent is to implement different semantics from that of strcpy, make the name more explicit and pass it the address of the destination pointer because realloc() is likely to return a different pointer. In your example, the original value of str is used after the call to myStrcpy(), which is incorrect.
Your second implementation is almost correct:
You should use size_t for the type of i. Strings can be longer than INT_MAX on current systems.
You should make the source strings const.
myStrncpy should stop when i reaches n, use && i < n
Testing i < n before dereferencing the str2 pointer allows an invalid pointer to be passed along with a zero value for n, which can be useful.
myStrcpy and myStrncpy should return the destination pointer.
Note that myStrncpy has much better semantics than strncpy. The standard function strncpy does not do what most programmers expect. Its semantics are error prone, it should not be used to avoid potential confusion.
Here is a simplified version:
#include <stdlib.h>
size_t myStrlen(const char *str) {
size_t len = 0;
while (str[len] != '\0') {
len++;
}
return len;
}
char *myStrcpy(char *str1, const char *str2) {
size_t i = 0;
while (str2[i] != '\0') {
str1[i] = str2[i];
i++;
}
str1[i] = '\0';
return str1;
}
char *myStrncpy(char *str1, const char *str2, size_t n) {
size_t i = 0;
while (i < n && str2[i] != '\0') {
str1[i] = str2[i];
i++;
}
str1[i] = '\0';
return str1;
}
A properly written strcpy and strncpy should not need any heap allocation.
char *mystrcpy(char *dest, char *src)
{
char *start = &dest[0];
while(*src) *dest++ = *src++;
return start;
}
char *mystrncpy(char *dest, const char *src, size_t n)
{
size_t i = 0;
char *start = &dest[0];
while(*src && i < n)
{
*dest++ = *src++;
i++;
}
for(; i < n; i++) *dest++ = 0;
return start;
}
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));