I want a space between every character of a string like I will give input "HELLO"
the result will be "H E L L O"
I need help in that
[Edit from comments]
I want it in a string
for (i = 0; i <= strlen(str); i++) {
printf("\n String is: %s", str[i]);
printf(" ");
}
The shorter, more general answer is that you need to bump characters back, and insert a ' ' in between them. What have you done so far? Does it need to be in place?
One (perhaps not optimal, but easy to follow solution) would be making a larger array, copying in alternating letters, something like (not guaranteed to work verbatim)
char foo[N]; // assuming this has N characters and you want to add a space in between all of them.
char bar[2*N];
for (int i = 0; i < N; i++) {
bar[2*i] = foo[i];
if (i != N - 1)
bar[2*i + 1] = ' ';
}
Of course, this new string is in bar, but functions as desired. At what point are you having issues?
try this
#include <stdio.h>
void add_spaces(char need_to_add[])
{
int len = strlen(need_to_add);
char with_spaces[len*2];
int space_index = 0;
for (int i=0 ; i<len ; i++)
{
with_spaces[space_index]=need_to_add[i];
with_spaces[++space_index]=' ';
space_index=space_index+1;
}
printf("%s\n", with_spaces);
}
int main()
{
char * a = "aaa";
add_spaces(a); // fraught with problems
return 1;
}
Related
Consider this simple program that concatenates all specified parameters and prints them in standard output. I used 2 for loops to append the strings, one to calculate the length of that string and one to concatenate the strings. Is there a way doing it with only one loop? It wouldn't be more efficient reallocating memory for each string to concatenate, would it? How would Java's StringBuilder be implemented in C? Would it loop twice as I did?
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(int argc, char** argv)
{
size_t len = 0;
// start for loop at i = 1 to skip the program name specified in argv
for(int i = 1; i < argc; i++)
len += strlen(argv[i]) + 1; // +1 for the space
char* toAppend = (char*)malloc(len * sizeof(char) + 1);
toAppend[0] = '\0'; // first string is empty and null terminated
for(int i = 1; i < argc; i++)
{
strcat(toAppend, argv[i]);
strcat(toAppend, " ");
}
printf(toAppend);
free(toAppend);
}
Your method of allocation is efficient, measuring the total length and allocating just once. But the concatenation loop repeatedly measures the length of the output buffer from the start to concatenate to it, resulting in quadratic runtime.
To fix it track your position as you go:
size_t pos = 0;
for(int i = 1; i < argc; i++) {
size_t len = strlen(argv[i]);
memcpy(toAppend+pos, argv[i], len);
pos += len;
toAppend[pos] = ' ';
pos++;
}
toAppend[pos] = 0;
This is the most efficient way to actually concatenate in memory, but the most efficient of all is not to concatenate. Instead:
for(int i = 1; i < argc; i++)
printf("%s ", argv[i]);
The whole reason stdio is buffered is so you don't have to build arbitrary-length in-memory buffers to do efficient output; instead it buffers up to a fixed size automatically and flushes when the buffer is full.
Note that your usage of printf is wrong and dangerous in the event that your input contains a % character anywhere; it should be printf("%s", toAppend);.
If you're writing to POSIX (or POSIX-ish) systems rather than just plain C, another option would be fmemopen, which would allow you to write the loop just as:
for(int i = 1; i < argc; i++)
fprintf(my_memfile, "%s ", argv[i]);
efficient way to concatenate strings in c
An efficient way is to calculate the string lengths - and remember them.
size_t sum = 1; // for \0
if (argc > 2) sum += argc - 2. // spaces
size_t length[argc]; // This is a VLA, available C99 and optionally in C11
for(int i = 1; i < argc; i++)
length[i] = strlen(argv[i]);
sum += length[i];
}
Then allocate, and then check for errors.
char *dest = malloc(sum);
if (dest == NULL) Handle_OutOfMemory();
Copy each string in turn
char *p = dest;
for(int i = 1; i < argc; i++)
// Use either memcpy() or strcpy().
// memcpy() tends to be faster for long strings than strcpy().
memcpy(p, argv[i], length[i]);
p += length[i]; // advance insertion point
if (i > 1) {
*p++ = ' '; // space separators
}
}
*p = '\0';
Now use dest[].
printf("<%s>\n", dest);
Free resources when done.
free(dest);
It wouldn't be more efficient reallocating memory for each string to concatenate, would it?
Usually repetitive re-allocations is best avoided, yet for small short strings it really makes scant difference. Focus on big O. My answer is O(n). Relocating in a loop tends to be O(n*n).
If performance was critical, try various approaches and profile for the intended system. The point being what is fast on one machine may differ on another. Usually it is best to first code a reasonable clear approach.
The most efficient way is probably to not use any str functions and copy the characters "by hand":
char* toAppend = malloc(len + 1);
size_t j = 0;
for(size_t i = 1; i < argc; i++)
{
for(size_t k = 0; argv[i][k]; k++)
toAppend[j++] = argv[i][k];
toAppend[j++] = ' ';
}
toAppend[j - 1] = '\0'; // Remove the last space and NULL-terminate the string
I made a code and my target is to put spacewhere the input word was found in a sentence.
i neet to replece the small word with space
like:
Three witches watched three watches
tch
output:
Three wi es wa ed three wa es
I made this code:
#include<stdio.h>
#define S 8
#define B 50
void main() {
char small[S] = {"ol"};
char big[B] = {"my older gradmom see my older sister"};
int i = 0, j = 0;
for (i = 0; i < B; i++)
{
for(j=0;j<S;j++)
{
if(small[j]!=big[i])
{
j=0;
break;
}
if(small[j]=='\0')
{
while (i-(j-1)!=i)
{
i = i - j;
big[i] = '\n';
i++;
}
}
}
}
puts(big);
}
First of all, in your exemple you work with newline '\n' and not with space.
Consider this simple example:
#include<stdio.h>
#define S 8
#define B 50
void main() {
char small[S] = {"ol"};
char big[B] = {"my older gradmom see my older sister"};
int i = 0, j = 0;
int cpt = 0;
int smallSize = 0;
// loop to retrieve smallSize
for (i = 0; i < S; i++)
{
if (small[i] != '\0')
smallSize++;
}
// main loop
for (i = 0; i < B; i++)
{
// stop if we hit the end of the string
if (big[i] == '\0')
break;
// increment the cpt and small index while the content of big and small are equal
if (big[i] == small[j])
{
cpt++;
j++;
}
// we didn't found the full small word
else
{
j = 0;
cpt = 0;
}
// test if we found the full word, if yes replace char in big by space
if (cpt == smallSize)
{
for (int k = 0; k < smallSize; k++)
{
big[i-k] = ' ';
}
j = 0;
cpt = 0;
}
}
puts(big);
}
You need first to retrieve the real size of the small array.
Once done, next step is to look inside "big" if there is the word small inside. If we find it, then replace all those char by spaces.
If you want to replace the whole small word with a single space, then you'll need to adapt this example !
I hope this help !
A possible way is to use to pointers to the string, one for reading and one for writing. This will allow to replace an arbitrary number of chars (the ones from small) with a single space. And you do not really want to nest loops but une only one to process every char from big.
Last but not least, void main() should never be used except in stand alone environment (kernel or embedded development). Code could become:
#include <stdio.h>
#define S 8
#define B 50
int main() { // void main is deprecated...
char small[S] = {"ol"};
char big[B] = {"my older gradmom see my older sister"};
int i = 0, j = 0;
int k = 0; // pointer to written back big
for (i = 0; i < B; i++)
{
if (big[i] == 0) break; // do not process beyond end of string
if(small[j]!=big[i])
{
for(int l=0; l<j; l++) big[k++] = small[l]; // copy an eventual partial small
big[k++] = big[i]; // copy the incoming character
j=0; // reset pointer to small
continue;
}
else if(small[++j] == 0) // reached end of small
{
big[k++] = ' '; // replace chars from small with a single space
j = 0; // reset pointer to small
}
}
big[k] = '\0';
puts(big);
return 0;
}
or even better (no need for fixed sizes of strings):
#include <stdio.h>
int main() { // void main is deprecated...
char small[] = {"ol"};
char big[] = {"my older gradmom see my older sister"};
int i = 0, j = 0;
int k = 0; // pointer to written back big
for (i = 0; i < sizeof(big); i++)
{
if(small[j]!=big[i])
...
In C strings are terminated with a null character '\0'. Your code defines a somehow random number at the beginning (B and S) and iterates over that much characters instead of the exact number of characters, the strings actually contain. You can use the fact that the string is terminated by testing the content of the string in a while loop.
i = 0;
while (str[i]) {
...
i = i + 1;
}
If you prefer for loops you can write it also as a for loop.
for (i = 0; str[i]; i++) {
...
}
Your code does not move the contents of the remaining string to the left. If you replace two characters ol with one character , you have to move the remaining characters to the left by one character. Otherwise you would have a hole in the string.
#include <stdio.h>
int main() {
char small[] = "ol";
char big[] = "my older gradmom see my older sister";
int s; // index, which loops through the small string
int b; // index, which loops through the big string
int m; // index, which loops through the characters to be modified
// The following loops through the big string up to the terminating
// null character in the big string.
b = 0;
while (big[b]) {
// The following loops through the small string up to the
// terminating null character, if the character in the small
// string matches the corresponding character in the big string.
s = 0;
while (small[s] && big[b+s] == small[s]) {
// In case of a match, continue with the next character in the
// small string.
s = s + 1;
}
// If we are at the end of the small string, we found in the
// big string.
if (small[s] == '\0') {
// Now we have to modify the big string. The modification
// starts at the current position in the big string.
m = b;
// First we have to put the space at the current position in the
// big string.
big[m] = ' ';
// And next the rest of the big string has to be moved left. The
// rest of the big string starts, where the match has ended.
while (big[b+s]) {
m = m + 1;
big[m] = big[b+s];
s = s + 1;
}
// Finally the big string has to be terminated by a null
// character.
big[m+1] = '\0';
}
// Continue at next character in big string.
b = b + 1;
}
puts(big);
return 0;
}
I was doing a program to copy all string words other than its first 2 words and putting a x at the end of it.
However i cant put x at its end. Please help!!!!
Below is my code.
#include<stdio.h>
#include<string.h>
int main()
{
char a[25], b[25];
int i, j, count = 0, l, k;
scanf("%[^\n]s", a);
i = strlen(a);
if (i > 20)
printf("Given Sentence is too long.");
else
{/* checking for first 2 words and counting 2 spaces*/
for (j = 0; j < i; j++)
{
if (a[j] == ' ')
count = count + 1;
if (count == 2)
{
k = j;
break;
}
}
/* copying remaining string into new one*/
for (j = 0; j < i - k; j++)
{
b[j] = a[j + k];
}
b[j + 1] = 'x';
printf("%s", b);
}
}
you are removing first two index. But you wrote k=j and if you check the current value j there it's 1. so you are updating wrongly k because you removed 2 indexes. So k value should be 2. So checked the below code
/* copying remaining string into new one*/
for (j = 0; j < i - 2; j++)
{
b[j] = a[j + 2];
}
b[j + 1] = 'x';
printf("%s", b);
Your index is off by one. After your second loop, the condition j < i-k was false, so j now is i-k. Therefore, the character after the end of what you copied is b[j], not b[j+1]. The correct line would therefore be b[j] = 'x';.
Just changing this would leave you with something that is not a string. A string is defined as a sequence of char, ending with a '\0' char. So you have to add b[j+1] = 0; as well.
After these changes, your code does what you intended, but still has undefined behavior.
One problem is that your scanf() will happily overflow your buffer -- use a field width here: scanf("%24[^\n]", a);. And by the way, the s at the and doesn't make any sense, you use either the s conversion or the [] conversion.
A somewhat sensible implementation would use functions suited for the job, like e.g. this:
#include<stdio.h>
#include<string.h>
int main(void)
{
// memory is *cheap* nowadays, these buffers are still somewhat tiny:
char a[256];
char b[256];
// read a line
if (!fgets(a, 256, stdin)) return 1;
// and strip the newline character if present
a[strcspn(a, "\n")] = 0;
// find first space
char *space = strchr(a, ' ');
// find second space
if (space) space = strchr(space+1, ' ');
if (space)
{
// have two spaces, copy the rest
strcpy(b, space+1);
// and append 'x':
strcat(b, "x");
}
else
{
// empty string:
b[0] = 0;
}
printf("%s",b);
return 0;
}
For functions you don't know, google for man <function>.
In C strings are array of chars as you know and the way C knows it is end of the string is '\0' character. In your example you are missing at the last few lines
/* copying remaining string into new one*/
for(j=0;j<i-k;j++)
{
b[j]=a[j+k];
}
b[j+1]='x';
printf("%s",b);
after the loop ends j is already increased 1 before it quits the loop.
So if your string before x is "test", it is like
't', 'e', 's', 't','\0' in char array, and since your j is increased more than it should have, it gets to the point just right of '\0', but characters after '\0' doesnt matter, because it is the end, so your x will not be added. Simple change to
b[j]='x';
I have a program I wrote to take a string of words and, based on the delimiter that appears, separate each word and add it to an array.
I've adjusted it to account for either a ' ' , '.' or '.'. Now the goal is to adjust for multiple delimiters appearing together (as in "the dog,,,was walking") and still only add the word. While my program works, and it doesn't print out extra delimiters, every time it encounters additional delimiters, it includes a space in the output instead of ignoring them.
int main(int argc, const char * argv[]) {
char *givenString = "USA,Canada,Mexico,Bermuda,Grenada,Belize";
int stringCharCount;
//get length of string to allocate enough memory for array
for (int i = 0; i < 1000; i++) {
if (givenString[i] == '\0') {
break;
}
else {
stringCharCount++;
}
}
// counting # of commas in the original string
int commaCount = 1;
for (int i = 0; i < stringCharCount; i++) {
if (givenString[i] == ',' || givenString[i] == '.' || givenString[i] == ' ') {
commaCount++;
}
}
//declare blank Array that is the length of commas (which is the number of elements in the original string)
//char *finalArray[commaCount];
int z = 0;
char *finalArray[commaCount] ;
char *wordFiller = malloc(stringCharCount);
int j = 0;
char current = ' ';
for (int i = 0; i <= stringCharCount; i++) {
if (((givenString[i] == ',' || givenString[i] == '\0' || givenString[i] == ',' || givenString[i] == ' ') && (current != (' ' | '.' | ',')))) {
finalArray[z] = wordFiller;
wordFiller = malloc(stringCharCount);
j=0;
z++;
current = givenString[i];
}
else {
wordFiller[j++] = givenString[i];
}
}
for (int i = 0; i < commaCount; i++) {
printf("%s\n", finalArray[i]);
}
return 0;
}
This program took me hours and hours to get together (with help from more experienced developers) and I can't help but get frustrated. I'm using the debugger to my best ability but definitely need more experience with it.
/////////
I went back to pad and paper and kind of rewrote my code. Now I'm trying to store delimiters in an array and compare the elements of that array to the current string value. If they are equal, then we have come across a new word and we add it to the final string array. I'm struggling to figure out the placement and content of the "for" loop that I would use for this.
char * original = "USA,Canada,Mexico,Bermuda,Grenada,Belize";
//creating two intialized variables to count the number of characters and elements to add to the array (so we can allocate enough mmemory)
int stringCharCount = 0;
//by setting elementCount to 1, we can account for the last word that comes after the last comma
int elementCount = 1;
//calculate value of stringCharCount and elementCount to allocate enough memory for temporary word storage and for final array
for (int i = 0; i < 1000; i++) {
if (original[i] == '\0') {
break;
}
else {
stringCharCount++;
if (original[i] == ',') {
elementCount++;
}
}
}
//account for the final element
elementCount = elementCount;
char *tempWord = malloc(stringCharCount);
char *finalArray[elementCount];
int a = 0;
int b = 0;
//int c = 0;
//char *delimiters[4] = {".", ",", " ", "\0"};
for (int i = 0; i <= stringCharCount; i++) {
if (original[i] == ',' || original[i] == '\0') {
finalArray[a] = tempWord;
tempWord = malloc(stringCharCount);
tempWord[b] = '\0';
b = 0;
a++;
}
else {
tempWord[b++] = original[i];
}
}
for (int i = 0; i < elementCount; i++) {
printf("%s\n", finalArray[i]);
}
return 0;
}
Many issues. Suggest dividing code into small pieces and debug those first.
--
Un-initialize data.
// int stringCharCount;
int stringCharCount = 0;
...
stringCharCount++;
Or
int stringCharCount = strlen(givenString);
Other problems too: finalArray[] is never assigned a terminarting null character yet printf("%s\n", finalArray[i]); used.
Unclear use of char *
char *wordFiller = malloc(stringCharCount);
wordFiller = malloc(stringCharCount);
There are more bugs than lines in your code.
I'd suggest you start with something much simpler.
Work through a basic programming book with excercises.
Edit
Or, if this is about learning to program, try another, simpler programming language:
In C# your task looks rather simple:
string givenString = "USA,Canada Mexico,Bermuda.Grenada,Belize";
string [] words = string.Split(new char[] {' ', ',', '.'});
foreach(word in words)
Console.WriteLine(word);
As you see, there are much issues to worry about:
No memory management (alloc/free) this is handeled by the Garbage Collector
no pointers, so nothing can go wrong with them
powerful builtin string capabilities like Split()
foreach makes loops much simpler
I can't write a workable code for a function that deletes N characters from the string S, starting from position P. How you guys would you write such a function?
void remove_substring(char *s, int p, int n) {
int i;
if(n == 0) {
printf("%s", s);
}
for (i = 0; i < p - 1; i++) {
printf("%c", s[i]);
}
for (i = strlen(s) - n; i < strlen(s); i++) {
printf("%c", s[i]);
}
}
Example:
s: "abcdefghi"
p: 4
n: 3
output:
abcghi
But for a case like n = 0 and p = 1 it's not working!
Thanks a lot!
A few people have shown you how to do this, but most of their solutions are highly condensed, use standard library functions or simply don't explain what's going on. Here's a version that includes not only some very basic error checking but some explanation of what's happening:
void remove_substr(char *s, size_t p, size_t n)
{
// p is 1-indexed for some reason... adjust it.
p--;
// ensure that we're not being asked to access
// memory past the current end of the string.
// Note that if p is already past the end of
// string then p + n will, necessarily, also be
// past the end of the string so this one check
// is sufficient.
if(p + n >= strlen(s))
return;
// Offset n to account for the data we will be
// skipping.
n += p;
// We copy one character at a time until we
// find the end-of-string character
while(s[n] != 0)
s[p++] = s[n++];
// And make sure our string is properly terminated.
s[p] = 0;
}
One caveat to watch out for: please don't call this function like this:
remove_substr("abcdefghi", 4, 3);
Or like this:
char *s = "abcdefghi";
remove_substr(s, 4, 3);
Doing so will result in undefined behavior, as string literals are read-only and modifying them is not allowed by the standard.
Strictly speaking, you didn't implement a removal of a substring: your code prints the original string with a range of characters removed.
Another thing to note is that according to your example, the index p is one-based, not zero-based like it is in C. Otherwise the output for "abcdefghi", 4, 3 would have been "abcdhi", not "abcghi".
With this in mind, let's make some changes. First, your math is a little off: the last loop should look like this:
for (i = p+n-1; i < strlen(s); i++) {
printf("%c", s[i]);
}
Demo on ideone.
If you would like to use C's zero-based indexing scheme, change your loops as follows:
for (i = 0; i < p; i++) {
printf("%c", s[i]);
}
for (i = p+n; i < strlen(s); i++) {
printf("%c", s[i]);
}
In addition, you should return from the if at the top, or add an else:
if(n == 0) {
printf("%s", s);
return;
}
or
if(n == 0) {
printf("%s", s);
} else {
// The rest of your code here
...
}
or remove the if altogether: it's only an optimization, your code is going to work fine without it, too.
Currently, you code would print the original string twice when n is 0.
If you would like to make your code remove the substring and return a result, you need to allocate the result, and replace printing with copying, like this:
char *remove_substring(char *s, int p, int n) {
// You need to do some checking before calling malloc
if (n == 0) return s;
size_t len = strlen(s);
if (n < 0 || p < 0 || p+n > len) return NULL;
size_t rlen = len-n+1;
char *res = malloc(rlen);
if (res == NULL) return NULL;
char *pt = res;
// Now let's use the two familiar loops,
// except printf("%c"...) will be replaced with *p++ = ...
for (int i = 0; i < p; i++) {
*pt++ = s[i];
}
for (int i = p+n; i < strlen(s); i++) {
*pt++ = s[i];
}
*pt='\0';
return res;
}
Note that this new version of your code returns dynamically allocated memory, which needs to be freed after use.
Here is a demo of this modified version on ideone.
Try copying the first part of the string, then the second
char result[10];
const char input[] = "abcdefg";
int n = 3;
int p = 4;
strncpy(result, input, p);
strncpy(result+p, input+p+n, length(input)-p-n);
printf("%s", result);
If you are looking to do this without the use of functions like strcpy or strncpy (which I see you said in a comment) then use a similar approach to how strcpy (or at least one possible variant) works under the hood:
void strnewcpy(char *dest, char *origin, int n, int p) {
while(p-- && *dest++ = *origin++)
;
origin += n;
while(*dest++ = *origin++)
;
}
metacode:
allocate a buffer for the destination
decalre a pointer s to your source string
advance the pointer "p-1" positions in your source string and copy them on the fly to destination
advance "n" positions
copy rest to destination
What did you try? Doesn't strcpy(s+p, s+p+n) work?
Edit: Fixed to not rely on undefined behaviour in strcpy:
void remove_substring(char *s, int p, int n)
{
p--; // 1 indexed - why?
memmove(s+p, s+p+n, strlen(s) - n);
}
If your heart's really set on it, you can also replace the memmove call with a loop:
char *dst = s + p;
char *src = s + p + n;
for (int i = 0; i < strlen(s) - n; i++)
*dst++ = *src++;
And if you do that, you can strip out the strlen call, too:
while ((*dst++ = *src++) != '\0);
But I'm not sure I recommend compressing it that much.