C Remove substring between two characters - c

Question

I have two possible strings that would be pointed to by a char *:
char *s = "this is a string";
char *s = "this is a string [this is more string]";
I want to be able to remove the brackets and their contents from the char if they exist. What is the best way to do that in C?
Thanks!

Answer 1

First, you must have access to memory you are permitted to modify. For hysterical raisins, C still allows you to declare pointers to constant character arrays as non-const... but you still can’t modify them.
Instead, declare yourself an array which is initialized by the constant data.
char s[] = "hello world";
That gives you a mutable array of 11+1 characters (a copy of the immutable source string).
Often, however, we want to be able to work on our strings, so we need to make sure there is enough room to play with them.
char s[100] = "hello world"; // string capacity == 99+1. In use == 11+1.
s[6] = '\0'; // now string contains "hello ". In use == 6+1.
strcat( s, "everyone" ); // now string contains "hello everyone". In use == 14+1.
To remove content from a string, you must first identify where the start and stop is:
char s[] = "this is a string [this is more string] original string";
char * first = strchr( s, '[' ); // find the '['
char * last = strchr( s, ']' ); // find the ']'
if (first and last) // if both were found...
// ...copy the end of the string over the part to be replaced
memmove( first, last+1, strlen(last+1)+1 );
// s now contains "this is a string original string"
Notice all those +1s in there? Be mindful:
of where you are indexing the string (you want to copy from after the ']' to the end of the string)
that strings must end with a nul-terminator — a zero-valued character (which we should also copy, so we added one to the strlen).
Remember, a string is just an array of characters, where the last used character is immediately followed by a zero-value character (the nul-terminator). All the methods you use to handle arrays of any other kind apply to handling strings (AKA arrays of characters).

Answer 2

This is one of those "fun" problems that attracts more than its share of answers.
Full credit to other answers for noting that attempts to modify a "string literal" triggers UB. (String literals are often stored in an immutable "read only" region of memory.) (Thanks to #chux-reinstatemonica for clarification.)
Taking this to the next level, below is a bit of code that handles both multiple regions bounded by a pair of delimiters ('[' & ']'), and handles nested instances, too. The code is simple enough, using cut as a counter that ensures nesting is accounted for. NB: It is presumed that the source string is "well formed".
#include <stdio.h>
int main( void ) {
char *wb =
"Once [upon] a [time] there [lived a] princess. "
"She [ was [a hottie]]. The end.";
char wo[100]; // big enough
int cut = 0;
size_t s = 0, d = 0;
for( ; wb[s]; s++ ) {
cut += ( wb[s] == '[' ); // entering region to cut?
if( !cut ) wo[ d++ ] = wb[ s ]; // not cutting, so copy
cut -= ( wb[s] == ']' ); // exiting region that was cut?
if( cut < 0 ) cut = 0; // in case of spurious ']'
}
wo[ d ] = '\0'; // terminate shortened string
puts( wb );
puts( wo );
return 0;
}
Once [upon] a [time] there [lived a] princess. She [ was [a hottie]]. The end.
Once a there princess. She . The end.
It now becomes a small challenge to, perhaps, remove multiple consecutive SPs in the output array. This could quite easily be done on the fly, and is left as an exercise.
One begins to see how something like this could be extended to make an "infix calculator program". Always something new!

Answer 3

Use the strchnul function defined in string.h
*(strchrnul(s, '[')) = '\0';
You may need to define the _GNU_SOURCE macro.
It won't work on string literals.

Answer 4

To remove the bracketed fragment(s) from the string, you cannot modify the string literal in place. You should either use an automatic array or an allocated block and construct the modified string there.
Here is an example:
#include <stdio.h>
#include <stdlib.h>
char *strip_brackets(const char *s) {
size_t i, j, len;
char *p;
// compute the modified string length
for (i = len = 0; s[i];) {
if (s[i] == '[') {
for (j = i + 1; s[j] && s[j] != ']'; j++)
continue;
if (s[j] == ']') {
i = j + 1;
continue;
}
}
len++;
i++;
}
p = malloc(len + 1);
if (p != NULL) {
// construct the modified string
for (i = len = 0; s[i];) {
if (s[i] == '[') {
for (j = i + 1; s[j] && s[j] != ']'; j++)
continue;
if (s[j] == ']') {
i = j + 1;
continue;
}
}
p[len++] = s[i++];
}
p[len] = '\0';
}
return p;
}
int main() {
const char *s = "this is a string [this is more string]";
char *s1 = strip_brackets(s);
printf("before: %s\n", s);
if (s1) {
printf("after : %s\n", s1);
free(s1);
}
return 0;
}