append a character from an array to a char pointer

Question

Ok, so I'm a person who usually writes Java/C++, and I've just started getting into writing C. I'm currently writing a lexical analyser, and I can't stand how strings work in C, since I can't perform string arithmetic. So here's my question:

char* buffer = "";
char* line = "hello, world";

int i;
for (i = 0; i < strlen(line); i++) {
    buffer += line[i];
}

How can I do that in C? Since the code above isn't valid C, how can I do something like that? Basically I'm looping though a string line, and I'm trying to append each character to the buffer string.

Answer 1

string literals are immutable in C. Modifying one causes Undefined Behavior.

If you use a char array (your buffer) big enough to hold your characters, you can still modify its content :

#include <stdio.h>

int main(void) {


    char * line = "hello, world";
    char buffer[32]; // ok, this array is big enough for our operation

    int i;
    for (i = 0; i < strlen(line) + 1; i++) 
    {
        buffer[i] = line[i];
    }

    printf("buffer : %s", buffer);

    return 0;
}

Answer 2

First off the buffer needs to have or exceed the length of the data being copied to it.

char a[length], b[] = "string";

Then the characters are copied to the buffer.

int i = 0;
while (i < length && b[i] != '\0') { a[i] = b[i]; i++; }
a[i] = '\0';

You can reverse the order if you need to, just start i at the smallest length value among the two strings, and decrement the value instead of increment. You can also use the heap, as others have suggested, ordinate towards an arbitrary or changing value of length. Furthermore, you can change up the snippet with pointers (and to give you a better idea of what is happening):

int i = 0;
char *j = a, *k = b;
while (j - a < length && *k) { *(j++) =  *(k++); }
*j = '\0';

Make sure to look up memcpy; and don't forget null terminators (oops).

Answer 3

#include <string.h>

//...

char *line = "hello, world";
char *buffer = ( char * ) malloc( strlen( line ) + 1 );

strcpy( buffer, line );

Though in C string literals have types of non-const arrays it is better to declare pointers initialized by string literals with qualifier const:

const char *line = "hello, world";

String literals in C/C++ are immutable.

If you want to append characters then the code can look the following way (each character of line is appended to buffer in a loop)

#include <string.h>

//...

char *line = "hello, world";
char *buffer = ( char * ) malloc( strlen( line ) + 1 );

buffer[0] = '\0';
char *p = Buffer;

for ( size_t i = 0; i < strlen( line ); i++ )
{
    *p++ = line[i];
    *p = '\0';
}

The general approach is that you find the pointer to the terminating zero substitute it for the target character advance the pointer and appenf the new terminating zero. The source buffer shall be large enough to accomodate one more character.