strcat(msg, ": ");
strcat(msg, buf);
Is there a way to do this in one line? I want to make my code cleaner and reduce clutter
Instead of doing multiple concatenations, try creating a formatted string. Try something like this:
#include <stdio.h>
#include <string.h>
int main()
{
char *before_colon = "Text before colon";
char *after_colon = "Text after colon";
// Make a string that is the size of both formatted strings, plus a
// character for the space, colon, and null character.
char final_string[strlen(before_colon) + strlen(after_colon) + 3];
// This works just like any other C formatted function (i.e printf, scanf)
sprintf(final_string, "%s: %s", before_colon, after_colon);
printf("%s\n", final_string);
}
output:
Text before colon: Text after colon
Here's a modified code of Charlie Sale with its own function to count characters in the string. Thus, StrLen is called in array declaration.
#include <stdio.h>
#include <string.h>
int StrLen(char* PtrFirstChar)
{
int n = 0;
while(*(PtrFirstChar++)) // will evaluate to FALSE when '\0' reached
n++;
return n;
}
int main()
{
char *before_colon = "Text before colon";
char *after_colon = "Text after colon";
// Make a string that is the size of both formatted strings, plus a
// character for the space, colon, and null character.
char final_string[StrLen(before_colon) + StrLen(after_colon) + 3];
// This works just like any other C formatted function (i.e printf, scanf)
sprintf(final_string, "%s: %s", before_colon, after_colon);
printf("%s\n", final_string);
}
You could write your own variant of strcat!
I'm going to use strncat as a basis because strcat is a really bad idea:
#include <stddef.h> /* for size_t */
#include <stdarg.h> /* for va_* */
char *
mstrncat(char *d, size_t maxlen, ...)
{
va_list ap;
va_start(ap, maxlen);
char *ret = d;
/* Fast-forward */
for (; *d && maxlen; ++d, --maxlen);
/* Reserve a space for the terminator */
if (maxlen)
--maxlen;
const char *p;
/* Concatenate parameters one by one */
while (maxlen && (p = va_arg(ap, const char *))) {
while (*p && maxlen--)
*d++ = *p++;
}
/* Terminate the string */
*d = 0;
va_end(ap);
return ret;
}
You can use it like this:
#include <stdio.h>
int
main()
{
char test[128]="test";
mstrncat(test, sizeof(test), "1", "two", "3", NULL);
puts(test);
return 0;
}
Related
I need the string "on" to be replaced with "in", strstr() function returns a pointer to a string so i figured assigning the new value to that pointer would work but it didn't
#include <stdio.h>
#include <string.h>
int main(void) {
char *m = "cat on couch";
*strstr(m, "on") = "in";
printf("%s\n", m);
}
Replacing a substring with another is easy if both substrings have the same length:
locate the position of the substring with strstr
if it is present, use memcpy to overwrite it with the new substring.
assigning the pointer with *strstr(m, "on") = "in"; is incorrect and should generate a compiler warning. You would avoid such mistakes with gcc -Wall -Werror.
note however that you cannot modify a string literal, you need to define an initialized array of char so you can modify it.
Here is a corrected version:
#include <stdio.h>
#include <string.h>
int main(void) {
char m[] = "cat on couch";
char *p = strstr(m, "on");
if (p != NULL) {
memcpy(p, "in", 2);
}
printf("%s\n", m);
return 0;
}
If the replacement is shorter, the code is a little more complicated:
#include <stdio.h>
#include <string.h>
int main(void) {
char m[] = "cat is out roaming";
char *p = strstr(m, "out");
if (p != NULL) {
memcpy(p, "in", 2);
memmove(p + 2, p + 3, strlen(p + 3) + 1);
}
printf("%s\n", m);
return 0;
}
In the generic case, it is even more complicated and the array must be large enough to accommodate for the length difference:
#include <stdio.h>
#include <string.h>
int main(void) {
char m[30] = "cat is inside the barn";
char *p = strstr(m, "inside");
if (p != NULL) {
memmove(p + 7, p + 6, strlen(p + 6) + 1);
memcpy(p, "outside", 7);
}
printf("%s\n", m);
return 0;
}
Here is a generic function that handles all cases:
#include <stdio.h>
#include <string.h>
char *strreplace(char *s, const char *s1, const char *s2) {
char *p = strstr(s, s1);
if (p != NULL) {
size_t len1 = strlen(s1);
size_t len2 = strlen(s2);
if (len1 != len2)
memmove(p + len2, p + len1, strlen(p + len1) + 1);
memcpy(p, s2, len2);
}
return s;
}
int main(void) {
char m[30] = "cat is inside the barn";
printf("%s\n", m);
printf("%s\n", strreplace(m, "inside", "in"));
printf("%s\n", strreplace(m, "in", "on"));
printf("%s\n", strreplace(m, "on", "outside"));
return 0;
}
There are a few problems with this approach. First, off, m is pointing to read-only memory, so attempting to overwrite the memory there it is undefined behavior.
Second, the line: strstr(m, "on") = "in" is not going to change the pointed-to string, but instead reassign the pointer.
Solution:
#include <stdio.h>
#include <string.h>
int main(void)
{
char m[] = "cat on couch";
memcpy(strstr(m, "on"), "in", 2);
printf("%s\n", m);
}
Note that if you had just used plain strcpy it would null-terminate after "cat in", so memcpy is necessary here. strncpy will also work, but you should read this discussion before using it.
It should also be known that if you are dealing with strings that are not hard-coded constants in your program, you should always check the return value of strstr, strchr, and related functions for NULL.
This function performs a generic pattern replace for all instances of a substring with a replacement string. It allocates a buffer of the correct size for the result. Behaviour is well defined for the case of the empty substring corresponding to the javascript replace() semantics. Where possible memcpy is used in place of strcpy.
/*
* strsub : substring and replace substring in strings.
*
* Function to replace a substring with a replacement string. Returns a
* buffer of the correct size containing the input string with all instances
* of the substring replaced by the replacement string.
*
* If the substring is empty the replace string is written before each character
* and at the end of the string.
*
* Returns NULL on error after setting the error number.
*
*/
char * strsub (char *input, char *substring, char *replace)
{
int number_of_matches = 0;
size_t substring_size = strlen(substring), replace_size = strlen(replace), buffer_size;
char *buffer, *bp, *ip;
/*
* Count the number of non overlapping substring occurences in the input string. This
* information is used to calculate the correct buffer size.
*/
if (substring_size)
{
ip = strstr(input, substring);
while (ip != NULL)
{
number_of_matches++;
ip = strstr(ip+substring_size, substring);
}
}
else
number_of_matches = strlen (input) + 1;
/*
* Allocate a buffer of the correct size for the output.
*/
buffer_size = strlen(input) + number_of_matches*(replace_size - substring_size) + 1;
if ((buffer = ((char *) malloc(buffer_size))) == NULL)
{
errno=ENOMEM;
return NULL;
}
/*
* Rescan the string replacing each occurence of a match with the replacement string.
* Take care to copy buffer content between matches or in the case of an empty find
* string one character.
*/
bp = buffer;
ip = strstr(input, substring);
while ((ip != NULL) && (*input != '\0'))
{
if (ip == input)
{
memcpy (bp, replace, replace_size+1);
bp += replace_size;
if (substring_size)
input += substring_size;
else
*(bp++) = *(input++);
ip = strstr(input, substring);
}
else
while (input != ip)
*(bp++) = *(input++);
}
/*
* Write any remaining suffix to the buffer, or in the case of an empty find string
* append the replacement pattern.
*/
if (substring_size)
strcpy (bp, input);
else
memcpy (bp, replace, replace_size+1);
return buffer;
}
For testing purposes I include a main program that uses the replacement function.
#define BUFSIZE 1024
char * read_string (const char * prompt)
{
char *buf, *bp;
if ((buf=(char *)malloc(BUFSIZE))==NULL)
{
error (0, ENOMEM, "Memory allocation failure in read_string");
return NULL;
}
else
bp=buf;
printf ("%s\n> ", prompt);
while ((*bp=getchar()) != '\n')bp++;
*bp = '\0';
return buf;
}
int main ()
{
char * input_string = read_string ("Please enter the input string");
char * pattern_string = read_string ("Please enter the test string");
char * replace_string = read_string ("Please enter the replacement string");
char * output_string = strsub (input_string, pattern_string, replace_string);
printf ("Result :\n> %s\n", output_string);
free (input_string);
free (pattern_string);
free (replace_string);
free (output_string);
exit(0);
}
I have the following string abcd1234 and I want to find a way to break this string into two different strings, abcd and 1234. I have tried the following code:
char buf[100],*str1,*str2;
int x;
fgets(buf,sizeof(buf),stdin);
str1=strtok(buf,"0123456789 \t\n");
str2=strtok(NULL," \n\t\0");
puts(str1);
puts(str2);
x=atoi(str2);
printf("x=%d", x);
but output is abcd 234. And if I try it with one letter and one number, e.g a2 I take only e on output and x is 0.
As per the man page of strtok()
Each call to strtok() returns a pointer to a null-terminated string containing the next token. This string does not include the delimiting byte. [...]
So, while using "0123456789 \t\n" as the delimiter for the first time, 1 will be treated as the actual delimiter and will not be considered in the subsequent parsing.
You may want to use strcspn() and/or strpbrk() to find out the index for the required sub-strings and parse accordingly.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
size_t extract(const char **sp, char *out, int (*test)(int ch));
int main(void){
char buf[100], str1[100], str2[100];
int x;
const char *p = buf;
//size_t len;
fgets(buf, sizeof(buf), stdin);
while(*p){
if(isalpha((unsigned char)*p)){
extract(&p, str1, isalpha);
puts(str1);
} else if(isdigit((unsigned char)*p)){
extract(&p, str2, isdigit);
x = atoi(str2);
printf("%s, x=%d\n", str2, x);
} else {
++p;//skip one char
}
}
return 0;
}
size_t extract(const char **sp, char *out, int (*test)(int ch)){
const char *p = *sp;
while(*p && test((unsigned char)*p)){
*out++ = *p++;
}
*out = '\0';
size_t len = p - *sp;
*sp = p;
return len;
}
Try below code.Hope this will help you.
#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main()
{
char string[]="abcd1234";
char digitStr[10];
char charStr[10];
int i,j = 0,k = 0;
for(i=0;string[i];i++)
{
if(isdigit(string[i]))
{
charStr[j++]=string[i];
}
else
{
digitStr[k++]=string[i];
}
}
charStr[j] = '\0';
digitStr[k] = '\0';
printf("%s %s\n",digitStr,charStr);
}
I realize I'm very late on this one, but this is for if anyone has a similar case
Assuming all input strings are like your example, this method will work.
char buf[100];
fgets(buf, sizeof(buf), stdin);
if (buf[strlen(buf) - 1] == '\n')
buf[strlen(buf) - 1] = '\0';
int x = atoi(strpbrk(buf, "0123456789"));
char letters[number - buf + 1];
memcpy(letters, sizeof(letters) - 1, buf);
letters[sizeof(letters) - 1] = '\0';
//letters is the word
//x is the number as an int, not a string
• Note the if statement after the fgets. This checks that the newline character was read by fgets, and turns it into a NUL character. (essentially truncating the string).
• As for strpbrk(), that's just a function that returns a pointer to the first occurence of any character in the second string inside the first string. I use it here to find the start of the digit sequence.
• I would also drop the atoi() for strtol() for safety.
• The letters[] array size is the return of strpbrk() (the address of the first number), minus the start of the array (giving the length of the letter string in bytes), plus one for the NUL character I add later.
So I want to be able to somehow change a string into hex like so: "ab.c2" --> "61622e6332". All the help I've found online shows how to do it by using print, but I don't want to use print because it doesn't store the hex value.
What I know so far is that if you take a char and cast it to an int you get the ascii value and with that I can get the hex, which is where I'm stumped.
Here's one way to do it, a complete program but the "meat" is in the tohex function:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char * tohex (unsigned char *s) {
size_t i, len = strlen (s) * 2;
// Allocate buffer for hex string result.
// Only output if allocation worked.
char *buff = malloc (len + 1);
if (buff != NULL) {
// Each char converted to hex digit string
// and put in correct place.
for (i = 0; i < len ; i += 2) {
sprintf (&(buff[i]), "%02x", *s++);
}
}
// Return allocated string (or NULL on failure).
return buff;
}
int main (void) {
char *input = "ab.c2";
char *hexbit = tohex (input);
printf ("[%s] -> [%s]\n", input, hexbit);
free (hexbit);
return 0;
}
There are of course other ways to achieve the same result, such as avoiding memory allocation if you can ensure you provide your own buffer that's big enough, something like:
#include <stdio.h>
void tohex (unsigned char *in, char *out) {
while (*in != '\0') {
sprintf (out, "%02x", *in++);
out += 2;
}
}
int main (void) {
char input[] = "ab.c2";
char output[sizeof(input) * 2 - 1];
tohex (input, output);
printf ("[%s] -> [%s]\n", input, output);
return 0;
}
I'm new to programming,and I have a small problem.
I have a file named questions.txt containing a string of questions, I want to read the string from the file then split it into array with each question having an index, for example a[i] = "Question i" etc.
I did so many tries, but it always ends up reading the last line in the file, when write a loop the program stops working.
This is what i came up with, it's all probably wrong:
char str[200];
char *ptr;
FILE * fp = fopen("questions.txt", "r");
while(fgets(str, 200, fp)!= NULL)
printf("%s", str);
ptr = strtok(str, "\n");
while(ptr != NULL)
{
ptr = strtok(str, "\n");
printf("%s\n", ptr);
ptr = strtok(NULL, "\n");
}
fclose(fp);
The file is:
what is your course?
who is your instructor?
Output i get is:
what is your course?
who is your instructor?
who is your instructor?
I want to read the string from the file then split it into an array with each question having an index...
Let me say, that you don't have a string to split into array.
You should better have a file with a one string of questions like this:
what is your course?:who is your instructor? // `:` is some kind of delimiter
I can suppose that you want to make a vector (one dimensional array) of the file. And in that vector, each element will contain a question from the file. Right?
I can share with you a function from my library I've made at the university. I'll write here a simple program. But it uses delimiters - :, for example. You can modify this function for working without delimiters -- this only depends on you.
In two words, this little program does the following:
// BEFORE: you have a string that ends with a null terminating character.
Question_1_abcbadsad:QUestion_2asasdasd:Question_3sldasdsa\n
^
here ^<< printing 'string' stops
// AFTER: an array of questions. Each of them ends with a null terminating character.
Question_1_abcbadsad\nQUestion_2asasdasd\nQuestion_3sldasdsa\n
^
^<< printing argz[0] will stop here
main.c
#include "argz.h"
int main()
{
error_t func;
char *argz; // pointer to a vector; to an array of questions
size_t argz_len;
// size of that vector (the size of the string you've got from the file)
// Consider `string` to be your `ptr`. You read a string from the file so
// `ptr` will point to the string.
char *string = "Question_1_abcbadsad:QUestion_2asasdasd:Question_3sldasdsa";
// Here `:` is a separate character.
func = argz_create_sep(string, ':', &argz, &argz_len);
if(func == OK)
argz_print(argz, argz_len);
else
printf("ERROR\n");
return 0;
}
argz.c
#include "argz.h"
error_t argz_create_sep (const char *string, int sep, char **argz, size_t *argz_len)
{
int i;
char *ch;
int len = strlen(string);
if(len==0)
return ENOMEM;
*argz = (char*) malloc (sizeof(char)*(len + 1));
strcpy(*argz, string);
*argz_len = strlen(*argz);
ch = *argz;
for(i = 0; i < len; i++) {
if(*ch == sep) *ch='\0';
ch++;
}
return OK;
}
void argz_print(const char *argz, size_t argz_len)
{
const char *ch;
int i;
ch = argz;
for(i = 0; i < argz_len; i++) {
if(*ch == '\0')
printf("\n");
else
printf("%c",*ch);
ch++;
}
printf("\n\n\n");
}
argz.h
#include <stddef.h> // for size_t
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
typedef enum {OK, ENOMEM} error_t;
/* function prototypes */
error_t argz_create_sep (const char *string, int sep, char **argz, size_t *argz_len);
void argz_print (const char *argz, size_t argz_len);
I think what you want is something like this:
#include <stdio.h>
int main(){
int i=0;
char str[200],s='1'; //s is give a random character
FILE * fp = fopen("questions.txt", "r");
while (s!=EOF){ //works until s= the end of file
s=getc(fp); //s starts to receive characters from text file
str[i]=s; //each character of text is placed into the string array
i++;
}
str[i]='\0'; //s reached EOF so lets indicate where we stopped in the string
fclose(fp);
printf("%s\n",str);
//EDIT: changing 1D str to 2D str2
char str2[10][200]; // 10 for max no. of questions, 200 - length of each question
int j=0,k=0;
i=0;
for(j=0;j<200;j++){
str2[i][k]=str[j];
k++;
if (str[j]=='\n'){
i++;
k=0;}
}
for(i=0;i<10;i++) //prints your 2D string array
printf("%s",str2[i]); //after the last question there will be junk
return 0;
}
How do I add two strings?
I tried name = "derp" + "herp";, but I got an error:
Expression must have integral or enum type
C does not have the support for strings that some other languages have. A string in C is just a pointer to an array of char that is terminated by the first null character. There is no string concatenation operator in C.
Use strcat to concatenate two strings. You could use the following function to do it:
#include <stdlib.h>
#include <string.h>
char* concat(const char *s1, const char *s2)
{
char *result = malloc(strlen(s1) + strlen(s2) + 1); // +1 for the null-terminator
// in real code you would check for errors in malloc here
strcpy(result, s1);
strcat(result, s2);
return result;
}
This is not the fastest way to do this, but you shouldn't be worrying about that now. Note that the function returns a block of heap allocated memory to the caller and passes on ownership of that memory. It is the responsibility of the caller to free the memory when it is no longer needed.
Call the function like this:
char* s = concat("derp", "herp");
// do things with s
free(s); // deallocate the string
If you did happen to be bothered by performance then you would want to avoid repeatedly scanning the input buffers looking for the null-terminator.
char* concat(const char *s1, const char *s2)
{
const size_t len1 = strlen(s1);
const size_t len2 = strlen(s2);
char *result = malloc(len1 + len2 + 1); // +1 for the null-terminator
// in real code you would check for errors in malloc here
memcpy(result, s1, len1);
memcpy(result + len1, s2, len2 + 1); // +1 to copy the null-terminator
return result;
}
If you are planning to do a lot of work with strings then you may be better off using a different language that has first class support for strings.
#include <stdio.h>
int main(){
char name[] = "derp" "herp";
printf("\"%s\"\n", name);//"derpherp"
return 0;
}
David Heffernan explained the issue in his answer, and I wrote the improved code. See below.
A generic function
We can write a useful variadic function to concatenate any number of strings:
#include <stdlib.h> // calloc
#include <stdarg.h> // va_*
#include <string.h> // strlen, strcpy
char* concat(int count, ...)
{
va_list ap;
int i;
// Find required length to store merged string
int len = 1; // room for NULL
va_start(ap, count);
for(i=0 ; i<count ; i++)
len += strlen(va_arg(ap, char*));
va_end(ap);
// Allocate memory to concat strings
char *merged = calloc(sizeof(char),len);
int null_pos = 0;
// Actually concatenate strings
va_start(ap, count);
for(i=0 ; i<count ; i++)
{
char *s = va_arg(ap, char*);
strcpy(merged+null_pos, s);
null_pos += strlen(s);
}
va_end(ap);
return merged;
}
Usage
#include <stdio.h> // printf
void println(char *line)
{
printf("%s\n", line);
}
int main(int argc, char* argv[])
{
char *str;
str = concat(0); println(str); free(str);
str = concat(1,"a"); println(str); free(str);
str = concat(2,"a","b"); println(str); free(str);
str = concat(3,"a","b","c"); println(str); free(str);
return 0;
}
Output:
// Empty line
a
ab
abc
Clean-up
Note that you should free up the allocated memory when it becomes unneeded to avoid memory leaks:
char *str = concat(2,"a","b");
println(str);
free(str);
I'll assume you need it for one-off things. I'll assume you're a PC developer.
Use the Stack, Luke. Use it everywhere. Don't use malloc / free for small allocations, ever.
#include <string.h>
#include <stdio.h>
#define STR_SIZE 10000
int main()
{
char s1[] = "oppa";
char s2[] = "gangnam";
char s3[] = "style";
{
char result[STR_SIZE] = {0};
snprintf(result, sizeof(result), "%s %s %s", s1, s2, s3);
printf("%s\n", result);
}
}
If 10 KB per string won't be enough, add a zero to the size and don't bother, - they'll release their stack memory at the end of the scopes anyway.
You should use strcat, or better, strncat. Google it (the keyword is "concatenating").
You cannot add string literals like that in C. You have to create a buffer of size of string literal one + string literal two + a byte for null termination character and copy the corresponding literals to that buffer and also make sure that it is null terminated. Or you can use library functions like strcat.
Concatenate Strings
Concatenating any two strings in C can be done in atleast 3 ways :-
1) By copying string 2 to the end of string 1
#include <stdio.h>
#include <string.h>
#define MAX 100
int main()
{
char str1[MAX],str2[MAX];
int i,j=0;
printf("Input string 1: ");
gets(str1);
printf("\nInput string 2: ");
gets(str2);
for(i=strlen(str1);str2[j]!='\0';i++) //Copying string 2 to the end of string 1
{
str1[i]=str2[j];
j++;
}
str1[i]='\0';
printf("\nConcatenated string: ");
puts(str1);
return 0;
}
2) By copying string 1 and string 2 to string 3
#include <stdio.h>
#include <string.h>
#define MAX 100
int main()
{
char str1[MAX],str2[MAX],str3[MAX];
int i,j=0,count=0;
printf("Input string 1: ");
gets(str1);
printf("\nInput string 2: ");
gets(str2);
for(i=0;str1[i]!='\0';i++) //Copying string 1 to string 3
{
str3[i]=str1[i];
count++;
}
for(i=count;str2[j]!='\0';i++) //Copying string 2 to the end of string 3
{
str3[i]=str2[j];
j++;
}
str3[i]='\0';
printf("\nConcatenated string : ");
puts(str3);
return 0;
}
3) By using strcat() function
#include <stdio.h>
#include <string.h>
#define MAX 100
int main()
{
char str1[MAX],str2[MAX];
printf("Input string 1: ");
gets(str1);
printf("\nInput string 2: ");
gets(str2);
strcat(str1,str2); //strcat() function
printf("\nConcatenated string : ");
puts(str1);
return 0;
}
Without GNU extension:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void) {
const char str1[] = "First";
const char str2[] = "Second";
char *res;
res = malloc(strlen(str1) + strlen(str2) + 1);
if (!res) {
fprintf(stderr, "malloc() failed: insufficient memory!\n");
return EXIT_FAILURE;
}
strcpy(res, str1);
strcat(res, str2);
printf("Result: '%s'\n", res);
free(res);
return EXIT_SUCCESS;
}
Alternatively with GNU extension:
#define _GNU_SOURCE
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void) {
const char str1[] = "First";
const char str2[] = "Second";
char *res;
if (-1 == asprintf(&res, "%s%s", str1, str2)) {
fprintf(stderr, "asprintf() failed: insufficient memory!\n");
return EXIT_FAILURE;
}
printf("Result: '%s'\n", res);
free(res);
return EXIT_SUCCESS;
}
See malloc, free and asprintf for more details.
#include <string.h>
#include <stdio.h>
int main()
{
int a,l;
char str[50],str1[50],str3[100];
printf("\nEnter a string: ");
scanf("%s",str);
str3[0]='\0';
printf("\nEnter the string which you want to concat with string one: ");
scanf("%s",str1);
strcat(str3,str);
strcat(str3,str1);
printf("\nThe string is %s\n",str3);
}
using memcpy
char *str1="hello";
char *str2=" world";
char *str3;
str3=(char *) malloc (11 *sizeof(char));
memcpy(str3,str1,5);
memcpy(str3+strlen(str1),str2,6);
printf("%s + %s = %s",str1,str2,str3);
free(str3);
my here use asprintf
sample code:
char* fileTypeToStr(mode_t mode) {
char * fileStrBuf = NULL;
asprintf(&fileStrBuf, "%s", "");
bool isFifo = (bool)S_ISFIFO(mode);
if (isFifo){
asprintf(&fileStrBuf, "%s %s,", fileStrBuf, "FIFO");
}
...
bool isSocket = (bool)S_ISSOCK(mode);
if (isSocket){
asprintf(&fileStrBuf, "%s %s,", fileStrBuf, "Socket");
}
return fileStrBuf;
}
In C, you don't really have strings, as a generic first-class object. You have to manage them as arrays of characters, which mean that you have to determine how you would like to manage your arrays. One way is to normal variables, e.g. placed on the stack. Another way is to allocate them dynamically using malloc.
Once you have that sorted, you can copy the content of one array to another, to concatenate two strings using strcpy or strcat.
Having said that, C do have the concept of "string literals", which are strings known at compile time. When used, they will be a character array placed in read-only memory. It is, however, possible to concatenate two string literals by writing them next to each other, as in "foo" "bar", which will create the string literal "foobar".