character concatenation in C - c

Question

I want to do in C, what can be achieved in Java as follows
String str = "hello";
System.out.println(str + 'a');
I have written the following.
1. It doesn't work
2. Is there an easier way to do this in C, something that can be achieved in java in a single line.
#include <stdio.h>
char* charcat(char str[], char c);
int main(void)
{
char str[] = "hello";
printf("%s\n",charcat(str,'a'));
}
char* charcat(char str[], char c)
{
char newstr[] = {c,'\0'};
char temp[20];
strcpy(temp,str);
strcat(temp,newstr);
return temp;
}
EDIT :
I have edited based on ameyCU's response.
char* charcat(char str[], char c);
int main(void)
{
char str[] = "hello";
printf("%s\n",charcat(str,'a'));
}
char* charcat(char str[], char c)
{
char* temp;
char newstr[] = {c,'\0'};
temp = malloc((strlen(str) + 1)* sizeof(char));
strcpy(temp,str);
return strcat(temp,newstr);
}
EDIT 2:
char* charcat(char str[], char c);
int main(void)
{
char str[] = "hello";
char temp[20];
strcpy(temp,str);
printf("%s\n",charcat(temp,'a'));
}
char* charcat(char str[], char c)
{
char newstr[] = {c,'\0'};
return strcat(str,newstr);
}

Answer 1

I think what you were trying to do was this:
char* charcat(char str[], char c)
{
char newstr[] = {c,'\0'};
char *temp=(char *)malloc((strlen(str)+1+1)*sizeof(char));
strcpy(temp,str);
strcat(temp,newstr);
return temp;
}
make sure you free() the pointer.

Answer 2

You can use strcat() function
char str1[20]="hello";
strcat(str1,"c");
printf("%s",str1);

Answer 3

Problem is that you return a local variable.
return temp;
temp is local variable and its scope is just inside the function it is declared.
After concatenation -strcat(temp,newstr);
You can do this -
strcpy(str,temp);
return str;
But this will also change the contents of original array.
EDIT
To keep original array intact assign a pointer to string in function and return the pointer .
And also to use functions like strcpy and strcat you need to include string.h header.

Answer 4

This uses snprintf() to get the required length for the target string. Memory is allocated and then snprintf() creates the target string.
#include<stdio.h>
#include<stdlib.h>
char* charcat(char str[], char c);
int main ( ) {
char str[] = "hello";
char *output = NULL;
printf ( "str-> %s\n\n", str);
if ( ( output = charcat ( str, 'a'))) {//successful return of pointer
printf ( "output-> %s\n", output);
free ( output);//release memory
}
return 0;
}
char* charcat(char str[], char c)
{
char *temp = NULL;
int length = 0;
length = snprintf ( NULL, 0, "%s%c", str, c);//get required length
if ( ( temp = malloc ( length + 1))) {//allocate plus one for '\0'
snprintf ( temp, length + 1, "%s%c", str, c);//malloc succeeds make target
}
return temp;
}

Answer 5

It is always better to use strncat() instead of strcat to avoid buffer overflows.
#include <cstdio>
#include <cstring>
int main ()
{
char str[20] = "hello";
strncat (str, "a", sizeof(str) - strlen(str) - 1);
printf("%s",str);
return 0;
}
Output:
helloa
RUN SUCCESSFUL (total time: 49ms)

Answer 6

Something like Java in a single line
// String str = "hello";
// System.out.println(str + 'a');
const char *str = "hello";
printf("%s%c\n", str, 'a');
Or is one wants to print a concatenated string, we need to do memory management.
char *charcatconst char *src, int ch) {
size_t len = strlen(src);
char *s = memcpy(malloc(len+2), src, len); // Out-of-memory check omitted.
s[len] = ch;
s[len+1] = '\0';
return s;
}
// simple usage, but a memory leak
puts(charcat("hello", 'a'));
// Better to free memory explicitly
char *s = charcat("hello", 'a');
puts(s);
free(s);