You could use strdup() to return a copy of a C-string, as in:

#include <string.h>

const char *stringA = "foo";
char *stringB = NULL;

stringB = strdup(stringA);
/* ... */
free(stringB);
stringB = NULL; 

You could also use strcpy(), but you need to allocate space first, which isn't hard to do but can lead to an overflow error, if not done correctly:

#include <string.h>

const char *stringA = "foo";
char *stringB = NULL;

/* you must add one to cover the byte needed for the terminating null character */
stringB = (char *) malloc( strlen(stringA) + 1 ); 
strcpy( stringB, stringA );
/* ... */
free(stringB);
stringB = NULL;

If you cannot use strdup(), I would recommend the use of strncpy() instead of strcpy(). The strncpy() function copies up to — and only up to — n bytes, which helps avoid overflow errors. If strlen(stringA) + 1 > n, however, you would need to terminate stringB, yourself. But, generally, you'll know what sizes you need for things:

#include <string.h>

const char *stringA = "foo";
char *stringB = NULL;

/* you must add one to cover the byte needed for the terminating null character */
stringB = (char *) malloc( strlen(stringA) + 1 ); 
strncpy( stringB, stringA, strlen(stringA) + 1 );
/* ... */
free(stringB);
stringB = NULL;

I think strdup() is cleaner, myself, so I try to use it where working with strings exclusively. I don't know if there are serious downsides to the POSIX/non-POSIX approach, performance-wise, but I am not a C or C++ expert.

Note that I cast the result of malloc() to char *. This is because your question is tagged as a c++ question. In C++, it is required to cast the result from malloc(). In C, however, you would not cast this.

EDIT

There you go, there's one complication: strdup() is not in C or C++. So use strcpy() or strncp() with a pre-sized array or a malloc-ed pointer. It's a good habit to use strncp() instead of strcpy(), wherever you might use that function. It will help reduce the potential for errors.

A C string is a nul-terminated character array.

The C language does not allow assigning the contents of an array to another
array. As noted by Barry, you must copy the individual characters one by one
from the source array to the destination array. e.g. -

#define _CRT_SECURE_NO_WARNINGS
#include 
#include 

int main()
{
    char str1[] = "Hello";
    char str2[10] = {0};

    for (int x = 0; x < strlen(str1); ++x)
    {
        str2[x] = str1[x];
    }

    printf("%s\n", str2);

    return 0;
}

To make this common task easier there are standard library functions provided
which will perform this operation. e.g. - memcpy(), etc.

memcpy(str2, str1, 6);

When the array contains a nul-terminated string of characters you can use
strcpy(), etc.

strcpy(str2, str1);

Caveat: Some of the above functions are considered unsafe as they do not guard
against buffer overruns of the source and destination arrays. There are safer
versions provided by the compiler.

Note that if and when you start learning C++ you will find that there you can
assign a C++ std::string object to another object of the same type. However,
even in C++ the same rules apply when working with C strings, "raw" character
arrays, etc.

• Wayne

To copy strings in C, you can use strcpy. Here is an example:

#include <stdio.h>
#include <string.h>

const char * my_str = "Content";
char * my_copy;
my_copy = malloc(sizeof(char) * (strlen(my_str) + 1));
strcpy(my_copy,my_str);

If you want to avoid accidental buffer overflows, use strncpy instead of strcpy. For example:

const char * my_str = "Content";
const size_t len_my_str = strlen(my_str) + 1;
char * my_copy = malloc(len_my_str);
strncpy(my_copy, my_str, len_my_str);