Assigning char to char* using pointers

char *str is a pointer to a char (or an array of chars), however, you never assigned it. As has been mentioned earlier a char * basically says "go there" but there is no there there, you never gave it a value. You first need to use malloc to create space to put things in. Here's an example

char *str = malloc(sizeof(char)*10) //allocate space for 10 chars
char c = 'a';
str[0] = c;

no error check was made to malloc which you should do in your own program. You can also do it as such

char str[10];
char c = 'a';
str[0] = c;

however with this method you will be restricted to 10 chars and you cannot change that amount, with the previous method you can use realloc to get more or less space in your array.

But it doesn't work.

char* str;

... is not initialized to anything, therefore dereferencing it is to undefined behaviour. If it where initialized, then in expression *str++ = c; str++ is a post-increment operator, which returns a copy of the pointer whilst incrementing the original. The effect is that the copy points to the previous, and therefore what is pointed to by the previous pointer is assigned c.

To which part that doesn't work are you referring?

EDIT:

As mentioned in one of the comments, a copy is not really returned but the value is increment in place after having been evaluated.

As a variable with automatic storage duration the pointer str has indeterminate value. If even it had the static storage duration its value would be NULL. So you may not use such a pointer to store data.

What you mean can look for example the following way

#include <stdio.h>

int main( void )
{
    char s[11];

    char *p = s;

    while (p != s + sizeof( s ) / sizeof( *s ) - 1 ) *p++ = 'a';
    *p = '\0';

    puts(s);

    return 0;
}

The program output is

aaaaaaaaaa

Here in the program the pointer p of the type char * is initialized by the address of the first character of the array s.

Thus this statement used in the loop

*p++ = 'a';

fills sequentially the array with the character 'a'.

The next example is more interesting

#include <stdio.h>

char * copy_string(char *dsn, const char *src)
{
    for (char *p = dsn; (*p++ = *src++) != '\0'; )
    {
        //  empty body
    }

    return dsn;
}

int main( void )
{
    char *src = "Hi QBl";
    char dsn[7];

    puts(copy_string(dsn, src));

    return 0;
}

The program output is

Hi QBl

Here is a demonstration of a function that copies one character array containing a string into another character array using pointers.

To assign each character of string 1 to that of string 2, an equivalent length should be allocated for the pointer variable of string 2.

#include <stdio.h>
#include <stdlib.h>

int main()
{   
    char *str = "Hello, world!";
    char *p = (char*)malloc(sizeof(char) * strlen(str));
    *p = '\0';
    char *pHead = p;
    printf("*str = %s\n", str);
    printf("*p = %s\n", pHead);

    while(*str != '\0')
    {
        *p = *str;
        str++;
        p++;
    }

    printf("*p = %s\n", pHead);
}

The output is:

*str = Hello, world!
*p = 
*p = Hello, world!