Simple c malloc

Question

this doesn't work:

void function(char* var)
{
    var = (char*) malloc (100);
}

int main()
{
    char* str;
    function(str);
    strcpy(str, "some random string");
    printf("%s\n", str);

    return 0;
}

this does:

void function(char* var)
{
    //var = (char*) malloc (100);
}

int main()
{
    char* str;
    //function(str);
    str = (char*) malloc (100);
    strcpy(str, "some random string");
    printf("%s\n", str);

    return 0;
}

Why?

Answer 1

You have to pass the address of the pointer to assign the address you want inside the function, otherwise you are just passing a copy of it:

void function(char** var)
{
    *var = malloc (100);
}

int main()
{
    char* str;
    function(&str);
    strcpy(str, "some random string");
    printf("%s\n", str);

    return 0;
}