why does strcpy copies more character to the variable than it is supposed to?

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I am currently working on with lots of strcpy' and calloc's. And then I heard that strncpy is safer to use. So what I did was create a function that will handle strcpy.. It is shown below.

void safeStrncpy(char * dest, char * src){
    //copy string 
    if(sizeof(dest) >= strlen(src) + 1){
        strncpy(dest, src, strlen(src));
    }else{
        if(realloc(dest, (strlen(src)) + 1) == NULL){
            printf("error");    
        }else{
            strncpy(dest, src, strlen(src));
        }
    }
}

You'll notice that I used sizeof(dest) . What I really want to do with that part is get the size of the memory allocated to dest so I'll know when to use realloc. But then I learned that you can't get the size of memory allocated to dest so I think of a workaround.

char * l = calloc(10,sizeof(char));
printf("%s", strcpy(l,"asdfghjkldfghasdfghjkl"));

The code shown above allocates 10 items to the pointer l. I thought that If I did that and copied more characters than what is allocated, it will just copy what fits to the size.. I was expecting that the value of l would be "asdfghjkld". But, to my surprise, it didn't. It copied the whole string which is this "asdfghjkldfghasdfghjkl".

Now, why use calloc if it will just be overridden? what happens in the background during char pointer declaration, memory allocation and strcpy? does c automatically reallocates the memory? Do I need to worry of this behavior, security issues maybe?

asked 28 secs ago

user3714598

17

why does strcpy copies more character to the variable than it is supposed to?