C language beginner [closed] - c

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I'm learning C language following Youtube video and I've got couple of questions.
I typed these below in Xcode
#include<stdio.h>
int main() // 1. why do we have to use this line?
{
char food[] = "tuna";
printf("the best food is %s",food);
strcpy(food,"bacon"); // error here
return 0;
}
When I typed these all, the error come out saying " implicity declaring library function 'strcpy' with type 'char*(char*,const char*)'"
I have no idea what this mean and why it happened? Sorry to bother with question guys but I need your help. Thank you

Answer 1

C is a super great language to learn but because of how low level it is in comparison to python, javascript, etc. There are many manual tasks you need to do including memory management.
But before we get into that, your initial problem is not including the string.h header. Once you include that header, you'll actually get Segmentation Fault.
Below is the explanation on why you'll get the Segmentation Fault.
So when you defined:
char food[] = "tuna"
C created char array with 5 bytes. Why 5? because all strings end with a NULL byte ('\0').
So your string in memory looks like "tuna\0".
Side note:
The importance of the NULL byte is to signify the end of a string. This is super important when you are using a function like printf. Because when printing a string, printf will look for the NULL byte to stop printing.
So this char array has the max size of 5 bytes, 1 being a null byte, so just 4 bytes (chars) of writeable memory.
Then you try to copy the word "bacon" into your 5 byte array.
So strcpy tries to copy the b, then the c, then the o, then the n, then when it tries to terminate the string with a NULL byte, it will seg fault because you're trying to access a byte of memory you don't have access to.
So in order to fix your issue, you could try to strcpy a string that is the same length or shorter than your original string. Or you can look into something like strdup and use pointers.
That code will look like:
#include <stdio.h>
#include <string.h>
int main()
{
char *food = "tuna";
printf("the best food is %s\n",food);
food = strdup("bacon");
printf("the best food is %s\n",food);
return 0;
}
Hope this helps!

Answer 2

I suggest you as exercise to explain me the behavior of this code. You have to understand how "strings" are managed by the C language into the memory.
The code I post below is not right, contains a violation of the space allocated for the variable food. In this case this code doesn't generate a Segmentation Fault, but It might generate such a error!
If you compile the code below with gcc -Wall --pedantic -std=c99 prgname.c
, the compiler doesn't signal warnings nor error, but the code is NOT RIGHT.
PAY ATTENTION! :)
#include <stdio.h>
#include <string.h>
int main() // 1. why do we have to use this line?
{ // --- 'cause is a standard!!! :)
char food[] = "tuna";
char b[]="prova";
printf("the best food is %s - %s - food-size %u\n",food,b,(unsigned int)sizeof(food));
strcpy(food,"fried or roasted bacon"); // error here
printf("the best food is %s - %s - food-size %u\n",food,b,(unsigned int)sizeof(food));
return 0;
}

Answer 3

We have to use main(),this is the function through which execution of C program starts.
For using strcpy()(A predefined function) you need to declare it first. and its declaration is present in string.h so just #include<string.h>
This will resolve your problem.implicit declaring library function 'strcpy' with type'char*(char*,const char*)``
There is one more problem when you declare
char food[]="tuna";
you get only 5 bytes allocated for you,four for character and one more for NULL. When you will try putting baconin it (using strcpy()),which is 6 bytes long. It will cause compiler to create segmentation fault.