Memcpy copies more data into less size

Question

I have declared 2 strings i.e string1 and string2. string1 is of size 8 and string2 is of size 200. Now, I am trying to copy string2 to string1. string1 has lesser size than string2. And I am copying number of bytes which is same as size of string2. My question is why memcpy is allowing to copy more number of bytes into a location where less number of bytes are allocated ? If I print the value of string1 now, then I am getting the same value as string2. But string1 does not have that much memory to accommodate string2. I want to know the reason behind this unusual behaviour.

// C program to demonstrate working of memcpy
#include <stdio.h>
#include <string.h>

int main()
{
    char str1[8] = "Hello\0";
    char str2[200] = "QuizLinkssjdjufuuejjhfgyyryryyryrynshshhsjakskkk\0";

    puts("str1 before memcpy ");
    puts(str1);

    // Copies contents of str2 to str1
    memcpy(str1, str2, sizeof(str2));

    puts("\nstr1 after memcpy ");
    puts(str1);

    return 0;
}

Below is the output.

str1 before memcpy 
Hello


str1 after memcpy 
QuizLinkssjdjufuuejjhfgyyryryyryrynshshhsjakskkk

Answer 1

why memcpy is allowing to copy more number of bytes into a location where less number of bytes are allocated

C doesn't perform any kind of bounds checking on arrays. That's part of what makes it fast. That also means that it's up to the programmer to not write past the bounds of the array. If you do so, you trigger undefined behavior in your code. What you're seeing is one of the ways undefined behavior can manifest.

The solution: don't read or write past the end of an array.

Answer 2

My question is why memcpy is allowing to copy more number of bytes into a location where less number of bytes are allocated ?

Allocation of memory is record-keeping within your process. Your process is granted memory by the operating system that it is allowed to use (and it can request more by various requests to the operating system). In modern systems, that is often a great deal of memory, used for static data, the hardware stack, dynamically allocated memory, program instructions, and more. Within that memory, it is your program’s responsibility to control how it is used.

The operating system and the hardware control attempts to use memory outside of what the operating system has granted you. Inside the memory you are granted, it is your program that determines how the memory is used (with some constraints, such as that some of the memory may be marked read-only or non-executable).

The C standard does not impose requirements beyond this: It does not require C implementations to stop your program from attempting to copy more data into str1 than space has been allocated for. You must know what the limits are and write your code to respect them.

It is possible to design programming languages and environments that track memory allocations and prevent programs from writing beyond the space allocated for an array, but this imposes extra cost on computing. It requires additional memory and processor instructions. C is designed to be fast and efficient, at the expense of burdening the programmer with writing code carefully.