Question

    char test[]={"abcde"};
    char* test1={"xyz"};
    memcpy(test+5,test1,3);
    printf("%s",test);

I’m trying to grasp how exactly memcpy works and this is the example I’ve written so far.
This gives output as abcdexyz&vjunkcharacters

and the following message.

*** stack smashing detected ***: ./testcode terminated
======= Backtrace: =========
/lib/i386-linux-gnu/libc.so.6(__fortify_fail+0x45)[0xb7656dd5]
/lib/i386-linux-gnu/libc.so.6(+0xffd8a)[0xb7656d8a]
./testcode[0x8048797]
/lib/i386-linux-gnu/libc.so.6(__libc_start_main+0xf3)[0xb75704d3]
./testcode[0x80483a1]

What are the reasons behind this situation?

Answer 1

Root Cause:

char test[]={"abcde"};

Allocates enough memory space to store 5 characters only.

memcpy(test+5,test1,3);  

Copies the data pointed by test1 beyond the allocated memory space.
Technically, writing beyond the bounds of an allocated memory in this fashion is Undefined Behavior, which means anything can happen.

What actually happens?

What actually happens here is memcpy copies characters beyond the allocated memory thus overwritting the NULL terminator which marks ends of your character array test.
Further, printf reads the contents from starting address of test till it encounters a random NULL thus printing out junk characters.

Solution:
You should ensure that destination buffer has enough memory allocated before you perform the memcpy. Since you intend to copy 3 characters, Your destination buffer test should be atleast:

5 + 3 + 1 byte for NULL terminator = 9 bytes

You can simply use:

char test[9]="abcde";