Consider the following poorly written C code:

int main() { int n = doCopy(); printf("%x ", n); } int doCopy() { int n = 1; char dest[8]; strcpy(dest, "01234567890123456789"); return n; } 

The gcc compiler (without turning on optimizations) was used to create an executable out of this code and then objdump -d was used to disassemble the executable. The output of the objdump command can be seen below:

0000000000400504 
: 400504: 55 push %rbp 400505: 48 89 e5 mov %rsp,%rbp 400508: 48 83 ec 10 sub $0x10,%rsp 40050c: b8 00 00 00 00 mov $0x0,%eax 400511: e8 24 00 00 00 callq 40053a  400516: 48 89 45 f8 mov %rax,-0x8(%rbp) 40051a: b8 68 06 40 00 mov $0x400668,%eax 40051f: 48 8b 55 f8 mov -0x8(%rbp),%rdx 400523: 48 89 d6 mov %rdx,%rsi 400526: 48 89 c7 mov %rax,%rdi 400529: b8 00 00 00 00 mov $0x0,%eax 40052e: e8 bd fe ff ff callq 4003f0  400533: b8 00 00 00 00 mov $0x0,%eax 400538: c9 leaveq 400539: c3 retq 000000000040053a : 40053a: 55 push %rbp 40053b: 48 89 e5 mov %rsp,%rbp 40053e: 48 83 ec 10 sub $0x10,%rsp 400542: 48 c7 45 f8 01 00 00 movq $0x1,-0x8(%rbp) 400549: 00 40054a: b9 6c 06 40 00 mov $0x40066c,%ecx 40054f: 48 8d 45 f0 lea -0x10(%rbp),%rax 400553: ba 15 00 00 00 mov $0x15,%edx 400558: 48 89 ce mov %rcx,%rsi 40055b: 48 89 c7 mov %rax,%rdi 40055e: e8 ad fe ff ff callq 400410  400563: 48 8b 45 f8 mov -0x8(%rbp),%rax 400567: c9 leaveq 400568: c3 retq 

Suppose that right before the doCopy function is called that the values of %rbp and %rsp are as follows:

%rbp: 0x7fffffffe200 %rsp: 0x7fffffffe1f0 

Now, assume that the memcpy function has been executed and the program is about to return from the doCopy function. What is the return address?

a. 1

b. 0x0000000039383736

c. 0x0000000000400516

d. 0x3736353433323130

e. 0x3534333231303938

f. 0x00007f0039383736

please explain for me