x86 - Looking for help: Explaining assembly code generated by C compiler -


i trying understand assembly instructions generated following c code:

#include<stdio.h> #include<unistd.h>  int main() {       char *happy[2];   happy[0] = "/bin/sh";   happy[1] = null;   execve(happy[0],happy,null);     }

unfortunately can't understand assembly code.
can explain code me? can figure out rax contains address of /bin/sh, don't know use of rsi register here. 

 dump of assembler code function main:    0x0000000000400a0d <+0>:     push   %rbp    0x0000000000400a0e <+1>:     mov    %rsp,%rbp    0x0000000000400a11 <+4>:     sub    $0x20,%rsp    0x0000000000400a15 <+8>:     mov    %fs:0x28,%rax    0x0000000000400a1e <+17>:    mov    %rax,-0x8(%rbp)    0x0000000000400a22 <+21>:    xor    %eax,%eax    0x0000000000400a24 <+23>:    lea    0x8c4d9(%rip),%rax        # 0x48cf04    0x0000000000400a2b <+30>:    mov    %rax,-0x20(%rbp)    0x0000000000400a2f <+34>:    movq   $0x0,-0x18(%rbp)    0x0000000000400a37 <+42>:    mov    -0x20(%rbp),%rax    0x0000000000400a3b <+46>:    lea    -0x20(%rbp),%rcx    0x0000000000400a3f <+50>:    mov    $0x0,%edx    0x0000000000400a44 <+55>:    mov    %rcx,%rsi    0x0000000000400a47 <+58>:    mov    %rax,%rdi    0x0000000000400a4a <+61>:    callq  0x432b10 <execve>    0x0000000000400a4f <+66>:    mov    $0x0,%eax    0x0000000000400a54 <+71>:    mov    -0x8(%rbp),%rdx    0x0000000000400a58 <+75>:    xor    %fs:0x28,%rdx    0x0000000000400a61 <+84>:    je     0x400a68 <main+91>    0x0000000000400a63 <+86>:    callq  0x435730 <__stack_chk_fail_local>    0x0000000000400a68 <+91>:    leaveq    0x0000000000400a69 <+92>:    retq end of assembler dump. `


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