ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的：

```asm
<itoa>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  test   %edi,%edi
  jns    L0
  movb   $0x2d,(%rsi)
  mov    $0x1,%ebx
L0:
  mov    %edi,0xc(%rsp)
  call   <nsize@plt>
  mov    0xc(%rsp),%edi
  add    %ebx,%eax
  movslq %eax,%rdx
  sub    $0x1,%eax
  cltq
  movb   $0x0,(%r12,%rdx,1)
  add    %rax,%r12
  nopl   0x0(%rax,%rax,1)
L8:
  movslq %edi,%rbx
  mov    %edi,%eax
  sub    $0x1,%r12
  imul   $0x66666667,%rbx,%rbx
  sar    $0x1f,%eax
  sar    $0x22,%rbx
  sub    %eax,%ebx
  lea    (%rbx,%rbx,4),%eax
  add    %eax,%eax
  sub    %eax,%edi
  call   <abs@plt>
  mov    %ebx,%edi
  add    $0x30,%eax
  mov    %al,0x1(%r12)
  test   %ebx,%ebx
  jne    L8
  add    $0x18,%rsp
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
void itoa(int n, char s[]) {
  int i, k;
  i = 0;
  if (n < 0) {
    s[i++] = '-';
  }
  k = nsize(n) + i;
  s[k--] = '\0';
  do {
    s[k--] = abs(n % 10) + '0';
  } while (n /= 10);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<itoa>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  test   %edi,%edi
  jns    L3
  movb   $0x2d,(%rsi)
  mov    $0x1,%ebx
L3:
  mov    %edi,0xc(%rsp)
  call   <nsize@plt>
  mov    0xc(%rsp),%edi
  add    %ebx,%eax
  movslq %eax,%rdx
  sub    $0x1,%eax
  cltq
  movb   $0x0,(%r12,%rdx,1)
  add    %rax,%r12
  nopl   0x0(%rax,%rax,1)
L13:
  movslq %edi,%rbx
  mov    %edi,%eax
  sub    $0x1,%r12
  imul   $0x66666667,%rbx,%rbx
  sar    $0x1f,%eax
  sar    $0x22,%rbx
  sub    %eax,%ebx
  lea    (%rbx,%rbx,4),%eax
  add    %eax,%eax
  sub    %eax,%edi
  call   <abs@plt>
  mov    %ebx,%edi
  add    $0x30,%eax
  mov    %al,0x1(%r12)
  test   %ebx,%ebx
  jne    L13
  add    $0x18,%rsp
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
void itoa(int n, char s[]) {
  int i, k;
  i = 0;
  if (n < 0) {
    s[i++] = '-';
  }
  k = nsize(n) + i;
  s[k--] = '\0';
  do {
    s[k--] = abs(n % 10) + '0';
  } while (n /= 10);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<insertStrAt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    %r8,-0x30(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x20(%rbp),%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x24(%rbp),%eax
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x30(%rbp),%rax
  add    %rax,%rcx
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x20(%rbp),%eax
  cltq
  add    %rax,%rsi
  mov    -0x30(%rbp),%rax
  add    %rsi,%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memmove@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L18
L6:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%ecx
  mov    -0x1c(%rbp),%edx
  add    %ecx,%edx
  movslq %edx,%rcx
  mov    -0x30(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x4(%rbp)
L18:
  mov    -0x4(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L6
  nop
  leave
  ret
```

assistant: ```c
insertStrAt(char s[], int insertAt, int slen, int sourceSIZE, char source[]) {
  int i;
  memmove(source + insertAt + slen, source + insertAt, sourceSIZE - (insertAt + slen));
  for (i = 0; i < slen; i++) {
    source[i + insertAt] = s[i];
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<insertStrAt>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %edx,%ebx
  movslq %esi,%rdi
  lea    (%r8,%rdi,1),%r12
  add    %edx,%esi
  sub    %esi,%ecx
  mov    %ecx,%edx
  movslq %ebx,%rax
  add    %rdi,%rax
  lea    (%r8,%rax,1),%rdi
  mov    %r12,%rsi
  call   <memmove@plt>
  test   %ebx,%ebx
  jle    L5
  mov    %ebx,%ecx
  mov    $0x0,%eax
L12:
  movzbl 0x0(%rbp,%rax,1),%edx
  mov    %dl,(%r12,%rax,1)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L12
L5:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
insertStrAt(char s[], int insertAt, int slen, int sourceSIZE, char source[]) {
  int i;
  memmove(source + insertAt + slen, source + insertAt, sourceSIZE - (insertAt + slen));
  for (i = 0; i < slen; i++) {
    source[i + insertAt] = s[i];
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<insertStrAt>:
  endbr64
  push   %r13
  movslq %esi,%rax
  push   %r12
  movslq %edx,%r12
  push   %rbp
  add    %r12d,%esi
  lea    (%r8,%rax,1),%rbp
  add    %r12,%rax
  push   %rbx
  sub    %esi,%ecx
  mov    %rdi,%rbx
  mov    %rbp,%rsi
  mov    %ecx,%edx
  lea    (%r8,%rax,1),%rdi
  sub    $0x8,%rsp
  call   <memmove@plt>
  test   %r12d,%r12d
  jle    L17
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L18:
  movzbl (%rbx,%rax,1),%edx
  mov    %dl,0x0(%rbp,%rax,1)
  add    $0x1,%rax
  cmp    %r12,%rax
  jne    L18
L17:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
insertStrAt(char s[], int insertAt, int slen, int sourceSIZE, char source[]) {
  int i;
  memmove(source + insertAt + slen, source + insertAt, sourceSIZE - (insertAt + slen));
  for (i = 0; i < slen; i++) {
    source[i + insertAt] = s[i];
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<insertStrAt>:
  endbr64
  push   %r14
  movslq %esi,%rax
  mov    %r8,%r14
  push   %r13
  mov    %rax,%r13
  push   %r12
  lea    (%r8,%rax,1),%r12
  push   %rbp
  mov    %rdi,%rbp
  movslq %edx,%rdi
  lea    (%rax,%rdi,1),%esi
  push   %rbx
  mov    %rdi,%rbx
  add    %rax,%rdi
  sub    %esi,%ecx
  add    %r8,%rdi
  mov    %r12,%rsi
  mov    %ecx,%edx
  call   <memmove@plt>
  test   %ebx,%ebx
  jle    L130
  lea    0x1(%rbp),%rsi
  mov    %r12,%rdx
  lea    -0x1(%rbx),%eax
  mov    %ebx,%ecx
  sub    %rsi,%rdx
  cmp    $0xe,%rdx
  jbe    L36
  cmp    $0x6,%eax
  jbe    L36
  cmp    $0xe,%eax
  jbe    L97
  mov    %ebx,%edx
  xor    %eax,%eax
  shr    $0x4,%edx
  shl    $0x4,%rdx
  nopl   (%rax)
L60:
  movdqu 0x0(%rbp,%rax,1),%xmm0
  movups %xmm0,(%r12,%rax,1)
  add    $0x10,%rax
  cmp    %rdx,%rax
  jne    L60
  mov    %ebx,%edx
  and    $0xfffffff0,%edx
  mov    %edx,%eax
  cmp    %edx,%ebx
  je     L130
  mov    %ebx,%ecx
  sub    %edx,%ecx
  lea    -0x1(%rcx),%esi
  cmp    $0x6,%esi
  jbe    L103
L129:
  mov    0x0(%rbp,%rdx,1),%rsi
  mov    %rsi,(%r12,%rdx,1)
  mov    %ecx,%edx
  and    $0xfffffff8,%edx
  add    %edx,%eax
  cmp    %edx,%ecx
  je     L130
L103:
  movslq %eax,%rdx
  movzbl 0x0(%rbp,%rdx,1),%ecx
  lea    0x0(%r13,%rax,1),%edx
  movslq %edx,%rdx
  mov    %cl,(%r14,%rdx,1)
  lea    0x1(%rax),%edx
  cmp    %edx,%ebx
  jle    L130
  movslq %edx,%rcx
  add    %r13d,%edx
  movzbl 0x0(%rbp,%rcx,1),%ecx
  movslq %edx,%rdx
  mov    %cl,(%r14,%rdx,1)
  lea    0x2(%rax),%edx
  cmp    %edx,%ebx
  jle    L130
  movslq %edx,%rcx
  add    %r13d,%edx
  movzbl 0x0(%rbp,%rcx,1),%ecx
  movslq %edx,%rdx
  mov    %cl,(%r14,%rdx,1)
  lea    0x3(%rax),%edx
  cmp    %edx,%ebx
  jle    L130
  movslq %edx,%rcx
  add    %r13d,%edx
  movzbl 0x0(%rbp,%rcx,1),%ecx
  movslq %edx,%rdx
  mov    %cl,(%r14,%rdx,1)
  lea    0x4(%rax),%edx
  cmp    %edx,%ebx
  jle    L130
  movslq %edx,%rcx
  add    %r13d,%edx
  movzbl 0x0(%rbp,%rcx,1),%ecx
  movslq %edx,%rdx
  mov    %cl,(%r14,%rdx,1)
  lea    0x5(%rax),%edx
  cmp    %edx,%ebx
  jle    L130
  movslq %edx,%rcx
  add    %r13d,%edx
  add    $0x6,%eax
  movzbl 0x0(%rbp,%rcx,1),%ecx
  movslq %edx,%rdx
  mov    %cl,(%r14,%rdx,1)
  cmp    %eax,%ebx
  jle    L130
  movslq %eax,%rdx
  add    %eax,%r13d
  movzbl 0x0(%rbp,%rdx,1),%edx
  movslq %r13d,%r13
  mov    %dl,(%r14,%r13,1)
L130:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  cs nopw 0x0(%rax,%rax,1)
L36:
  mov    %ebx,%ebx
  xor    %eax,%eax
  nopl   0x0(%rax)
L4:
  movzbl 0x0(%rbp,%rax,1),%edx
  mov    %dl,(%r12,%rax,1)
  add    $0x1,%rax
  cmp    %rbx,%rax
  jne    L4
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L97:
  xor    %edx,%edx
  xor    %eax,%eax
  jmp    L129
```

assistant: ```c
insertStrAt(char s[], int insertAt, int slen, int sourceSIZE, char source[]) {
  int i;
  memmove(source + insertAt + slen, source + insertAt, sourceSIZE - (insertAt + slen));
  for (i = 0; i < slen; i++) {
    source[i + insertAt] = s[i];
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<tas>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  mov    %rdi,-0x10(%rbp)
  mov    $0x1,%ebx
  mov    -0x10(%rbp),%rcx
  mov    -0x10(%rbp),%rdx
  mov    %ebx,%eax
  lock xchg %al,(%rcx)
  mov    %eax,%ebx
  movsbl %bl,%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int tas(volatile char *lock) {
  register char res = 1;
  __asm__ __volatile__("lock xchgb %0, %1\n" : "+q"(res), "+m"(*lock) : : "memory", "cc");
  return res;
}
```
user: Translate the below assembly code into C programming language:

```asm
<tas>:
  endbr64
  mov    $0x1,%eax
  lock xchg %al,(%rdi)
  movsbl %al,%eax
  ret
```

assistant: ```c
int tas(volatile char *lock) {
  register char res = 1;
  __asm__ __volatile__("lock xchgb %0, %1\n" : "+q"(res), "+m"(*lock) : : "memory", "cc");
  return res;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<tas>:
  endbr64
  mov    $0x1,%eax
  lock xchg %al,(%rdi)
  movsbl %al,%eax
  ret
```

assistant: ```c
int tas(volatile char *lock) {
  register char res = 1;
  __asm__ __volatile__("lock xchgb %0, %1\n" : "+q"(res), "+m"(*lock) : : "memory", "cc");
  return res;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<tas>:
  endbr64
  mov    $0x1,%eax
  lock xchg %al,(%rdi)
  movsbl %al,%eax
  ret
```

assistant: ```c
int tas(volatile char *lock) {
  register char res = 1;
  __asm__ __volatile__("lock xchgb %0, %1\n" : "+q"(res), "+m"(*lock) : : "memory", "cc");
  return res;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<getCollisionalTransitions>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %r9,0x18(%rsp)
  mov    0x78(%rsp),%r15
  movsd  %xmm0,0x20(%rsp)
  test   %ecx,%ecx
  jle    L2
  mov    %rdx,%r13
  mov    %r8,%r14
  lea    0x1(%rcx),%eax
  mov    %rax,0x28(%rsp)
  mov    $0x1,%r12d
L11:
  mov    0x18(%rsp),%rax
  movslq (%rax,%r12,4),%rax
  lea    0x0(,%rax,8),%rbp
  mov    0x70(%rsp),%rdi
  movslq (%rdi,%r12,4),%rdx
  lea    0x0(,%rdx,8),%rbx
  mov    0x80(%rsp),%rcx
  movsd  (%rcx,%r12,8),%xmm2
  movsd  (%r14,%rdx,8),%xmm0
  divsd  (%r14,%rax,8),%xmm0
  movsd  %xmm2,0x10(%rsp)
  mulsd  %xmm2,%xmm0
  movsd  %xmm0,0x8(%rsp)
  movsd  (%r15,%rdx,8),%xmm0
  subsd  (%r15,%rax,8),%xmm0
  xorpd  D0(%rip),%xmm0
  divsd  0x20(%rsp),%xmm0
  call   <exp@plt>
  mulsd  0x8(%rsp),%xmm0
  lea    0x0(%r13,%rbp,1),%rax
  mov    %rbp,%rdx
  add    (%rax),%rdx
  movsd  (%rdx),%xmm1
  subsd  %xmm0,%xmm1
  movsd  %xmm1,(%rdx)
  mov    %rbx,%rsi
  add    (%rax),%rsi
  movsd  0x10(%rsp),%xmm2
  movapd %xmm2,%xmm1
  addsd  (%rsi),%xmm1
  movsd  %xmm1,(%rsi)
  lea    0x0(%r13,%rbx,1),%rax
  add    (%rax),%rbp
  addsd  0x0(%rbp),%xmm0
  movsd  %xmm0,0x0(%rbp)
  add    (%rax),%rbx
  movsd  (%rbx),%xmm0
  subsd  %xmm2,%xmm0
  movsd  %xmm0,(%rbx)
  add    $0x1,%r12
  cmp    0x28(%rsp),%r12
  jne    L11
L2:
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D0
type_infered: byte[16]

tool: {"D0": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
void getCollisionalTransitions(const int levels, double *F, double **DFDX, const int transitions, const double *weights, const int *lower, const int *upper, const double *energy, const double *CulArr, const double T) {
  double lw, uw;
  int li, ui;
  int i;
  double Cul, Clu;
  for (i = 1; i <= transitions; i++) {
    li = lower[i];
    lw = weights[li];
    ui = upper[i];
    uw = weights[ui];
    Cul = CulArr[i];
    Clu = (uw / lw) * Cul * exp(-(energy[ui] - energy[li]) / T);
    DFDX[li][li] -= Clu;
    DFDX[li][ui] += Cul;
    DFDX[ui][li] += Clu;
    DFDX[ui][ui] -= Cul;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getCollisionalTransitions>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %r9,0x28(%rsp)
  mov    0x88(%rsp),%r12
  movsd  %xmm0,0x30(%rsp)
  test   %ecx,%ecx
  jle    L17
  lea    0x1(%rcx),%eax
  mov    %rdx,%r15
  mov    %r8,%rbp
  mov    $0x1,%r14d
  mov    %rax,0x38(%rsp)
  nopl   0x0(%rax)
L12:
  mov    0x28(%rsp),%rax
  mov    0x80(%rsp),%rsi
  mov    0x90(%rsp),%rcx
  movslq (%rax,%r14,4),%rax
  movslq (%rsi,%r14,4),%rdi
  movsd  (%rcx,%r14,8),%xmm1
  add    $0x1,%r14
  movsd  0x0(%rbp,%rdi,8),%xmm0
  divsd  0x0(%rbp,%rax,8),%xmm0
  mov    %rdi,0x18(%rsp)
  lea    0x0(,%rax,8),%rbx
  mulsd  %xmm1,%xmm0
  lea    0x0(,%rdi,8),%r13
  mov    %rax,0x10(%rsp)
  movsd  %xmm1,0x20(%rsp)
  movsd  %xmm0,0x8(%rsp)
  movsd  (%r12,%rdi,8),%xmm0
  subsd  (%r12,%rax,8),%xmm0
  xorpd  D3(%rip),%xmm0
  divsd  0x30(%rsp),%xmm0
  call   <exp@plt>
  mov    0x10(%rsp),%rax
  mulsd  0x8(%rsp),%xmm0
  movsd  0x20(%rsp),%xmm1
  mov    0x18(%rsp),%rdi
  mov    (%r15,%rax,8),%rax
  lea    (%rax,%rbx,1),%r9
  add    %r13,%rax
  movsd  (%r9),%xmm2
  subsd  %xmm0,%xmm2
  movsd  %xmm2,(%r9)
  movsd  (%rax),%xmm2
  addsd  %xmm1,%xmm2
  movsd  %xmm2,(%rax)
  mov    (%r15,%rdi,8),%rax
  lea    (%rax,%rbx,1),%rdx
  add    %r13,%rax
  addsd  (%rdx),%xmm0
  movsd  %xmm0,(%rdx)
  movsd  (%rax),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  cmp    %r14,0x38(%rsp)
  jne    L12
L17:
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: read_data
arguments: data_tag: D3
type: byte[16]

tool: {"D3": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
void getCollisionalTransitions(const int levels, double *F, double **DFDX, const int transitions, const double *weights, const int *lower, const int *upper, const double *energy, const double *CulArr, const double T) {
  double lw, uw;
  int li, ui;
  int i;
  double Cul, Clu;
  for (i = 1; i <= transitions; i++) {
    li = lower[i];
    lw = weights[li];
    ui = upper[i];
    uw = weights[ui];
    Cul = CulArr[i];
    Clu = (uw / lw) * Cul * exp(-(energy[ui] - energy[li]) / T);
    DFDX[li][li] -= Clu;
    DFDX[li][ui] += Cul;
    DFDX[ui][li] += Clu;
    DFDX[ui][ui] -= Cul;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getCollisionalTransitions>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %r9,0x28(%rsp)
  mov    0x88(%rsp),%r12
  movsd  %xmm0,0x30(%rsp)
  test   %ecx,%ecx
  jle    L0
  lea    0x1(%rcx),%eax
  mov    %rdx,%r15
  mov    %r8,%rbp
  mov    $0x1,%r14d
  mov    %rax,0x38(%rsp)
  nopl   0x0(%rax)
L1:
  mov    0x28(%rsp),%rax
  mov    0x80(%rsp),%rsi
  mov    0x90(%rsp),%rcx
  movslq (%rax,%r14,4),%rax
  movslq (%rsi,%r14,4),%rdi
  movsd  (%rcx,%r14,8),%xmm1
  add    $0x1,%r14
  movsd  0x0(%rbp,%rdi,8),%xmm0
  divsd  0x0(%rbp,%rax,8),%xmm0
  mov    %rdi,0x18(%rsp)
  lea    0x0(,%rax,8),%rbx
  mulsd  %xmm1,%xmm0
  lea    0x0(,%rdi,8),%r13
  mov    %rax,0x10(%rsp)
  movsd  %xmm1,0x20(%rsp)
  movsd  %xmm0,0x8(%rsp)
  movsd  (%r12,%rdi,8),%xmm0
  subsd  (%r12,%rax,8),%xmm0
  xorpd  D4(%rip),%xmm0
  divsd  0x30(%rsp),%xmm0
  call   <exp@plt>
  mov    0x10(%rsp),%rax
  mulsd  0x8(%rsp),%xmm0
  movsd  0x20(%rsp),%xmm1
  mov    0x18(%rsp),%rdi
  mov    (%r15,%rax,8),%rax
  lea    (%rax,%rbx,1),%r9
  add    %r13,%rax
  movsd  (%r9),%xmm2
  subsd  %xmm0,%xmm2
  movsd  %xmm2,(%r9)
  movsd  (%rax),%xmm2
  addsd  %xmm1,%xmm2
  movsd  %xmm2,(%rax)
  mov    (%r15,%rdi,8),%rax
  lea    (%rax,%rbx,1),%rdx
  add    %r13,%rax
  addsd  (%rdx),%xmm0
  movsd  %xmm0,(%rdx)
  movsd  (%rax),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  cmp    %r14,0x38(%rsp)
  jne    L1
L0:
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D4
type: byte[16]

tool: {"D4": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
void getCollisionalTransitions(const int levels, double *F, double **DFDX, const int transitions, const double *weights, const int *lower, const int *upper, const double *energy, const double *CulArr, const double T) {
  double lw, uw;
  int li, ui;
  int i;
  double Cul, Clu;
  for (i = 1; i <= transitions; i++) {
    li = lower[i];
    lw = weights[li];
    ui = upper[i];
    uw = weights[ui];
    Cul = CulArr[i];
    Clu = (uw / lw) * Cul * exp(-(energy[ui] - energy[li]) / T);
    DFDX[li][li] -= Clu;
    DFDX[li][ui] += Cul;
    DFDX[ui][li] += Clu;
    DFDX[ui][ui] -= Cul;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x2,%esi
  lea    D53(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x4,%esi
  lea    D79(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D82(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x4,%esi
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x10,%esi
  lea    D74(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x30(%rbp),%rax
  mov    %rax,-0x38(%rbp)
  mov    $0x28,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x8,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D53
type_infered: string
name: unpack_data
arguments: data_label: D79
type_infered: string
name: unpack_data
arguments: data_label: D82
type_infered: string
name: unpack_data
arguments: data_label: D20
type_infered: string
name: unpack_data
arguments: data_label: D48
type_infered: string
name: unpack_data
arguments: data_label: D28
type_infered: string
name: unpack_data
arguments: data_label: D74
type_infered: string
name: unpack_data
arguments: data_label: D9
type_infered: string
name: unpack_data
arguments: data_label: D4
type_infered: string

tool: D53: short: %d


tool: D79: int: %d


tool: D82: long: %d


tool: D20: long long: %d


tool: D48: float: %d


tool: D28: double: %d


tool: D74: long double: %d


tool: D9: sizeof a:%d

tool: D4: sizeof b:%d

assistant: ```c
int main() {
  printf("short: %d\n", sizeof(short));
  printf("int: %d\n", sizeof(int));
  printf("long: %d\n", sizeof(long));
  printf("long long: %d\n", sizeof(long long));
  printf("float: %d\n", sizeof(float));
  printf("double: %d\n", sizeof(double));
  printf("long double: %d\n", sizeof(long double));
  int a[10];
  int *b = a;
  printf("sizeof a:%d", sizeof(a));
  printf("\n");
  printf("sizeof b:%d", sizeof(b));
  printf("\n");
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%esi
  lea    D61(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x4,%esi
  lea    D70(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D46(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D53(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x4,%esi
  lea    D35(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x10,%esi
  lea    D10(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x28,%esi
  lea    D66(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x8,%esi
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D61
type_infered: string
name: parse_data
arguments: data_label: D70
type_infered: string
name: parse_data
arguments: data_label: D46
type_infered: string
name: parse_data
arguments: data_label: D53
type_infered: string
name: parse_data
arguments: data_label: D35
type_infered: string
name: parse_data
arguments: data_label: D3
type_infered: string
name: parse_data
arguments: data_label: D10
type_infered: string
name: parse_data
arguments: data_label: D66
type_infered: string
name: parse_data
arguments: data_label: D1
type_infered: string

tool: D61: short: %d


tool: D70: int: %d


tool: D46: long: %d


tool: D53: long long: %d


tool: D35: float: %d


tool: D3: double: %d


tool: D10: long double: %d


tool: D66: sizeof a:%d

tool: D1: sizeof b:%d

assistant: ```c
int main() {
  printf("short: %d\n", sizeof(short));
  printf("int: %d\n", sizeof(int));
  printf("long: %d\n", sizeof(long));
  printf("long long: %d\n", sizeof(long long));
  printf("float: %d\n", sizeof(float));
  printf("double: %d\n", sizeof(double));
  printf("long double: %d\n", sizeof(long double));
  int a[10];
  int *b = a;
  printf("sizeof a:%d", sizeof(a));
  printf("\n");
  printf("sizeof b:%d", sizeof(b));
  printf("\n");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%esi
  lea    D87(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x4,%esi
  lea    D5(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D60(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D74(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x4,%esi
  lea    D56(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D70(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x10,%esi
  lea    D8(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x28,%esi
  lea    D39(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x8,%esi
  lea    D32(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: label: D87
type_guessed: string
name: read_data
arguments: label: D5
type_guessed: string
name: read_data
arguments: label: D60
type_guessed: string
name: read_data
arguments: label: D74
type_guessed: string
name: read_data
arguments: label: D56
type_guessed: string
name: read_data
arguments: label: D70
type_guessed: string
name: read_data
arguments: label: D8
type_guessed: string
name: read_data
arguments: label: D39
type_guessed: string
name: read_data
arguments: label: D32
type_guessed: string

tool: D87: short: %d


tool: D5: int: %d


tool: D60: long: %d


tool: D74: long long: %d


tool: D56: float: %d


tool: D70: double: %d


tool: D8: long double: %d


tool: D39: sizeof a:%d

tool: D32: sizeof b:%d

assistant: ```c
int main() {
  printf("short: %d\n", sizeof(short));
  printf("int: %d\n", sizeof(int));
  printf("long: %d\n", sizeof(long));
  printf("long long: %d\n", sizeof(long long));
  printf("float: %d\n", sizeof(float));
  printf("double: %d\n", sizeof(double));
  printf("long double: %d\n", sizeof(long double));
  int a[10];
  int *b = a;
  printf("sizeof a:%d", sizeof(a));
  printf("\n");
  printf("sizeof b:%d", sizeof(b));
  printf("\n");
  return 0;
}
```