ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<undflt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x20(%rbp)
  mov    $0x0,%eax
  call   <undflt_init@plt>
  mov    $0x1,%esi
  mov    $0x17f8,%edi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L31
  mov    $0xffffffff,%eax
  jmp    L0
L31:
  mov    -0x8(%rbp),%rax
  mov    -0x18(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,0x8(%rax)
  mov    -0x8(%rbp),%rax
  mov    -0x28(%rbp),%rdx
  mov    %rdx,0x18(%rax)
  mov    -0x8(%rbp),%rax
  mov    -0x28(%rbp),%rdx
  mov    %rdx,0x20(%rax)
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,0x28(%rax)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <undflt_func@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  cmpl   $0x0,-0xc(%rbp)
  jne    L38
  mov    $0xffffffff,%eax
  jmp    L0
L38:
  mov    -0x20(%rbp),%eax
L0:
  leave
  ret
```

assistant: ```c
int undflt(unsigned char *in, int insz, unsigned char *out, int outsz) {
  typedef struct {
    void *input;
    void *input_end;
    int dummy1;
    int dummy2;
    void *ptr;
    void *output;
    void *output_end;
    char data[0x17c8];
  } undflt_ctx_t;
  undflt_ctx_t *ctx;
  int ret;
  undflt_init();
  ctx = calloc(sizeof(undflt_ctx_t), 1);
  if (!ctx)
    return (-1);
  ctx->input = in;
  ctx->input_end = in + insz;
  ctx->ptr = out;
  ctx->output = out;
  ctx->output_end = out + outsz;
  ret = undflt_func(ctx);
  free(ctx);
  if (!ret)
    return (-1);
  return (outsz);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<undflt>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%r13d
  mov    %rdx,%rbp
  mov    %ecx,%r14d
  mov    $0x0,%eax
  call   <undflt_init@plt>
  mov    $0x1,%esi
  mov    $0x17f8,%edi
  call   <calloc@plt>
  test   %rax,%rax
  je     L16
  mov    %rax,%rbx
  mov    %r12,(%rax)
  movslq %r13d,%r13
  add    %r13,%r12
  mov    %r12,0x8(%rax)
  mov    %rbp,0x18(%rax)
  mov    %rbp,0x20(%rax)
  movslq %r14d,%rax
  add    %rax,%rbp
  mov    %rbp,0x28(%rbx)
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <undflt_func@plt>
  mov    %eax,%ebp
  mov    %rbx,%rdi
  call   <free@plt>
  test   %ebp,%ebp
  mov    $0xffffffff,%eax
  cmove  %eax,%r14d
L19:
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L16:
  mov    $0xffffffff,%r14d
  jmp    L19
```

assistant: ```c
int undflt(unsigned char *in, int insz, unsigned char *out, int outsz) {
  typedef struct {
    void *input;
    void *input_end;
    int dummy1;
    int dummy2;
    void *ptr;
    void *output;
    void *output_end;
    char data[0x17c8];
  } undflt_ctx_t;
  undflt_ctx_t *ctx;
  int ret;
  undflt_init();
  ctx = calloc(sizeof(undflt_ctx_t), 1);
  if (!ctx)
    return (-1);
  ctx->input = in;
  ctx->input_end = in + insz;
  ctx->ptr = out;
  ctx->output = out;
  ctx->output_end = out + outsz;
  ret = undflt_func(ctx);
  free(ctx);
  if (!ret)
    return (-1);
  return (outsz);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<undflt>:
  endbr64
  push   %r14
  xor    %eax,%eax
  mov    %rdi,%r14
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  movslq %esi,%r12
  push   %rbp
  push   %rbx
  mov    %rdx,%rbx
  call   <undflt_init@plt>
  mov    $0x1,%esi
  mov    $0x17f8,%edi
  call   <calloc@plt>
  test   %rax,%rax
  je     L15
  mov    %r14,(%rax)
  add    %r12,%r14
  mov    %rax,%rbp
  mov    %rbx,0x18(%rax)
  mov    %rbp,%rdi
  mov    %rbx,0x20(%rax)
  mov    %r14,0x8(%rax)
  movslq %r13d,%rax
  add    %rax,%rbx
  xor    %eax,%eax
  mov    %rbx,0x28(%rbp)
  call   <undflt_func@plt>
  mov    %rbp,%rdi
  mov    %eax,%ebx
  call   <free@plt>
  test   %ebx,%ebx
  je     L15
L0:
  pop    %rbx
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  cs nopw 0x0(%rax,%rax,1)
L15:
  mov    $0xffffffff,%r13d
  jmp    L0
```

assistant: ```c
int undflt(unsigned char *in, int insz, unsigned char *out, int outsz) {
  typedef struct {
    void *input;
    void *input_end;
    int dummy1;
    int dummy2;
    void *ptr;
    void *output;
    void *output_end;
    char data[0x17c8];
  } undflt_ctx_t;
  undflt_ctx_t *ctx;
  int ret;
  undflt_init();
  ctx = calloc(sizeof(undflt_ctx_t), 1);
  if (!ctx)
    return (-1);
  ctx->input = in;
  ctx->input_end = in + insz;
  ctx->ptr = out;
  ctx->output = out;
  ctx->output_end = out + outsz;
  ret = undflt_func(ctx);
  free(ctx);
  if (!ret)
    return (-1);
  return (outsz);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<undflt>:
  endbr64
  push   %r14
  movq   %rdx,%xmm1
  xor    %eax,%eax
  mov    %ecx,%r14d
  push   %r13
  punpcklqdq %xmm1,%xmm1
  mov    %rdi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  push   %rbx
  movslq %esi,%rbx
  sub    $0x10,%rsp
  movaps %xmm1,(%rsp)
  call   <undflt_init@plt>
  mov    $0x1,%esi
  mov    $0x17f8,%edi
  call   <calloc@plt>
  test   %rax,%rax
  je     L13
  add    %r13,%rbx
  movdqa (%rsp),%xmm1
  movq   %r13,%xmm0
  mov    %rax,%rbp
  movq   %rbx,%xmm2
  mov    %rbp,%rdi
  punpcklqdq %xmm2,%xmm0
  movups %xmm1,0x18(%rax)
  movups %xmm0,(%rax)
  movslq %r14d,%rax
  add    %rax,%r12
  xor    %eax,%eax
  mov    %r12,0x28(%rbp)
  call   <undflt_func@plt>
  mov    %rbp,%rdi
  mov    %eax,%ebx
  call   <free@plt>
  test   %ebx,%ebx
  je     L13
L28:
  add    $0x10,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  cs nopw 0x0(%rax,%rax,1)
L13:
  mov    $0xffffffff,%r14d
  jmp    L28
```

assistant: ```c
int undflt(unsigned char *in, int insz, unsigned char *out, int outsz) {
  typedef struct {
    void *input;
    void *input_end;
    int dummy1;
    int dummy2;
    void *ptr;
    void *output;
    void *output_end;
    char data[0x17c8];
  } undflt_ctx_t;
  undflt_ctx_t *ctx;
  int ret;
  undflt_init();
  ctx = calloc(sizeof(undflt_ctx_t), 1);
  if (!ctx)
    return (-1);
  ctx->input = in;
  ctx->input_end = in + insz;
  ctx->ptr = out;
  ctx->output = out;
  ctx->output_end = out + outsz;
  ret = undflt_func(ctx);
  free(ctx);
  if (!ret)
    return (-1);
  return (outsz);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<int_to_leds_m$leds$green_toggle>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   1050 <leds$leds$green_toggle@plt>
  mov    %al,-0x1(%rbp)
  movzbl -0x1(%rbp),%eax
  leave
  ret
```

assistant: ```c
char int_to_leds_m$leds$green_toggle(void) {
  char result;
  result = leds$leds$green_toggle();
  return result;
}
```
user: What is the c source code of the assembly code below?

```asm
<int_to_leds_m$leds$green_toggle>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   1050 <leds$leds$green_toggle@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
char int_to_leds_m$leds$green_toggle(void) {
  char result;
  result = leds$leds$green_toggle();
  return result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<int_to_leds_m$leds$green_toggle>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   1050 <leds$leds$green_toggle@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
char int_to_leds_m$leds$green_toggle(void) {
  char result;
  result = leds$leds$green_toggle();
  return result;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<int_to_leds_m$leds$green_toggle>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   1050 <leds$leds$green_toggle@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
char int_to_leds_m$leds$green_toggle(void) {
  char result;
  result = leds$leds$green_toggle();
  return result;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<H>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x130,%rsp
  mov    %rdi,-0x128(%rbp)
  mov    %rsi,-0x130(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x130(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <getroundkey@plt>
  cltq
  mov    %rax,%rcx
  lea    -0x110(%rbp),%rax
  mov    $0x40,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movl   $0x0,-0x114(%rbp)
  jmp    L19
L29:
  mov    -0x114(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x128(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%ecx
  mov    -0x114(%rbp),%eax
  cltq
  mov    -0x110(%rbp,%rax,4),%edx
  mov    -0x114(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rsi
  mov    -0x128(%rbp),%rax
  add    %rsi,%rax
  xor    %ecx,%edx
  mov    %edx,(%rax)
  addl   $0x1,-0x114(%rbp)
L19:
  cmpl   $0x3f,-0x114(%rbp)
  jle    L29
  mov    -0x128(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L23
  call   <__stack_chk_fail@plt>
L23:
  leave
  ret
```

assistant: ```c
int H(int *block, int *key) {
  int roundkey[64];
  memcpy(roundkey, getroundkey(key), 64);
  int i;
  for (i = 0; i < 64; i++) {
    block[i] = block[i] ^ roundkey[i];
  }
  return block;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<H>:
  endbr64
  push   %rbx
  sub    $0x110,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  call   <getroundkey@plt>
  cltq
  movdqu (%rax),%xmm0
  movaps %xmm0,(%rsp)
  movdqu 0x10(%rax),%xmm1
  movaps %xmm1,0x10(%rsp)
  movdqu 0x20(%rax),%xmm2
  movaps %xmm2,0x20(%rsp)
  movdqu 0x30(%rax),%xmm3
  movaps %xmm3,0x30(%rsp)
  mov    $0x0,%eax
L18:
  mov    (%rsp,%rax,1),%edx
  xor    %edx,(%rbx,%rax,1)
  add    $0x4,%rax
  cmp    $0x100,%rax
  jne    L18
  mov    %ebx,%eax
  mov    0x108(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L19
  add    $0x110,%rsp
  pop    %rbx
  ret
L19:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int H(int *block, int *key) {
  int roundkey[64];
  memcpy(roundkey, getroundkey(key), 64);
  int i;
  for (i = 0; i < 64; i++) {
    block[i] = block[i] ^ roundkey[i];
  }
  return block;
}
```
user: What is the c source code of the assembly code below:

```asm
<H>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  call   <getroundkey@plt>
  mov    %rsp,%rcx
  cltq
  movdqu (%rax),%xmm0
  movdqu 0x10(%rax),%xmm1
  movdqu 0x20(%rax),%xmm2
  movdqu 0x30(%rax),%xmm3
  xor    %eax,%eax
  movaps %xmm0,(%rsp)
  movaps %xmm1,0x10(%rsp)
  movaps %xmm2,0x20(%rsp)
  movaps %xmm3,0x30(%rsp)
  nopw   0x0(%rax,%rax,1)
L4:
  mov    (%rcx,%rax,1),%edx
  xor    %edx,(%rbx,%rax,1)
  add    $0x4,%rax
  cmp    $0x100,%rax
  jne    L4
  mov    %ebx,%eax
  mov    0x108(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L18
  add    $0x110,%rsp
  pop    %rbx
  ret
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int H(int *block, int *key) {
  int roundkey[64];
  memcpy(roundkey, getroundkey(key), 64);
  int i;
  for (i = 0; i < 64; i++) {
    block[i] = block[i] ^ roundkey[i];
  }
  return block;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<H>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  call   <getroundkey@plt>
  movdqu (%rbx),%xmm4
  cltq
  movdqu (%rax),%xmm0
  movdqu 0x10(%rax),%xmm3
  movdqu 0x20(%rax),%xmm2
  movdqu 0x30(%rax),%xmm1
  movaps %xmm0,(%rsp)
  pxor   %xmm4,%xmm0
  movups %xmm0,(%rbx)
  movdqu 0x10(%rbx),%xmm0
  movaps %xmm3,0x10(%rsp)
  pxor   %xmm3,%xmm0
  movaps %xmm2,0x20(%rsp)
  movups %xmm0,0x10(%rbx)
  movdqu 0x20(%rbx),%xmm0
  movaps %xmm1,0x30(%rsp)
  pxor   %xmm2,%xmm0
  movups %xmm0,0x20(%rbx)
  movdqu 0x30(%rbx),%xmm0
  pxor   %xmm1,%xmm0
  movups %xmm0,0x30(%rbx)
  movdqu 0x40(%rbx),%xmm0
  pxor   0x40(%rsp),%xmm0
  movups %xmm0,0x40(%rbx)
  movdqu 0x50(%rbx),%xmm0
  pxor   0x50(%rsp),%xmm0
  movups %xmm0,0x50(%rbx)
  movdqu 0x60(%rbx),%xmm0
  pxor   0x60(%rsp),%xmm0
  movups %xmm0,0x60(%rbx)
  movdqu 0x70(%rbx),%xmm0
  pxor   0x70(%rsp),%xmm0
  movups %xmm0,0x70(%rbx)
  movdqu 0x80(%rbx),%xmm0
  pxor   0x80(%rsp),%xmm0
  movups %xmm0,0x80(%rbx)
  movdqu 0x90(%rbx),%xmm0
  pxor   0x90(%rsp),%xmm0
  movups %xmm0,0x90(%rbx)
  movdqu 0xa0(%rbx),%xmm0
  pxor   0xa0(%rsp),%xmm0
  movups %xmm0,0xa0(%rbx)
  movdqu 0xb0(%rbx),%xmm0
  pxor   0xb0(%rsp),%xmm0
  movups %xmm0,0xb0(%rbx)
  movdqu 0xc0(%rbx),%xmm0
  pxor   0xc0(%rsp),%xmm0
  movups %xmm0,0xc0(%rbx)
  movdqu 0xd0(%rbx),%xmm0
  pxor   0xd0(%rsp),%xmm0
  movups %xmm0,0xd0(%rbx)
  movdqu 0xe0(%rbx),%xmm0
  pxor   0xe0(%rsp),%xmm0
  movups %xmm0,0xe0(%rbx)
  movdqu 0xf0(%rbx),%xmm0
  pxor   0xf0(%rsp),%xmm0
  movups %xmm0,0xf0(%rbx)
  mov    0x108(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x110,%rsp
  mov    %ebx,%eax
  pop    %rbx
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int H(int *block, int *key) {
  int roundkey[64];
  memcpy(roundkey, getroundkey(key), 64);
  int i;
  for (i = 0; i < 64; i++) {
    block[i] = block[i] ^ roundkey[i];
  }
  return block;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ratioFixed2_DWTAnlz>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm1
  divss  %xmm1,%xmm0
  mov    -0x28(%rbp),%rax
  movss  %xmm0,(%rax)
  movl   $0x1,-0x4(%rbp)
  jmp    L4
L16:
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x2,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm1
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  divss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L4:
  mov    -0x4(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L16
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void ratioFixed2_DWTAnlz(float *dwt, int n, float *ratio, int nRatio) {
  ratio[0] = dwt[1] / dwt[2];
  for (int i = 1; i < nRatio; i++)
    ratio[i] = dwt[i + 2] / dwt[2];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ratioFixed2_DWTAnlz>:
  endbr64
  movss  0x4(%rdi),%xmm0
  divss  0x8(%rdi),%xmm0
  movss  %xmm0,(%rdx)
  cmp    $0x1,%ecx
  jle    L3
  mov    %ecx,%ecx
  mov    $0x1,%eax
L4:
  movss  0x8(%rdi,%rax,4),%xmm0
  divss  0x8(%rdi),%xmm0
  movss  %xmm0,(%rdx,%rax,4)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L4
L3:
  ret
```

assistant: ```c
void ratioFixed2_DWTAnlz(float *dwt, int n, float *ratio, int nRatio) {
  ratio[0] = dwt[1] / dwt[2];
  for (int i = 1; i < nRatio; i++)
    ratio[i] = dwt[i + 2] / dwt[2];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ratioFixed2_DWTAnlz>:
  endbr64
  movss  0x4(%rdi),%xmm0
  divss  0x8(%rdi),%xmm0
  movss  %xmm0,(%rdx)
  cmp    $0x1,%ecx
  jle    L4
  movslq %ecx,%rcx
  mov    $0x1,%eax
  nop
L12:
  movss  0x8(%rdi,%rax,4),%xmm0
  divss  0x8(%rdi),%xmm0
  movss  %xmm0,(%rdx,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%rcx
  jne    L12
L4:
  ret
```

assistant: ```c
void ratioFixed2_DWTAnlz(float *dwt, int n, float *ratio, int nRatio) {
  ratio[0] = dwt[1] / dwt[2];
  for (int i = 1; i < nRatio; i++)
    ratio[i] = dwt[i + 2] / dwt[2];
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ratioFixed2_DWTAnlz>:
  endbr64
  mov    %rdi,%rsi
  mov    %ecx,%edi
  movss  0x4(%rsi),%xmm0
  divss  0x8(%rsi),%xmm0
  movss  %xmm0,(%rdx)
  cmp    $0x1,%ecx
  jle    L31
  lea    -0x1(%rcx),%r8d
  mov    %ecx,%ecx
  lea    0x8(%rsi),%r10
  lea    (%rdx,%rcx,4),%r9
  lea    0x4(%rdx),%rax
  cmp    %r9,%r10
  lea    0xc(%rsi),%r10
  setae  %r9b
  cmp    %r10,%rax
  setae  %r10b
  or     %r10b,%r9b
  je     L75
  lea    0x10(%rsi),%r9
  sub    %r9,%rax
  cmp    $0x8,%rax
  lea    -0x2(%rdi),%eax
  seta   %r9b
  cmp    $0x2,%eax
  seta   %al
  test   %al,%r9b
  je     L75
  mov    %r8d,%ecx
  movss  0x8(%rsi),%xmm1
  xor    %eax,%eax
  shr    $0x2,%ecx
  shl    $0x4,%rcx
  shufps $0x0,%xmm1,%xmm1
  nopl   0x0(%rax,%rax,1)
L44:
  movups 0xc(%rsi,%rax,1),%xmm0
  divps  %xmm1,%xmm0
  movups %xmm0,0x4(%rdx,%rax,1)
  add    $0x10,%rax
  cmp    %rcx,%rax
  jne    L44
  mov    %r8d,%eax
  and    $0xfffffffc,%eax
  and    $0x3,%r8d
  je     L31
  lea    0x3(%rax),%r9d
  add    $0x2,%eax
  movslq %r9d,%r8
  lea    0x0(,%r8,4),%rcx
  movss  (%rsi,%r8,4),%xmm0
  divss  0x8(%rsi),%xmm0
  movss  %xmm0,-0x8(%rdx,%rcx,1)
  cmp    %eax,%edi
  jle    L31
  movss  0x4(%rsi,%rcx,1),%xmm0
  divss  0x8(%rsi),%xmm0
  movss  %xmm0,-0x4(%rdx,%rcx,1)
  cmp    %r9d,%edi
  jle    L31
  movss  0x8(%rsi,%rcx,1),%xmm0
  divss  0x8(%rsi),%xmm0
  movss  %xmm0,(%rdx,%r8,4)
  ret
  nopl   0x0(%rax)
L75:
  mov    $0x1,%eax
  nopl   (%rax)
L0:
  movss  0x8(%rsi,%rax,4),%xmm0
  divss  0x8(%rsi),%xmm0
  movss  %xmm0,(%rdx,%rax,4)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L0
L31:
  ret
```

assistant: ```c
void ratioFixed2_DWTAnlz(float *dwt, int n, float *ratio, int nRatio) {
  ratio[0] = dwt[1] / dwt[2];
  for (int i = 1; i < nRatio; i++)
    ratio[i] = dwt[i + 2] / dwt[2];
}
```
user: Translate the below assembly code into C programming language:

```asm
<puthex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  lea    D5(%rip),%rax
  mov    %rax,-0x8(%rbp)
  cmpl   $0xfff,-0x14(%rbp)
  jle    L28
  mov    -0x14(%rbp),%eax
  sar    $0xc,%eax
  cltq
  and    $0xf,%eax
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  lea    (%rdx,%rax,1),%rsi
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rcx
  mov    -0x20(%rbp),%rdx
  mov    %rcx,(%rdx)
  movzbl (%rsi),%edx
  mov    %dl,(%rax)
L28:
  cmpl   $0xff,-0x14(%rbp)
  jle    L2
  mov    -0x14(%rbp),%eax
  sar    $0x8,%eax
  cltq
  and    $0xf,%eax
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  lea    (%rdx,%rax,1),%rsi
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rcx
  mov    -0x20(%rbp),%rdx
  mov    %rcx,(%rdx)
  movzbl (%rsi),%edx
  mov    %dl,(%rax)
L2:
  cmpl   $0xf,-0x14(%rbp)
  jle    L0
  mov    -0x14(%rbp),%eax
  sar    $0x4,%eax
  cltq
  and    $0xf,%eax
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  lea    (%rdx,%rax,1),%rsi
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rcx
  mov    -0x20(%rbp),%rdx
  mov    %rcx,(%rdx)
  movzbl (%rsi),%edx
  mov    %dl,(%rax)
L0:
  mov    -0x14(%rbp),%eax
  cltq
  and    $0xf,%eax
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  lea    (%rdx,%rax,1),%rsi
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rcx
  mov    -0x20(%rbp),%rdx
  mov    %rcx,(%rdx)
  movzbl (%rsi),%edx
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D5
type: string

tool: D5: 0123456789abcdef

assistant: ```c
void puthex(uint16_t word, char **posx) {
  const char *digits = "0123456789abcdef";
  if (word >= 0x1000)
    *(*posx)++ = digits[(word >> 12) & 0xf];
  if (word >= 0x100)
    *(*posx)++ = digits[(word >> 8) & 0xf];
  if (word >= 0x10)
    *(*posx)++ = digits[(word >> 4) & 0xf];
  *(*posx)++ = digits[word & 0xf];
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<puthex>:
  endbr64
  mov    %edi,%eax
  mov    %rsi,%rdx
  cmp    $0xfff,%edi
  jle    L38
  mov    (%rsi),%rsi
  lea    0x1(%rsi),%rcx
  mov    %rcx,(%rdx)
  mov    %edi,%ecx
  sar    $0xc,%ecx
  and    $0xf,%ecx
  lea    D3(%rip),%rdi
  movzbl (%rdi,%rcx,1),%ecx
  mov    %cl,(%rsi)
L24:
  mov    (%rdx),%rsi
  lea    0x1(%rsi),%rcx
  mov    %rcx,(%rdx)
  mov    %eax,%ecx
  sar    $0x8,%ecx
  and    $0xf,%ecx
  lea    D3(%rip),%rdi
  movzbl (%rdi,%rcx,1),%ecx
  mov    %cl,(%rsi)
L0:
  mov    (%rdx),%rsi
  lea    0x1(%rsi),%rcx
  mov    %rcx,(%rdx)
  mov    %eax,%ecx
  sar    $0x4,%ecx
  and    $0xf,%ecx
  lea    D3(%rip),%rdi
  movzbl (%rdi,%rcx,1),%ecx
  mov    %cl,(%rsi)
L2:
  mov    (%rdx),%rcx
  lea    0x1(%rcx),%rsi
  mov    %rsi,(%rdx)
  and    $0xf,%eax
  lea    D3(%rip),%rdx
  movzbl (%rdx,%rax,1),%eax
  mov    %al,(%rcx)
  ret
L38:
  cmp    $0xff,%edi
  jg     L24
  cmp    $0xf,%edi
  jle    L2
  jmp    L0
```

tool_calls: name: read_data
arguments: tag: D3
type_guessed: string

tool: D3: 0123456789abcdef

assistant: ```c
void puthex(uint16_t word, char **posx) {
  const char *digits = "0123456789abcdef";
  if (word >= 0x1000)
    *(*posx)++ = digits[(word >> 12) & 0xf];
  if (word >= 0x100)
    *(*posx)++ = digits[(word >> 8) & 0xf];
  if (word >= 0x10)
    *(*posx)++ = digits[(word >> 4) & 0xf];
  *(*posx)++ = digits[word & 0xf];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<puthex>:
  endbr64
  mov    %edi,%eax
  mov    %rsi,%rdx
  cmp    $0xfff,%edi
  jle    L12
  mov    (%rsi),%rdi
  lea    0x1(%rdi),%rcx
  mov    %rcx,(%rsi)
  mov    %eax,%esi
  lea    D5(%rip),%rcx
  sar    $0xc,%esi
  and    $0xf,%esi
  movzbl (%rcx,%rsi,1),%esi
  mov    %sil,(%rdi)
L23:
  mov    (%rdx),%rdi
  lea    0x1(%rdi),%rsi
  mov    %rsi,(%rdx)
  mov    %eax,%esi
  sar    $0x8,%esi
  and    $0xf,%esi
  movzbl (%rcx,%rsi,1),%esi
  mov    %sil,(%rdi)
L1:
  mov    (%rdx),%rdi
  lea    0x1(%rdi),%rsi
  mov    %rsi,(%rdx)
  mov    %eax,%esi
  sar    $0x4,%esi
  and    $0xf,%esi
  movzbl (%rcx,%rsi,1),%esi
  mov    %sil,(%rdi)
L32:
  mov    (%rdx),%rsi
  and    $0xf,%eax
  movzbl (%rcx,%rax,1),%eax
  lea    0x1(%rsi),%rdi
  mov    %rdi,(%rdx)
  mov    %al,(%rsi)
  ret
  nopw   0x0(%rax,%rax,1)
L12:
  lea    D5(%rip),%rcx
  cmp    $0xff,%edi
  jg     L23
  cmp    $0xf,%edi
  jle    L32
  jmp    L1
```

tool_calls: name: unpack_data
arguments: tag: D5
type_guessed: string

tool: D5: 0123456789abcdef

assistant: ```c
void puthex(uint16_t word, char **posx) {
  const char *digits = "0123456789abcdef";
  if (word >= 0x1000)
    *(*posx)++ = digits[(word >> 12) & 0xf];
  if (word >= 0x100)
    *(*posx)++ = digits[(word >> 8) & 0xf];
  if (word >= 0x10)
    *(*posx)++ = digits[(word >> 4) & 0xf];
  *(*posx)++ = digits[word & 0xf];
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<puthex>:
  endbr64
  mov    %edi,%eax
  mov    %rsi,%rdx
  cmp    $0xfff,%edi
  jle    L3
  mov    (%rsi),%rdi
  lea    0x1(%rdi),%rcx
  mov    %rcx,(%rsi)
  mov    %eax,%esi
  lea    D2(%rip),%rcx
  sar    $0xc,%esi
  and    $0xf,%esi
  movzbl (%rcx,%rsi,1),%esi
  mov    %sil,(%rdi)
L1:
  mov    (%rdx),%rdi
  lea    0x1(%rdi),%rsi
  mov    %rsi,(%rdx)
  mov    %eax,%esi
  sar    $0x8,%esi
  and    $0xf,%esi
  movzbl (%rcx,%rsi,1),%esi
  mov    %sil,(%rdi)
L27:
  mov    (%rdx),%rdi
  lea    0x1(%rdi),%rsi
  mov    %rsi,(%rdx)
  mov    %eax,%esi
  sar    $0x4,%esi
  and    $0xf,%esi
  movzbl (%rcx,%rsi,1),%esi
  mov    %sil,(%rdi)
L9:
  mov    (%rdx),%rsi
  and    $0xf,%eax
  movzbl (%rcx,%rax,1),%eax
  lea    0x1(%rsi),%rdi
  mov    %rdi,(%rdx)
  mov    %al,(%rsi)
  ret
  nopw   0x0(%rax,%rax,1)
L3:
  lea    D2(%rip),%rcx
  cmp    $0xff,%edi
  jg     L1
  cmp    $0xf,%edi
  jle    L9
  jmp    L27
```

tool_calls: name: read_data
arguments: data_label: D2
type_guessed: string

tool: D2: 0123456789abcdef

assistant: ```c
void puthex(uint16_t word, char **posx) {
  const char *digits = "0123456789abcdef";
  if (word >= 0x1000)
    *(*posx)++ = digits[(word >> 12) & 0xf];
  if (word >= 0x100)
    *(*posx)++ = digits[(word >> 8) & 0xf];
  if (word >= 0x10)
    *(*posx)++ = digits[(word >> 4) & 0xf];
  *(*posx)++ = digits[word & 0xf];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x6,%esi
  mov    $0x64,%edi
  call   <sushi2@plt>
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D3
type_infered: string

tool: D3: %d


assistant: ```c
int main(int argc, char const *argv[]) {
  printf("%d\n", sushi2(100, 6));
  return 0;
}
```