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
<json_parse_literal>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  je     L7
  mov    $0x0,%eax
  jmp    L48
L7:
  cmpq   $0x0,-0x28(%rbp)
  je     L35
  mov    -0x30(%rbp),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x30(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L35
  mov    $0x0,%eax
  jmp    L48
L35:
  mov    -0x8(%rbp),%rax
L48:
  leave
  ret
```

assistant: ```c
size_t json_parse_literal(const char *str, const char *literal, char *buf, size_t size) {
  const size_t len = strlen(literal);
  if (strncmp(str, literal, len) != 0)
    return 0;
  if (buf) {
    strncpy(buf, literal, size);
    if (buf[size - 1] != '\0')
      return 0;
  }
  return len;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<json_parse_literal>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbx
  mov    %rdx,%rbp
  mov    %rcx,%r13
  mov    %rsi,%rdi
  call   <strlen@plt>
  mov    %rax,%rdx
  mov    %rax,%r14
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  jne    L18
  test   %rbp,%rbp
  je     L10
  mov    %r13,%rdx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <strncpy@plt>
  cmpb   $0x0,-0x1(%rbp,%r13,1)
  mov    $0x0,%eax
  cmovne %rax,%r14
L10:
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L18:
  mov    $0x0,%r14d
  jmp    L10
```

assistant: ```c
size_t json_parse_literal(const char *str, const char *literal, char *buf, size_t size) {
  const size_t len = strlen(literal);
  if (strncmp(str, literal, len) != 0)
    return 0;
  if (buf) {
    strncpy(buf, literal, size);
    if (buf[size - 1] != '\0')
      return 0;
  }
  return len;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<json_parse_literal>:
  endbr64
  push   %r14
  mov    %rcx,%r14
  push   %r13
  mov    %rdi,%r13
  mov    %rsi,%rdi
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  call   <strlen@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  mov    %rax,%rdx
  mov    %rax,%r12
  call   <strncmp@plt>
  test   %rax,%rax
  jne    L5
  test   %rbx,%rbx
  je     L3
  mov    %r14,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strncpy@plt>
  cmpb   $0x0,-0x1(%rbx,%r14,1)
  jne    L5
L3:
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L5:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
size_t json_parse_literal(const char *str, const char *literal, char *buf, size_t size) {
  const size_t len = strlen(literal);
  if (strncmp(str, literal, len) != 0)
    return 0;
  if (buf) {
    strncpy(buf, literal, size);
    if (buf[size - 1] != '\0')
      return 0;
  }
  return len;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<json_parse_literal>:
  endbr64
  push   %r14
  mov    %rcx,%r14
  push   %r13
  mov    %rdi,%r13
  mov    %rsi,%rdi
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  call   <strlen@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  mov    %rax,%rdx
  mov    %rax,%r12
  call   <strncmp@plt>
  test   %rax,%rax
  jne    L14
  test   %rbx,%rbx
  je     L16
  mov    %r14,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strncpy@plt>
  cmpb   $0x0,-0x1(%rbx,%r14,1)
  jne    L14
L16:
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L14:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
size_t json_parse_literal(const char *str, const char *literal, char *buf, size_t size) {
  const size_t len = strlen(literal);
  if (strncmp(str, literal, len) != 0)
    return 0;
  if (buf) {
    strncpy(buf, literal, size);
    if (buf[size - 1] != '\0')
      return 0;
  }
  return len;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<productExceptSelf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    -0x2c(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x38(%rbp),%rax
  mov    -0xc(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movl   $0x1,(%rax)
  movl   $0x1,-0x18(%rbp)
  jmp    L11
L36:
  mov    -0x18(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%ecx
  mov    -0x18(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    -0x18(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rsi
  mov    -0x8(%rbp),%rdx
  add    %rsi,%rdx
  imul   %ecx,%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x18(%rbp)
L11:
  mov    -0x18(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L36
  movl   $0x1,-0x14(%rbp)
  mov    -0xc(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x10(%rbp)
  jmp    L31
L18:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  imul   -0x14(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    -0x14(%rbp),%edx
  imul   %edx,%eax
  mov    %eax,-0x14(%rbp)
  subl   $0x1,-0x10(%rbp)
L31:
  cmpl   $0x0,-0x10(%rbp)
  jns    L18
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int *productExceptSelf(int *nums, int numsSize, int *returnSize) {
  int size = numsSize;
  *returnSize = size;
  int *res = (int *)malloc(size * sizeof(int));
  res[0] = 1;
  for (int i = 1; i < size; i++) {
    res[i] = res[i - 1] * nums[i - 1];
  }
  int prod = 1;
  for (int i = size - 1; i >= 0; i--) {
    res[i] *= prod;
    prod *= nums[i];
  }
  return res;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<productExceptSelf>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  mov    %esi,(%rdx)
  lea    0x0(,%rsi,4),%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  movl   $0x1,(%rax)
  cmp    $0x1,%ebp
  jle    L38
  lea    -0x1(%rbp),%eax
  lea    (%rdi,%rax,4),%r8
  mov    %rdi,%rax
  mov    %rbx,%rsi
  sub    %rdi,%rsi
L35:
  mov    (%rax),%edx
  mov    (%rsi,%rax,1),%ecx
  add    $0x4,%rax
  imul   %ecx,%edx
  mov    %edx,(%rax)
  cmp    %r8,%rax
  jne    L35
  lea    -0x1(%rbp),%edx
L23:
  movslq %edx,%rax
  lea    (%rdi,%rax,4),%rax
  mov    $0x1,%ecx
  sub    %rdi,%rbx
L22:
  mov    %ecx,%esi
  imul   (%rax),%esi
  mov    %esi,(%rax)
  imul   (%rbx,%rax,1),%ecx
  sub    $0x4,%rax
  sub    $0x1,%edx
  jns    L22
L20:
  mov    %rdi,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L38:
  mov    %ebp,%edx
  sub    $0x1,%edx
  js     L20
  jmp    L23
```

assistant: ```c
int *productExceptSelf(int *nums, int numsSize, int *returnSize) {
  int size = numsSize;
  *returnSize = size;
  int *res = (int *)malloc(size * sizeof(int));
  res[0] = 1;
  for (int i = 1; i < size; i++) {
    res[i] = res[i - 1] * nums[i - 1];
  }
  int prod = 1;
  for (int i = size - 1; i >= 0; i--) {
    res[i] *= prod;
    prod *= nums[i];
  }
  return res;
}
```
user: Translate the below assembly code into C programming language:

```asm
<productExceptSelf>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x0(,%rsi,4),%edi
  sub    $0x8,%rsp
  mov    %esi,(%rdx)
  call   <malloc@plt>
  lea    -0x1(%rbp),%ecx
  movl   $0x1,(%rax)
  cmp    $0x1,%ebp
  jle    L36
  mov    %ecx,%esi
  mov    %rbx,%rdi
  lea    0x4(%rax),%rdx
  lea    (%rax,%rsi,4),%r8
  sub    %rax,%rdi
  mov    $0x1,%esi
  jmp    L16
  nopw   0x0(%rax,%rax,1)
L30:
  add    $0x4,%rdx
L16:
  imul   -0x4(%rdi,%rdx,1),%esi
  mov    %esi,(%rdx)
  cmp    %r8,%rdx
  jne    L30
L38:
  movslq %ecx,%rdx
  sub    %rax,%rbx
  mov    $0x1,%esi
  lea    (%rax,%rdx,4),%rdx
  mov    %rbx,%r8
  nopw   0x0(%rax,%rax,1)
L22:
  mov    (%rdx),%edi
  sub    $0x1,%ecx
  imul   %esi,%edi
  mov    %edi,(%rdx)
  imul   (%r8,%rdx,1),%esi
  sub    $0x4,%rdx
  cmp    $0xffffffff,%ecx
  jne    L22
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nop
L36:
  test   %ecx,%ecx
  jns    L38
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int *productExceptSelf(int *nums, int numsSize, int *returnSize) {
  int size = numsSize;
  *returnSize = size;
  int *res = (int *)malloc(size * sizeof(int));
  res[0] = 1;
  for (int i = 1; i < size; i++) {
    res[i] = res[i - 1] * nums[i - 1];
  }
  int prod = 1;
  for (int i = size - 1; i >= 0; i--) {
    res[i] *= prod;
    prod *= nums[i];
  }
  return res;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<productExceptSelf>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x0(,%rsi,4),%edi
  sub    $0x8,%rsp
  mov    %esi,(%rdx)
  call   <malloc@plt>
  lea    -0x1(%rbp),%ecx
  movl   $0x1,(%rax)
  cmp    $0x1,%ebp
  jle    L27
  lea    -0x2(%rbp),%esi
  mov    %rbx,%rdi
  lea    0x4(%rax),%rdx
  lea    0x8(%rax,%rsi,4),%r8
  sub    %rax,%rdi
  mov    $0x1,%esi
  nopw   0x0(%rax,%rax,1)
L15:
  imul   -0x4(%rdi,%rdx,1),%esi
  add    $0x4,%rdx
  mov    %esi,-0x4(%rdx)
  cmp    %r8,%rdx
  jne    L15
L33:
  movslq %ecx,%rdx
  sub    %rax,%rbx
  mov    $0x1,%esi
  lea    (%rax,%rdx,4),%rdx
  mov    %rbx,%r8
  nopl   0x0(%rax,%rax,1)
L37:
  mov    (%rdx),%edi
  sub    $0x1,%ecx
  imul   %esi,%edi
  mov    %edi,(%rdx)
  imul   (%r8,%rdx,1),%esi
  sub    $0x4,%rdx
  cmp    $0xffffffff,%ecx
  jne    L37
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nop
L27:
  test   %ecx,%ecx
  jns    L33
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int *productExceptSelf(int *nums, int numsSize, int *returnSize) {
  int size = numsSize;
  *returnSize = size;
  int *res = (int *)malloc(size * sizeof(int));
  res[0] = 1;
  for (int i = 1; i < size; i++) {
    res[i] = res[i - 1] * nums[i - 1];
  }
  int prod = 1;
  for (int i = size - 1; i >= 0; i--) {
    res[i] *= prod;
    prod *= nums[i];
  }
  return res;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sum_vec>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L13
L10:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L13:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L10
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double sum_vec(double *vec, int length) {
  int i;
  double r = 0.0;
  for (i = 0; i < length; i++)
    r += vec[i];
  return (r);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<sum_vec>:
  endbr64
  test   %esi,%esi
  jle    L15
  mov    %rdi,%rax
  lea    -0x1(%rsi),%edx
  lea    0x8(%rdi,%rdx,8),%rdx
  pxor   %xmm0,%xmm0
L18:
  addsd  (%rax),%xmm0
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L18
  ret
L15:
  pxor   %xmm0,%xmm0
  ret
```

assistant: ```c
double sum_vec(double *vec, int length) {
  int i;
  double r = 0.0;
  for (i = 0; i < length; i++)
    r += vec[i];
  return (r);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sum_vec>:
  endbr64
  test   %esi,%esi
  jle    L1
  lea    -0x1(%rsi),%eax
  pxor   %xmm0,%xmm0
  lea    0x8(%rdi,%rax,8),%rax
  nopl   0x0(%rax)
L8:
  addsd  (%rdi),%xmm0
  add    $0x8,%rdi
  cmp    %rax,%rdi
  jne    L8
  ret
  cs nopw 0x0(%rax,%rax,1)
L1:
  pxor   %xmm0,%xmm0
  ret
```

assistant: ```c
double sum_vec(double *vec, int length) {
  int i;
  double r = 0.0;
  for (i = 0; i < length; i++)
    r += vec[i];
  return (r);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sum_vec>:
  endbr64
  test   %esi,%esi
  jle    L30
  cmp    $0x1,%esi
  je     L48
  mov    %esi,%edx
  mov    %rdi,%rax
  pxor   %xmm0,%xmm0
  shr    %edx
  sub    $0x1,%edx
  shl    $0x4,%rdx
  lea    0x10(%rdi,%rdx,1),%rdx
  nopl   0x0(%rax)
L29:
  movsd  (%rax),%xmm1
  add    $0x10,%rax
  addsd  %xmm1,%xmm0
  movsd  -0x8(%rax),%xmm1
  addsd  %xmm1,%xmm0
  cmp    %rdx,%rax
  jne    L29
  mov    %esi,%eax
  and    $0xfffffffe,%eax
  and    $0x1,%esi
  je     L42
L1:
  cltq
  addsd  (%rdi,%rax,8),%xmm0
  ret
  nopl   0x0(%rax)
L42:
  ret
  nopl   0x0(%rax)
L30:
  pxor   %xmm0,%xmm0
  ret
L48:
  pxor   %xmm0,%xmm0
  xor    %eax,%eax
  jmp    L1
```

assistant: ```c
double sum_vec(double *vec, int length) {
  int i;
  double r = 0.0;
  for (i = 0; i < length; i++)
    r += vec[i];
  return (r);
}
```
user: Translate the below assembly code into C programming language:

```asm
<strchr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  jmp    L8
L9:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L38
  mov    $0x0,%eax
  jmp    L26
L38:
  addq   $0x1,-0x8(%rbp)
L8:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  mov    -0xc(%rbp),%edx
  cmp    %dl,%al
  jne    L9
  mov    -0x8(%rbp),%rax
L26:
  pop    %rbp
  ret
```

assistant: ```c
char *strchr(const char *s, int c) {
  while (*s != (char)c) {
    if (!*s)
      return ((void *)0);
    s++;
  }
  return (char *)s;
}
```
user: What is the c source code of the assembly code below:

```asm
<strchr>:
  endbr64
  movzbl (%rdi),%eax
  mov    %esi,%edx
  cmp    %sil,%al
  je     L29
L0:
  test   %al,%al
  je     L9
  add    $0x1,%rdi
  movzbl (%rdi),%eax
  cmp    %dl,%al
  jne    L0
  mov    %rdi,%rax
  ret
L29:
  mov    %rdi,%rax
  ret
L9:
  mov    $0x0,%eax
  ret
```

assistant: ```c
char *strchr(const char *s, int c) {
  while (*s != (char)c) {
    if (!*s)
      return ((void *)0);
    s++;
  }
  return (char *)s;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strchr>:
  endbr64
  movzbl (%rdi),%eax
  mov    %esi,%edx
  cmp    %sil,%al
  jne    L1
  jmp    L25
L5:
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  cmp    %dl,%al
  je     L25
L1:
  test   %al,%al
  jne    L5
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L25:
  mov    %rdi,%rax
  ret
```

assistant: ```c
char *strchr(const char *s, int c) {
  while (*s != (char)c) {
    if (!*s)
      return ((void *)0);
    s++;
  }
  return (char *)s;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<strchr>:
  endbr64
  movzbl (%rdi),%eax
  mov    %esi,%edx
  cmp    %sil,%al
  jne    L38
  jmp    L37
L24:
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  cmp    %dl,%al
  je     L37
L38:
  test   %al,%al
  jne    L24
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L37:
  mov    %rdi,%rax
  ret
```

assistant: ```c
char *strchr(const char *s, int c) {
  while (*s != (char)c) {
    if (!*s)
      return ((void *)0);
    s++;
  }
  return (char *)s;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<mystrlen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L5
L1:
  addl   $0x1,-0x4(%rbp)
L5:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L1
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int mystrlen(char *s) {
  int i = 0;
  while (s[i]) {
    i++;
  }
  return i;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<mystrlen>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L6
  mov    $0x1,%eax
L10:
  mov    %eax,%edx
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rdi,%rax,1)
  jne    L10
L12:
  mov    %edx,%eax
  ret
L6:
  mov    $0x0,%edx
  jmp    L12
```

assistant: ```c
int mystrlen(char *s) {
  int i = 0;
  while (s[i]) {
    i++;
  }
  return i;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<mystrlen>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L10
  mov    $0x1,%eax
  xchg   %ax,%ax
L11:
  mov    %eax,%r8d
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rdi,%rax,1)
  jne    L11
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L10:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int mystrlen(char *s) {
  int i = 0;
  while (s[i]) {
    i++;
  }
  return i;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<mystrlen>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L10
  mov    $0x1,%eax
  xchg   %ax,%ax
L2:
  mov    %eax,%r8d
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rdi,%rax,1)
  jne    L2
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L10:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int mystrlen(char *s) {
  int i = 0;
  while (s[i]) {
    i++;
  }
  return i;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<get_eta>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movss  %xmm0,-0x14(%rbp)
  movss  %xmm1,-0x18(%rbp)
  pxor   %xmm1,%xmm1
  cvtss2sd -0x14(%rbp),%xmm1
  movsd  D8(%rip),%xmm0
  subsd  %xmm1,%xmm0
  pxor   %xmm2,%xmm2
  cvtss2sd -0x14(%rbp),%xmm2
  movsd  D8(%rip),%xmm1
  addsd  %xmm2,%xmm1
  divsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  movd   %eax,%xmm0
  call   <sqrt@plt>
  movd   %xmm0,%eax
  mov    %eax,-0x8(%rbp)
  pxor   %xmm1,%xmm1
  cvtss2sd -0x8(%rbp),%xmm1
  movsd  D8(%rip),%xmm0
  addsd  %xmm1,%xmm0
  pxor   %xmm2,%xmm2
  cvtss2sd -0x8(%rbp),%xmm2
  movsd  D8(%rip),%xmm1
  subsd  %xmm2,%xmm1
  divsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,-0x4(%rbp)
  movss  -0x18(%rbp),%xmm1
  movss  -0xc(%rbp),%xmm0
  mov    -0x4(%rbp),%eax
  movaps %xmm1,%xmm2
  movaps %xmm0,%xmm1
  movd   %eax,%xmm0
  call   <mix@plt>
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D8
type_infered: f64

tool: D8: 1

assistant: ```c
float get_eta(float f0, float f90) {
  float eta_min = (1.0 - f0) / (1.0 + f0);
  float sqrt_f0 = sqrt(f0);
  float eta_max = (1.0 + sqrt_f0) / (1.0 - sqrt_f0);
  return mix(eta_max, eta_min, f90);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_eta>:
  endbr64
  sub    $0x18,%rsp
  movss  %xmm1,0xc(%rsp)
  pxor   %xmm4,%xmm4
  cvtss2sd %xmm0,%xmm4
  movsd  %xmm4,(%rsp)
  call   <sqrt@plt>
  pxor   %xmm3,%xmm3
  cvtss2sd %xmm0,%xmm3
  movsd  D2(%rip),%xmm2
  movapd %xmm2,%xmm5
  movsd  (%rsp),%xmm4
  subsd  %xmm4,%xmm5
  movapd %xmm5,%xmm1
  addsd  %xmm2,%xmm4
  divsd  %xmm4,%xmm1
  movapd %xmm3,%xmm0
  addsd  %xmm2,%xmm0
  subsd  %xmm3,%xmm2
  divsd  %xmm2,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  0xc(%rsp),%xmm2
  cvtsd2ss %xmm1,%xmm1
  call   <mix@plt>
  add    $0x18,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D2
type_infered: f64

tool: D2: 1

assistant: ```c
float get_eta(float f0, float f90) {
  float eta_min = (1.0 - f0) / (1.0 + f0);
  float sqrt_f0 = sqrt(f0);
  float eta_max = (1.0 + sqrt_f0) / (1.0 - sqrt_f0);
  return mix(eta_max, eta_min, f90);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<get_eta>:
  endbr64
  sub    $0x18,%rsp
  pxor   %xmm5,%xmm5
  movss  %xmm1,0xc(%rsp)
  cvtss2sd %xmm0,%xmm5
  movsd  %xmm5,(%rsp)
  call   <sqrt@plt>
  movsd  (%rsp),%xmm5
  pxor   %xmm4,%xmm4
  movsd  D7(%rip),%xmm3
  cvtss2sd %xmm0,%xmm4
  movapd %xmm4,%xmm0
  movss  0xc(%rsp),%xmm2
  pxor   %xmm1,%xmm1
  movapd %xmm3,%xmm6
  addsd  %xmm3,%xmm0
  add    $0x18,%rsp
  subsd  %xmm5,%xmm6
  addsd  %xmm3,%xmm5
  subsd  %xmm4,%xmm3
  divsd  %xmm5,%xmm6
  divsd  %xmm3,%xmm0
  cvtsd2ss %xmm6,%xmm1
  cvtsd2ss %xmm0,%xmm0
  jmp    1070 <mix@plt>
```

tool_calls: name: read_data
arguments: label: D7
type_infered: f64

tool: D7: 1

assistant: ```c
float get_eta(float f0, float f90) {
  float eta_min = (1.0 - f0) / (1.0 + f0);
  float sqrt_f0 = sqrt(f0);
  float eta_max = (1.0 + sqrt_f0) / (1.0 - sqrt_f0);
  return mix(eta_max, eta_min, f90);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<get_eta>:
  endbr64
  sub    $0x18,%rsp
  pxor   %xmm5,%xmm5
  movss  %xmm1,0xc(%rsp)
  cvtss2sd %xmm0,%xmm5
  movsd  %xmm5,(%rsp)
  call   <sqrt@plt>
  movsd  (%rsp),%xmm5
  pxor   %xmm4,%xmm4
  movsd  D1(%rip),%xmm3
  cvtss2sd %xmm0,%xmm4
  movapd %xmm4,%xmm0
  movss  0xc(%rsp),%xmm2
  pxor   %xmm1,%xmm1
  movapd %xmm3,%xmm6
  addsd  %xmm3,%xmm0
  add    $0x18,%rsp
  subsd  %xmm5,%xmm6
  addsd  %xmm3,%xmm5
  subsd  %xmm4,%xmm3
  divsd  %xmm5,%xmm6
  divsd  %xmm3,%xmm0
  cvtsd2ss %xmm6,%xmm1
  cvtsd2ss %xmm0,%xmm0
  jmp    1070 <mix@plt>
```

tool_calls: name: unpack_data
arguments: tag: D1
datatype: f64

tool: D1: 1

assistant: ```c
float get_eta(float f0, float f90) {
  float eta_min = (1.0 - f0) / (1.0 + f0);
  float sqrt_f0 = sqrt(f0);
  float eta_max = (1.0 + sqrt_f0) / (1.0 - sqrt_f0);
  return mix(eta_max, eta_min, f90);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<div_trunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %edx,-0x2c(%rbp)
  mov    %rcx,-0x38(%rbp)
  mov    $0x3f,%edx
  mov    -0x24(%rbp),%eax
  mov    %edx,%ecx
  sar    %cl,%eax
  mov    %eax,-0x1c(%rbp)
  mov    $0x3f,%edx
  mov    -0x2c(%rbp),%eax
  mov    %edx,%ecx
  sar    %cl,%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x1c(%rbp),%eax
  xor    -0x18(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x28(%rbp),%eax
  xor    -0x1c(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x24(%rbp),%eax
  xor    -0x1c(%rbp),%eax
  mov    %eax,%edx
  cmpl   $0x0,-0x10(%rbp)
  sete   %al
  movzbl %al,%eax
  and    -0x1c(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x2c(%rbp),%eax
  xor    -0x18(%rbp),%eax
  sub    -0x18(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x38(%rbp),%rcx
  mov    -0x8(%rbp),%edx
  mov    -0x10(%rbp),%esi
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <divu@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xffffffff,-0x4(%rbp)
  jne    L8
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  jmp    L34
L8:
  mov    -0x38(%rbp),%rax
  mov    (%rax),%eax
  xor    -0x14(%rbp),%eax
  sub    -0x14(%rbp),%eax
  mov    %eax,%edx
  mov    -0x38(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%eax
  xor    -0x14(%rbp),%eax
  test   %eax,%eax
  jns    L16
  mov    -0x38(%rbp),%rax
  mov    (%rax),%eax
  test   %eax,%eax
  je     L16
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  jmp    L34
L16:
  mov    -0x4(%rbp),%eax
  xor    -0x1c(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
L34:
  leave
  ret
```

assistant: ```c
int64_t div_trunc(int64_t nhi, int64_t nlo, int64_t m, int64_t *q) {
  int64_t n_neg = nhi >> 63;
  int64_t m_neg = m >> 63;
  int64_t diff = n_neg ^ m_neg;
  uint64_t ulo = (uint64_t)((nlo ^ n_neg) - n_neg);
  uint64_t uhi = (uint64_t)((nhi ^ n_neg) + (n_neg & !ulo));
  uint64_t v = (uint64_t)((m ^ m_neg) - m_neg);
  int64_t r = (int64_t)divu(uhi, ulo, v, (uint64_t *)q);
  if (r == -1)
    return INT64_MIN;
  *q = (*q ^ diff) - diff;
  if ((*q ^ diff) < 0 && *q)
    return INT64_MIN;
  r = (r ^ n_neg) - n_neg;
  return r;
}
```