ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<a_ctz_l>:
  endbr64
  mov    $0x20,%eax
  test   %rdi,%rdi
  je     L6
  test   $0x1,%dil
  jne    L17
  mov    $0x0,%eax
L16:
  add    $0x1,%eax
  shr    %rdi
  test   $0x1,%dil
  je     L16
  ret
L17:
  mov    $0x0,%eax
L6:
  ret
```

assistant: ```c
int a_ctz_l(unsigned long x) {
  if (x == 0)
    return 32;
  int nTrailingZeros = 0;
  while (!(x & 1)) {
    ++nTrailingZeros;
    x >>= 1;
  }
  return nTrailingZeros;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<a_ctz_l>:
  endbr64
  mov    $0x20,%eax
  test   %rdi,%rdi
  je     L3
  xor    %eax,%eax
  test   $0x1,%dil
  jne    L3
  cs nopw 0x0(%rax,%rax,1)
L5:
  shr    %rdi
  add    $0x1,%eax
  test   $0x1,%dil
  je     L5
  ret
  nopl   (%rax)
L3:
  ret
```

assistant: ```c
int a_ctz_l(unsigned long x) {
  if (x == 0)
    return 32;
  int nTrailingZeros = 0;
  while (!(x & 1)) {
    ++nTrailingZeros;
    x >>= 1;
  }
  return nTrailingZeros;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<a_ctz_l>:
  endbr64
  mov    $0x20,%eax
  test   %rdi,%rdi
  je     L25
  xor    %eax,%eax
  test   $0x1,%dil
  jne    L25
  cs nopw 0x0(%rax,%rax,1)
L13:
  shr    %rdi
  add    $0x1,%eax
  test   $0x1,%dil
  je     L13
  ret
  nopl   (%rax)
L25:
  ret
```

assistant: ```c
int a_ctz_l(unsigned long x) {
  if (x == 0)
    return 32;
  int nTrailingZeros = 0;
  while (!(x & 1)) {
    ++nTrailingZeros;
    x >>= 1;
  }
  return nTrailingZeros;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<visitor_counter>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int visitor_counter(const char *key, void *data, void *arg) {
  (void)key;
  (void)data;
  size_t *count = arg;
  count[0]++;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<visitor_counter>:
  endbr64
  addq   $0x1,(%rdx)
  mov    $0x0,%eax
  ret
```

assistant: ```c
int visitor_counter(const char *key, void *data, void *arg) {
  (void)key;
  (void)data;
  size_t *count = arg;
  count[0]++;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<visitor_counter>:
  endbr64
  addq   $0x1,(%rdx)
  xor    %eax,%eax
  ret
```

assistant: ```c
int visitor_counter(const char *key, void *data, void *arg) {
  (void)key;
  (void)data;
  size_t *count = arg;
  count[0]++;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<visitor_counter>:
  endbr64
  addq   $0x1,(%rdx)
  xor    %eax,%eax
  ret
```

assistant: ```c
int visitor_counter(const char *key, void *data, void *arg) {
  (void)key;
  (void)data;
  size_t *count = arg;
  count[0]++;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<scnprintf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xe0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xd0(%rbp)
  mov    %rdx,-0xd8(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L7
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L7:
  mov    -0xd0(%rbp),%rax
  mov    %eax,-0xbc(%rbp)
  mov    -0xd8(%rbp),%rdx
  mov    -0xb8(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xb8(%rbp),%ecx
  mov    -0xd8(%rbp),%rdx
  mov    -0xd0(%rbp),%rsi
  mov    -0xc8(%rbp),%rax
  mov    %rax,%rdi
  call   <vsnprintf@plt>
  mov    %eax,-0xb4(%rbp)
  mov    -0xb8(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    -0xb4(%rbp),%eax
  cmp    -0xbc(%rbp),%eax
  jl     L29
  mov    -0xbc(%rbp),%eax
  sub    $0x1,%eax
  jmp    L3
L29:
  mov    -0xb4(%rbp),%eax
L3:
  leave
  ret
```

assistant: ```c
int scnprintf(char *buf, size_t size, const char *fmt, ...) {
  ssize_t ssize = size;
  va_list args;
  int i;
  va_start(args, fmt);
  i = vsnprintf(buf, size, fmt, args);
  va_end(args);
  return (i >= ssize) ? (ssize - 1) : i;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<scnprintf>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%rbp
  mov    %rdx,%rbx
  mov    %rdx,%rsi
  mov    $0x0,%r12d
  mov    %r12d,%edi
  call   <va_start@plt>
  mov    %r12d,%ecx
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <vsnprintf@plt>
  mov    %eax,%ebx
  mov    %r12d,%edi
  call   <va_end@plt>
  lea    -0x1(%rbp),%eax
  cmp    %ebp,%ebx
  cmovge %eax,%ebx
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int scnprintf(char *buf, size_t size, const char *fmt, ...) {
  ssize_t ssize = size;
  va_list args;
  int i;
  va_start(args, fmt);
  i = vsnprintf(buf, size, fmt, args);
  va_end(args);
  return (i >= ssize) ? (ssize - 1) : i;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<scnprintf>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %ebp,%edi
  mov    %rsi,%rbx
  mov    %rdx,%rsi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %ebp,%ecx
  mov    %r12,%rdx
  mov    %rbx,%rsi
  mov    %r13,%rdi
  call   <vsnprintf@plt>
  mov    %ebp,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  lea    -0x1(%rbx),%eax
  cmp    %ebx,%r12d
  cmovge %eax,%r12d
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int scnprintf(char *buf, size_t size, const char *fmt, ...) {
  ssize_t ssize = size;
  va_list args;
  int i;
  va_start(args, fmt);
  i = vsnprintf(buf, size, fmt, args);
  va_end(args);
  return (i >= ssize) ? (ssize - 1) : i;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<scnprintf>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %ebp,%edi
  mov    %rsi,%rbx
  mov    %rdx,%rsi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %ebp,%ecx
  mov    %r12,%rdx
  mov    %rbx,%rsi
  mov    %r13,%rdi
  call   <vsnprintf@plt>
  mov    %ebp,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  lea    -0x1(%rbx),%eax
  cmp    %ebx,%r12d
  cmovge %eax,%r12d
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int scnprintf(char *buf, size_t size, const char *fmt, ...) {
  ssize_t ssize = size;
  va_list args;
  int i;
  va_start(args, fmt);
  i = vsnprintf(buf, size, fmt, args);
  va_end(args);
  return (i >= ssize) ? (ssize - 1) : i;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<col_means>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  mov    -0x28(%rbp),%eax
  add    $0x1,%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L39
L31:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  pxor   %xmm0,%xmm0
  movsd  %xmm0,(%rax)
  movl   $0x0,-0xc(%rbp)
  jmp    L20
L35:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm2
  subsd  %xmm2,%xmm0
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  pxor   %xmm2,%xmm2
  cvtsi2sd %eax,%xmm2
  divsd  %xmm2,%xmm0
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0xc(%rbp)
L20:
  mov    -0xc(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L35
  addl   $0x1,-0x10(%rbp)
L39:
  mov    -0x10(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jl     L31
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  pxor   %xmm0,%xmm0
  movsd  %xmm0,(%rax)
  movl   $0x0,-0xc(%rbp)
  jmp    L21
L50:
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm2
  subsd  %xmm2,%xmm0
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  pxor   %xmm2,%xmm2
  cvtsi2sd %eax,%xmm2
  divsd  %xmm2,%xmm0
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0xc(%rbp)
L21:
  mov    -0xc(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L50
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
double *col_means(double **X, double *Y, int r, int c) {
  int i, j;
  double *res;
  res = malloc((c + 1) * sizeof(double));
  for (i = 0; i < c; ++i) {
    res[i] = 0;
    for (j = 0; j < r; ++j)
      res[i] += (X[j][i] - res[i]) / (j + 1);
  }
  res[c] = 0;
  for (j = 0; j < r; ++j)
    res[c] += (Y[j] - res[c]) / (j + 1);
  return res;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<col_means>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%r12
  mov    %edx,%ebp
  mov    %ecx,%r13d
  lea    0x8(,%rcx,8),%edi
  call   <malloc@plt>
  mov    %rax,%r8
  test   %r13d,%r13d
  jle    L57
  mov    %rax,%r9
  mov    %r13d,%eax
  lea    (%r8,%rax,8),%r10
  mov    $0x0,%esi
  lea    0x1(%rbp),%edi
L17:
  mov    %r9,%rdx
  movq   $0x0,(%r9)
  test   %ebp,%ebp
  jle    L5
  mov    $0x1,%eax
L18:
  movsd  (%rdx),%xmm1
  mov    -0x8(%rbx,%rax,8),%rcx
  movsd  (%rcx,%rsi,1),%xmm0
  subsd  %xmm1,%xmm0
  pxor   %xmm2,%xmm2
  cvtsi2sd %eax,%xmm2
  divsd  %xmm2,%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rdx)
  add    $0x1,%rax
  cmp    %rdi,%rax
  jne    L18
L5:
  add    $0x8,%r9
  add    $0x8,%rsi
  cmp    %r10,%r9
  jne    L17
L57:
  movslq %r13d,%r13
  lea    (%r8,%r13,8),%rdx
  movq   $0x0,(%rdx)
  test   %ebp,%ebp
  jle    L23
  lea    0x1(%rbp),%ecx
  mov    $0x1,%eax
L48:
  movsd  (%rdx),%xmm1
  movsd  -0x8(%r12,%rax,8),%xmm0
  subsd  %xmm1,%xmm0
  pxor   %xmm2,%xmm2
  cvtsi2sd %eax,%xmm2
  divsd  %xmm2,%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rdx)
  add    $0x1,%rax
  cmp    %rax,%rcx
  jne    L48
L23:
  mov    %r8,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
double *col_means(double **X, double *Y, int r, int c) {
  int i, j;
  double *res;
  res = malloc((c + 1) * sizeof(double));
  for (i = 0; i < c; ++i) {
    res[i] = 0;
    for (j = 0; j < r; ++j)
      res[i] += (X[j][i] - res[i]) / (j + 1);
  }
  res[c] = 0;
  for (j = 0; j < r; ++j)
    res[c] += (Y[j] - res[c]) / (j + 1);
  return res;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<col_means>:
  endbr64
  push   %r13
  movslq %ecx,%r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x8(,%r13,8),%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r8
  test   %r13d,%r13d
  jle    L69
  mov    %rax,%rcx
  mov    %r13d,%eax
  pxor   %xmm3,%xmm3
  xor    %esi,%esi
  lea    (%r8,%rax,8),%r9
  lea    0x1(%r12),%edi
  nopl   0x0(%rax,%rax,1)
L36:
  movq   $0x0,(%rcx)
  mov    $0x1,%eax
  movapd %xmm3,%xmm1
  test   %r12d,%r12d
  jle    L53
  nopl   (%rax)
L35:
  mov    -0x8(%rbx,%rax,8),%rdx
  pxor   %xmm2,%xmm2
  cvtsi2sd %eax,%xmm2
  add    $0x1,%rax
  movsd  (%rdx,%rsi,1),%xmm0
  subsd  %xmm1,%xmm0
  divsd  %xmm2,%xmm0
  addsd  %xmm0,%xmm1
  movsd  %xmm1,(%rcx)
  cmp    %rax,%rdi
  jne    L35
L53:
  add    $0x8,%rcx
  add    $0x8,%rsi
  cmp    %rcx,%r9
  jne    L36
L7:
  lea    (%r8,%r13,8),%rdx
  movq   $0x0,(%rdx)
  test   %r12d,%r12d
  jle    L66
  lea    0x1(%r12),%ecx
  mov    $0x1,%eax
  movapd %xmm3,%xmm1
  xchg   %ax,%ax
L20:
  movsd  -0x8(%rbp,%rax,8),%xmm0
  pxor   %xmm2,%xmm2
  cvtsi2sd %eax,%xmm2
  add    $0x1,%rax
  subsd  %xmm1,%xmm0
  divsd  %xmm2,%xmm0
  addsd  %xmm0,%xmm1
  movsd  %xmm1,(%rdx)
  cmp    %rcx,%rax
  jne    L20
L66:
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L69:
  pxor   %xmm3,%xmm3
  jmp    L7
```

assistant: ```c
double *col_means(double **X, double *Y, int r, int c) {
  int i, j;
  double *res;
  res = malloc((c + 1) * sizeof(double));
  for (i = 0; i < c; ++i) {
    res[i] = 0;
    for (j = 0; j < r; ++j)
      res[i] += (X[j][i] - res[i]) / (j + 1);
  }
  res[c] = 0;
  for (j = 0; j < r; ++j)
    res[c] += (Y[j] - res[c]) / (j + 1);
  return res;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<col_means>:
  endbr64
  push   %r13
  movslq %ecx,%r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x8(,%r13,8),%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r8
  test   %r13d,%r13d
  jle    L3
  mov    %rax,%rcx
  mov    %r13d,%eax
  pxor   %xmm3,%xmm3
  xor    %esi,%esi
  lea    (%r8,%rax,8),%r9
  lea    0x1(%r12),%edi
  nopl   0x0(%rax,%rax,1)
L47:
  movq   $0x0,(%rcx)
  mov    $0x1,%eax
  movapd %xmm3,%xmm1
  test   %r12d,%r12d
  jle    L40
  nopl   (%rax)
L38:
  mov    -0x8(%rbx,%rax,8),%rdx
  pxor   %xmm2,%xmm2
  cvtsi2sd %eax,%xmm2
  add    $0x1,%rax
  movsd  (%rdx,%rsi,1),%xmm0
  subsd  %xmm1,%xmm0
  divsd  %xmm2,%xmm0
  addsd  %xmm0,%xmm1
  movsd  %xmm1,(%rcx)
  cmp    %rax,%rdi
  jne    L38
L40:
  add    $0x8,%rcx
  add    $0x8,%rsi
  cmp    %rcx,%r9
  jne    L47
L51:
  lea    (%r8,%r13,8),%rdx
  movq   $0x0,(%rdx)
  test   %r12d,%r12d
  jle    L12
  lea    0x1(%r12),%ecx
  mov    $0x1,%eax
  movapd %xmm3,%xmm1
  xchg   %ax,%ax
L68:
  movsd  -0x8(%rbp,%rax,8),%xmm0
  pxor   %xmm2,%xmm2
  cvtsi2sd %eax,%xmm2
  add    $0x1,%rax
  subsd  %xmm1,%xmm0
  divsd  %xmm2,%xmm0
  addsd  %xmm0,%xmm1
  movsd  %xmm1,(%rdx)
  cmp    %rax,%rcx
  jne    L68
L12:
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  pxor   %xmm3,%xmm3
  jmp    L51
```

assistant: ```c
double *col_means(double **X, double *Y, int r, int c) {
  int i, j;
  double *res;
  res = malloc((c + 1) * sizeof(double));
  for (i = 0; i < c; ++i) {
    res[i] = 0;
    for (j = 0; j < r; ++j)
      res[i] += (X[j][i] - res[i]) / (j + 1);
  }
  res[c] = 0;
  for (j = 0; j < r; ++j)
    res[c] += (Y[j] - res[c]) / (j + 1);
  return res;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<solcon_dobldobl_drop_coordinate_by_index>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    -0x8(%rbp),%rcx
  mov    -0x10(%rbp),%rdx
  lea    -0x24(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x15c,%edi
  call   <_ada_use_c2phc@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  leave
  ret
```

assistant: ```c
int solcon_dobldobl_drop_coordinate_by_index(int k) {
  int fail, *b;
  double *c;
  fail = _ada_use_c2phc(348, &k, b, c);
  return fail;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<solcon_dobldobl_drop_coordinate_by_index>:
  endbr64
  sub    $0x18,%rsp
  mov    %edi,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x15c,%edi
  call   <_ada_use_c2phc@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int solcon_dobldobl_drop_coordinate_by_index(int k) {
  int fail, *b;
  double *c;
  fail = _ada_use_c2phc(348, &k, b, c);
  return fail;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<solcon_dobldobl_drop_coordinate_by_index>:
  endbr64
  sub    $0x18,%rsp
  xor    %ecx,%ecx
  xor    %edx,%edx
  mov    %edi,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  mov    $0x15c,%edi
  call   <_ada_use_c2phc@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int solcon_dobldobl_drop_coordinate_by_index(int k) {
  int fail, *b;
  double *c;
  fail = _ada_use_c2phc(348, &k, b, c);
  return fail;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<solcon_dobldobl_drop_coordinate_by_index>:
  endbr64
  sub    $0x18,%rsp
  xor    %ecx,%ecx
  xor    %edx,%edx
  mov    %edi,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  mov    $0x15c,%edi
  call   <_ada_use_c2phc@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int solcon_dobldobl_drop_coordinate_by_index(int k) {
  int fail, *b;
  double *c;
  fail = _ada_use_c2phc(348, &k, b, c);
  return fail;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<rdrand_long>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    $0xa,%eax
  mov    %eax,%edx
L0:
  rdrand %rax
  jb     L15
  dec    %edx
  jne    L0
L15:
  mov    %rax,%rcx
  mov    %edx,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rcx,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  pop    %rbp
  ret
```

assistant: ```c
long rdrand_long(unsigned long *v) {
  int ok;
  asm volatile("1: "
               ".byte 0x48,0x0f,0xc7,0xf0"
               "\n\t"
               "jc 2f\n\t"
               "decl %0\n\t"
               "jnz 1b\n\t"
               "2:"
               : "=r"(ok), "=a"(*v)
               : "0"(10));
  return ok;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<rdrand_long>:
  endbr64
  mov    $0xa,%edx
L10:
  rdrand %rax
  jb     L9
  dec    %edx
  jne    L10
L9:
  mov    %rax,(%rdi)
  movslq %edx,%rax
  ret
```

assistant: ```c
long rdrand_long(unsigned long *v) {
  int ok;
  asm volatile("1: "
               ".byte 0x48,0x0f,0xc7,0xf0"
               "\n\t"
               "jc 2f\n\t"
               "decl %0\n\t"
               "jnz 1b\n\t"
               "2:"
               : "=r"(ok), "=a"(*v)
               : "0"(10));
  return ok;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<rdrand_long>:
  endbr64
  mov    $0xa,%edx
L4:
  rdrand %rax
  jb     L15
  dec    %edx
  jne    L4
L15:
  mov    %rax,(%rdi)
  movslq %edx,%rax
  ret
```

assistant: ```c
long rdrand_long(unsigned long *v) {
  int ok;
  asm volatile("1: "
               ".byte 0x48,0x0f,0xc7,0xf0"
               "\n\t"
               "jc 2f\n\t"
               "decl %0\n\t"
               "jnz 1b\n\t"
               "2:"
               : "=r"(ok), "=a"(*v)
               : "0"(10));
  return ok;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<rdrand_long>:
  endbr64
  mov    $0xa,%edx
L5:
  rdrand %rax
  jb     L10
  dec    %edx
  jne    L5
L10:
  mov    %rax,(%rdi)
  movslq %edx,%rax
  ret
```

assistant: ```c
long rdrand_long(unsigned long *v) {
  int ok;
  asm volatile("1: "
               ".byte 0x48,0x0f,0xc7,0xf0"
               "\n\t"
               "jc 2f\n\t"
               "decl %0\n\t"
               "jnz 1b\n\t"
               "2:"
               : "=r"(ok), "=a"(*v)
               : "0"(10));
  return ok;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<bound_int16>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  cmp    %rax,-0x10(%rbp)
  jge    L20
  mov    -0x8(%rbp),%rax
  mov    -0x10(%rbp),%rdx
  mov    %rdx,(%rax)
  jmp    L0
L20:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  cmp    %rax,-0x18(%rbp)
  jle    L0
  mov    -0x8(%rbp),%rax
  mov    -0x18(%rbp),%rdx
  mov    %rdx,(%rax)
L0:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void bound_int16(int16_t *number, int16_t max, int16_t min) {
  if (*number > max)
    *number = max;
  else if (*number < min)
    *number = min;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<bound_int16>:
  endbr64
  mov    (%rdi),%rax
  cmp    %rsi,%rax
  jle    L10
  mov    %rsi,(%rdi)
  ret
L10:
  cmp    %rdx,%rax
  jge    L6
  mov    %rdx,(%rdi)
L6:
  ret
```

assistant: ```c
void bound_int16(int16_t *number, int16_t max, int16_t min) {
  if (*number > max)
    *number = max;
  else if (*number < min)
    *number = min;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<bound_int16>:
  endbr64
  mov    (%rdi),%rax
  cmp    %rsi,%rax
  jle    L10
  mov    %rsi,(%rdi)
  ret
L10:
  cmp    %rdx,%rax
  jge    L8
  mov    %rdx,(%rdi)
L8:
  ret
```

assistant: ```c
void bound_int16(int16_t *number, int16_t max, int16_t min) {
  if (*number > max)
    *number = max;
  else if (*number < min)
    *number = min;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<bound_int16>:
  endbr64
  mov    (%rdi),%rax
  cmp    %rsi,%rax
  jle    L9
  mov    %rsi,(%rdi)
  ret
L9:
  cmp    %rdx,%rax
  jge    L12
  mov    %rdx,(%rdi)
L12:
  ret
```

assistant: ```c
void bound_int16(int16_t *number, int16_t max, int16_t min) {
  if (*number > max)
    *number = max;
  else if (*number < min)
    *number = min;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<backward_zero_eater>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x2e,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L23
  addq   $0x1,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  cltq
  mov    %rax,-0x8(%rbp)
L9:
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x30,%al
  jne    L40
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  subq   $0x1,-0x8(%rbp)
  jmp    L9
L40:
  nop
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  jne    L2
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L2
L23:
  nop
L2:
  leave
  ret
```

assistant: ```c
void backward_zero_eater(char *str) {
  str = strchr(str, '.');
  if (str == NULL)
    return;
  str++;
  size_t fraction = strlen(str) - 1;
  while (true) {
    if (str[fraction] == '0')
      str[fraction] = '\0';
    else
      break;
    fraction--;
  }
  if (str[fraction] == '.')
    str[fraction] = '\0';
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<backward_zero_eater>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x2e,%esi
  call   <strchr@plt>
  test   %rax,%rax
  je     L5
  mov    %rax,%rbx
  lea    0x1(%rax),%rbp
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%edx
  movslq %edx,%rdx
  lea    0x0(%rbp,%rdx,1),%rdi
  movzbl (%rdi),%esi
  cmp    $0x30,%sil
  jne    L41
  add    %rbx,%rdx
  mov    %rdi,%rcx
  cltq
  lea    -0x1(%rbx,%rax,1),%rax
  sub    %rdi,%rax
L4:
  movb   $0x0,(%rcx)
  add    %rax,%rcx
  sub    $0x1,%rdx
  movzbl 0x1(%rdx),%esi
  cmp    $0x30,%sil
  je     L4
L52:
  cmp    $0x2e,%sil
  je     L2
L5:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L41:
  mov    %rdi,%rcx
  jmp    L52
L2:
  movb   $0x0,(%rcx)
  jmp    L5
```

assistant: ```c
void backward_zero_eater(char *str) {
  str = strchr(str, '.');
  if (str == NULL)
    return;
  str++;
  size_t fraction = strlen(str) - 1;
  while (true) {
    if (str[fraction] == '0')
      str[fraction] = '\0';
    else
      break;
    fraction--;
  }
  if (str[fraction] == '.')
    str[fraction] = '\0';
}
```