ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  mov    $0x32,%ecx
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1a8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x198(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  rep stos %rax,%es:(%rdi)
  mov    $0xf,%edi
  call   <fibonacci_serise@plt>
  test   %eax,%eax
  jle    L38
  lea    -0x1(%rax),%r13d
  movslq %eax,%r12
  lea    D10(%rip),%r14
  xor    %ebx,%ebx
  lea    D35(%rip),%r15
  jmp    L100
L45:
  mov    %r14,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  je     L38
L100:
  mov    0x0(%rbp,%rbx,4),%esi
  cmp    %ebx,%r13d
  jne    L45
  mov    %r15,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L100
  nopl   0x0(%rax,%rax,1)
L38:
  mov    $0x190,%edx
  xor    %esi,%esi
  mov    %rbp,%rdi
  xor    %ebx,%ebx
  call   <memset@plt>
  mov    %rbp,%rsi
  mov    $0x64,%edi
  lea    D10(%rip),%r14
  call   <fibonacci_serise_upto@plt>
  lea    D35(%rip),%r15
  lea    -0x1(%rax),%r13d
  movslq %eax,%r12
  test   %eax,%eax
  jg     L9
  jmp    L81
  nopl   0x0(%rax)
L102:
  mov    %r14,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  je     L81
L9:
  mov    0x0(%rbp,%rbx,4),%esi
  cmp    %ebx,%r13d
  jne    L102
  mov    %r15,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L9
  nopl   0x0(%rax,%rax,1)
L81:
  mov    0x198(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  add    $0x1a8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L32:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D10
type_guessed: string
name: disassemble_data
arguments: tag: D35
type_guessed: string

tool: D10: %d,

tool: D35: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int serise[100] = {0x00};
  int max_term = fibonacci_serise(15, serise);
  for (int i = 0; i < max_term; i++) {
    if ((max_term - 1) == i) {
      printf("%d\n", serise[i]);
    } else {
      printf("%d,", serise[i]);
    }
  }
  memset(serise, 0x00, sizeof(serise));
  max_term = fibonacci_serise_upto(100, serise);
  for (int i = 0; i < max_term; i++) {
    if ((max_term - 1) == i) {
      printf("%d\n", serise[i]);
    } else {
      printf("%d,", serise[i]);
    }
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<__itt_fsplit>:
  endbr64
  test   %rdi,%rdi
  je     L235
  push   %rbx
  mov    %rdi,%r10
  mov    %rsi,%r9
  mov    %rdx,%rbx
  mov    %rcx,%r11
  test   %rsi,%rsi
  je     L266
  test   %rdx,%rdx
  je     L212
  test   %rcx,%rcx
  je     L212
  movzbl (%rdi),%esi
  test   %sil,%sil
  je     L46
  movzbl (%r9),%edx
  mov    %rdi,%r8
  mov    $0x0,%edi
  jmp    L65
L46:
  mov    $0x0,%r9d
  jmp    L202
L266:
  mov    $0x0,%r9d
  jmp    L202
L212:
  mov    $0x0,%r9d
  jmp    L202
L183:
  mov    $0x0,%r9d
  jmp    L202
L283:
  add    %rdi,%r9
  add    $0x1,%r8
  movzbl -0x1(%r8),%ecx
  test   %cl,%cl
  je     L202
L0:
  mov    %rbx,%rdx
  mov    %esi,%eax
L165:
  cmp    %cl,%al
  je     L283
  add    $0x1,%rdx
  movzbl -0x1(%rdx),%eax
  test   %al,%al
  jne    L165
L202:
  mov    %r9,%rax
  pop    %rbx
  ret
L79:
  add    $0x1,%edi
  addl   $0x1,(%r11)
  add    $0x1,%r8
  movzbl -0x1(%r8),%ecx
  test   %cl,%cl
  je     L36
L168:
  movzbl (%r9),%esi
  mov    %rbx,%rdx
  mov    %esi,%eax
  test   %sil,%sil
  je     L79
L126:
  cmp    %cl,%al
  je     L206
  add    $0x1,%rdx
  movzbl -0x1(%rdx),%eax
  test   %al,%al
  jne    L126
  jmp    L79
L206:
  movslq %edi,%rdi
  lea    (%r10,%rdi,1),%rax
  movzbl (%rax),%ecx
  test   %cl,%cl
  je     L267
  lea    0x1(%r10,%rdi,1),%r8
  mov    %rax,%r9
  mov    %r8,%rdi
  sub    %rax,%rdi
  jmp    L0
L267:
  mov    %rax,%r9
  jmp    L202
L234:
  add    $0x1,%edi
  add    $0x1,%r8
  movzbl (%r8),%esi
  test   %sil,%sil
  je     L227
L65:
  test   %dl,%dl
  je     L187
  lea    0x1(%r9),%rcx
  mov    %edx,%eax
L63:
  cmp    %sil,%al
  je     L234
  add    $0x1,%rcx
  movzbl -0x1(%rcx),%eax
  test   %al,%al
  jne    L63
L187:
  test   %sil,%sil
  je     L183
  movl   $0x0,(%r11)
  mov    %r8,(%rbx)
  movslq %edi,%rax
  movzbl (%r10,%rax,1),%ecx
  lea    0x1(%r10,%rax,1),%r8
  lea    0x1(%r9),%rbx
  test   %cl,%cl
  jne    L168
L36:
  movslq %edi,%r9
  add    %r10,%r9
  jmp    L202
L227:
  mov    $0x0,%r9d
  jmp    L202
L235:
  mov    $0x0,%r9d
  mov    %r9,%rax
  ret
```

assistant: ```c
const char *__itt_fsplit(const char *s, const char *sep, const char **out, int *len) {
  int i;
  int j;
  if (!s || !sep || !out || !len)
    return 0;
  for (i = 0; s[i]; i++) {
    int b = 0;
    for (j = 0; sep[j]; j++)
      if (s[i] == sep[j]) {
        b = 1;
        break;
      }
    if (!b)
      break;
  }
  if (!s[i])
    return 0;
  *len = 0;
  *out = s + i;
  for (; s[i]; i++, (*len)++) {
    int b = 0;
    for (j = 0; sep[j]; j++)
      if (s[i] == sep[j]) {
        b = 1;
        break;
      }
    if (b)
      break;
  }
  for (; s[i]; i++) {
    int b = 0;
    for (j = 0; sep[j]; j++)
      if (s[i] == sep[j]) {
        b = 1;
        break;
      }
    if (!b)
      break;
  }
  return s + i;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<__itt_fsplit>:
  endbr64
  test   %rdi,%rdi
  je     L91
  mov    %rsi,%r11
  test   %rsi,%rsi
  je     L91
  test   %rdx,%rdx
  je     L91
  push   %rbp
  push   %rbx
  mov    %rcx,%rbx
  test   %rcx,%rcx
  je     L168
  movzbl (%rdi),%esi
  mov    %rdi,%r9
  test   %sil,%sil
  je     L168
  movzbl (%r11),%r8d
  xor    %r10d,%r10d
  xchg   %ax,%ax
L127:
  test   %r8b,%r8b
  je     L112
  lea    0x1(%r11),%rcx
  mov    %r8d,%eax
  jmp    L18
  xchg   %ax,%ax
L45:
  movzbl (%rcx),%eax
  add    $0x1,%rcx
  test   %al,%al
  je     L112
L18:
  cmp    %al,%sil
  jne    L45
  movzbl 0x1(%rdi),%esi
  add    $0x1,%rdi
  add    $0x1,%r10d
  test   %sil,%sil
  jne    L127
L168:
  xor    %r8d,%r8d
L180:
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L112:
  movslq %r10d,%rax
  movl   $0x0,(%rbx)
  lea    0x1(%r11),%rbp
  lea    (%r9,%rax,1),%r8
  mov    %rdi,(%rdx)
  lea    0x1(%r9,%rax,1),%rsi
  movzbl (%r8),%ecx
  sub    %esi,%r10d
  test   %cl,%cl
  je     L180
  nopl   0x0(%rax)
L52:
  movzbl (%r11),%edi
  lea    (%r10,%rsi,1),%r8d
  mov    %rbp,%rdx
  mov    %edi,%eax
  test   %dil,%dil
  jne    L95
  jmp    L104
  nopl   0x0(%rax)
L129:
  movzbl (%rdx),%eax
  add    $0x1,%rdx
  test   %al,%al
  je     L104
L95:
  cmp    %al,%cl
  jne    L129
  movslq %r8d,%rax
  lea    (%r9,%rax,1),%r8
  movzbl (%r8),%ecx
  test   %cl,%cl
  je     L180
  lea    0x1(%r9,%rax,1),%rsi
  mov    %r8,%r9
L209:
  mov    %r9,%r8
  mov    %rbp,%rdx
  mov    %edi,%eax
  jmp    L199
L109:
  movzbl (%rdx),%eax
  add    $0x1,%rdx
  test   %al,%al
  je     L180
L199:
  cmp    %cl,%al
  jne    L109
  movzbl (%rsi),%ecx
  mov    %rsi,%r8
  add    $0x1,%r9
  add    $0x1,%rsi
  test   %cl,%cl
  jne    L209
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L104:
  mov    %rsi,%r8
  add    $0x1,%rsi
  addl   $0x1,(%rbx)
  movzbl -0x1(%rsi),%ecx
  test   %cl,%cl
  jne    L52
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
L91:
  xor    %eax,%eax
  ret
```

assistant: ```c
const char *__itt_fsplit(const char *s, const char *sep, const char **out, int *len) {
  int i;
  int j;
  if (!s || !sep || !out || !len)
    return 0;
  for (i = 0; s[i]; i++) {
    int b = 0;
    for (j = 0; sep[j]; j++)
      if (s[i] == sep[j]) {
        b = 1;
        break;
      }
    if (!b)
      break;
  }
  if (!s[i])
    return 0;
  *len = 0;
  *out = s + i;
  for (; s[i]; i++, (*len)++) {
    int b = 0;
    for (j = 0; sep[j]; j++)
      if (s[i] == sep[j]) {
        b = 1;
        break;
      }
    if (b)
      break;
  }
  for (; s[i]; i++) {
    int b = 0;
    for (j = 0; sep[j]; j++)
      if (s[i] == sep[j]) {
        b = 1;
        break;
      }
    if (!b)
      break;
  }
  return s + i;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<__itt_fsplit>:
  endbr64
  test   %rdi,%rdi
  je     L46
  mov    %rsi,%r11
  test   %rsi,%rsi
  je     L46
  test   %rdx,%rdx
  je     L46
  push   %r13
  push   %r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  test   %rcx,%rcx
  je     L194
  movzbl (%rdi),%esi
  mov    %rdi,%r9
  test   %sil,%sil
  je     L194
  movzbl (%r11),%r10d
  xor    %r8d,%r8d
  nopw   0x0(%rax,%rax,1)
L58:
  test   %r10b,%r10b
  je     L183
  lea    0x1(%r11),%rcx
  mov    %r10d,%eax
  jmp    L172
  cs nopw 0x0(%rax,%rax,1)
L40:
  movzbl (%rcx),%eax
  add    $0x1,%rcx
  test   %al,%al
  je     L183
L172:
  cmp    %al,%sil
  jne    L40
  movzbl 0x1(%rdi),%esi
  add    $0x1,%rdi
  add    $0x1,%r8d
  test   %sil,%sil
  jne    L58
L194:
  xor    %r10d,%r10d
L7:
  pop    %rbx
  mov    %r10,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L183:
  movslq %r8d,%rax
  movl   $0x0,0x0(%rbp)
  lea    (%r9,%rax,1),%r10
  mov    %rdi,(%rdx)
  lea    0x1(%r9,%rax,1),%rdi
  movzbl (%r10),%ecx
  test   %cl,%cl
  je     L7
  sub    %edi,%r8d
  mov    $0x1,%r13d
  lea    0x1(%r11),%rbx
  mov    %r8d,%r12d
  sub    %edi,%r13d
  nopl   0x0(%rax)
L53:
  movzbl (%r11),%r8d
  mov    %edi,%r10d
  lea    (%r12,%rdi,1),%esi
  mov    %rbx,%rdx
  mov    %r8d,%eax
  test   %r8b,%r8b
  jne    L11
  jmp    L70
L129:
  movzbl (%rdx),%eax
  add    $0x1,%rdx
  test   %al,%al
  je     L70
L11:
  cmp    %al,%cl
  jne    L129
  movslq %esi,%rsi
  lea    0x1(%r9,%rsi,1),%rdi
  add    %r9,%rsi
L68:
  mov    %rsi,%r10
  mov    %rbx,%rdx
  mov    %r8d,%eax
  jmp    L43
  nopl   (%rax)
L156:
  movzbl (%rdx),%eax
  add    $0x1,%rdx
  test   %al,%al
  je     L7
L43:
  cmp    %cl,%al
  jne    L156
  movzbl (%rdi),%ecx
  mov    %rdi,%r10
  add    $0x1,%rsi
  add    $0x1,%rdi
  test   %cl,%cl
  jne    L68
  pop    %rbx
  mov    %r10,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L70:
  add    %r13d,%r10d
  mov    %r10d,0x0(%rbp)
  mov    %rdi,%r10
  movzbl (%rdi),%ecx
  add    $0x1,%rdi
  test   %cl,%cl
  jne    L53
  pop    %rbx
  mov    %r10,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L46:
  xor    %eax,%eax
  ret
```

assistant: ```c
const char *__itt_fsplit(const char *s, const char *sep, const char **out, int *len) {
  int i;
  int j;
  if (!s || !sep || !out || !len)
    return 0;
  for (i = 0; s[i]; i++) {
    int b = 0;
    for (j = 0; sep[j]; j++)
      if (s[i] == sep[j]) {
        b = 1;
        break;
      }
    if (!b)
      break;
  }
  if (!s[i])
    return 0;
  *len = 0;
  *out = s + i;
  for (; s[i]; i++, (*len)++) {
    int b = 0;
    for (j = 0; sep[j]; j++)
      if (s[i] == sep[j]) {
        b = 1;
        break;
      }
    if (b)
      break;
  }
  for (; s[i]; i++) {
    int b = 0;
    for (j = 0; sep[j]; j++)
      if (s[i] == sep[j]) {
        b = 1;
        break;
      }
    if (!b)
      break;
  }
  return s + i;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<readstring>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x200,%edx
  mov    %rax,%rsi
  mov    $0x1,%edi
  mov    $0x0,%eax
  call   <read@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jg     L9
  mov    $0x0,%edi
  call   <exit@plt>
L9:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int readstring(readbuffer)
char *readbuffer;
{
  int c_max;
  c_max = read(1, readbuffer, 512);
  if (c_max <= 0) {
    exit(0);
  }
  return c_max;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<readstring>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  mov    $0x200,%edx
  mov    $0x1,%edi
  mov    $0x0,%eax
  call   <read@plt>
  test   %eax,%eax
  jle    L8
  add    $0x8,%rsp
  ret
L8:
  mov    $0x0,%edi
  call   <exit@plt>
```

assistant: ```c
int readstring(readbuffer)
char *readbuffer;
{
  int c_max;
  c_max = read(1, readbuffer, 512);
  if (c_max <= 0) {
    exit(0);
  }
  return c_max;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<readstring>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  xor    %eax,%eax
  mov    $0x200,%edx
  mov    $0x1,%edi
  call   <read@plt>
  test   %eax,%eax
  jle    L3
  add    $0x8,%rsp
  ret
  nopl   (%rax)
L3:
  xor    %edi,%edi
  call   <exit@plt>
```

assistant: ```c
int readstring(readbuffer)
char *readbuffer;
{
  int c_max;
  c_max = read(1, readbuffer, 512);
  if (c_max <= 0) {
    exit(0);
  }
  return c_max;
}
```
user: What is the c source code of the assembly code below?

```asm
<readstring>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  xor    %eax,%eax
  mov    $0x200,%edx
  mov    $0x1,%edi
  call   <read@plt>
  test   %eax,%eax
  jle    L9
  add    $0x8,%rsp
  ret
  nopl   (%rax)
L9:
  xor    %edi,%edi
  call   <exit@plt>
```

assistant: ```c
int readstring(readbuffer)
char *readbuffer;
{
  int c_max;
  c_max = read(1, readbuffer, 512);
  if (c_max <= 0) {
    exit(0);
  }
  return c_max;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<XTL_SubVars_IncrementValue>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <XTL_SubVars_GetValueInt@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xffffd8f1,-0x4(%rbp)
  jne    L2
  mov    $0xffffffff,%eax
  jmp    L17
L2:
  addl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <XTL_SubVars_UpdateInt@plt>
L17:
  leave
  ret
```

assistant: ```c
int XTL_SubVars_IncrementValue(char *key) {
  int retval;
  retval = XTL_SubVars_GetValueInt(key);
  if (retval == -9999)
    return -1;
  ++retval;
  return XTL_SubVars_UpdateInt(key, retval);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<XTL_SubVars_IncrementValue>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <XTL_SubVars_GetValueInt@plt>
  cmp    $0xffffd8f1,%eax
  je     L15
  lea    0x1(%rax),%esi
  mov    %rbx,%rdi
  call   <XTL_SubVars_UpdateInt@plt>
L12:
  pop    %rbx
  ret
L15:
  mov    $0xffffffff,%eax
  jmp    L12
```

assistant: ```c
int XTL_SubVars_IncrementValue(char *key) {
  int retval;
  retval = XTL_SubVars_GetValueInt(key);
  if (retval == -9999)
    return -1;
  ++retval;
  return XTL_SubVars_UpdateInt(key, retval);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<XTL_SubVars_IncrementValue>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <XTL_SubVars_GetValueInt@plt>
  cmp    $0xffffd8f1,%eax
  je     L6
  mov    %rbp,%rdi
  lea    0x1(%rax),%esi
  pop    %rbp
  jmp    1070 <XTL_SubVars_UpdateInt@plt>
L6:
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
int XTL_SubVars_IncrementValue(char *key) {
  int retval;
  retval = XTL_SubVars_GetValueInt(key);
  if (retval == -9999)
    return -1;
  ++retval;
  return XTL_SubVars_UpdateInt(key, retval);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<XTL_SubVars_IncrementValue>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <XTL_SubVars_GetValueInt@plt>
  cmp    $0xffffd8f1,%eax
  je     L4
  mov    %rbp,%rdi
  lea    0x1(%rax),%esi
  pop    %rbp
  jmp    1070 <XTL_SubVars_UpdateInt@plt>
L4:
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
int XTL_SubVars_IncrementValue(char *key) {
  int retval;
  retval = XTL_SubVars_GetValueInt(key);
  if (retval == -9999)
    return -1;
  ++retval;
  return XTL_SubVars_UpdateInt(key, retval);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xread_partial>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
L25:
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jns    L43
  mov    D15(%rip),%rax
  mov    (%rax),%rdx
  mov    D10(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  je     L11
  mov    D15(%rip),%rax
  mov    (%rax),%rdx
  mov    D34(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  je     L11
L43:
  mov    -0x8(%rbp),%rax
  jmp    L42
L11:
  nop
  jmp    L25
L42:
  leave
  ret
```

assistant: ```c
ssize_t xread_partial(int fd, void *buf, size_t len) {
  ssize_t n;
  while (1) {
    n = read(fd, buf, len);
    if ((n < 0) && (errno == EAGAIN || errno == EINTR))
      continue;
    return n;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<xread_partial>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r12d
  mov    %rsi,%rbp
  mov    %rdx,%rbx
L15:
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <read@plt>
  test   %rax,%rax
  jns    L17
  mov    D8(%rip),%rdx
  mov    (%rdx),%rdx
  mov    D7(%rip),%rcx
  cmp    (%rcx),%rdx
  je     L15
  mov    D14(%rip),%rcx
  cmp    (%rcx),%rdx
  je     L15
L17:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
ssize_t xread_partial(int fd, void *buf, size_t len) {
  ssize_t n;
  while (1) {
    n = read(fd, buf, len);
    if ((n < 0) && (errno == EAGAIN || errno == EINTR))
      continue;
    return n;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xread_partial>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
L5:
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <read@plt>
  test   %rax,%rax
  jns    L19
  mov    D9(%rip),%rdx
  mov    D20(%rip),%rcx
  mov    (%rdx),%rdx
  cmp    (%rcx),%rdx
  je     L5
  mov    D29(%rip),%rcx
  cmp    (%rcx),%rdx
  je     L5
L19:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
ssize_t xread_partial(int fd, void *buf, size_t len) {
  ssize_t n;
  while (1) {
    n = read(fd, buf, len);
    if ((n < 0) && (errno == EAGAIN || errno == EINTR))
      continue;
    return n;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<xread_partial>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
L2:
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <read@plt>
  test   %rax,%rax
  jns    L11
  mov    D7(%rip),%rdx
  mov    D6(%rip),%rcx
  mov    (%rdx),%rdx
  cmp    (%rcx),%rdx
  je     L2
  mov    D25(%rip),%rcx
  cmp    (%rcx),%rdx
  je     L2
L11:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
ssize_t xread_partial(int fd, void *buf, size_t len) {
  ssize_t n;
  while (1) {
    n = read(fd, buf, len);
    if ((n < 0) && (errno == EAGAIN || errno == EINTR))
      continue;
    return n;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<max>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    %r8d,-0x28(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x24(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jge    L1
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L1
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x4(%rbp)
L1:
  mov    -0x28(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jge    L6
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L6
  mov    -0x28(%rbp),%eax
  mov    %eax,-0x4(%rbp)
L6:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int max(int *a, int n, int i, int j, int k) {
  int m = i;
  if (j < n && a[j] > a[m])
    m = j;
  if (k < n && a[k] > a[m])
    m = k;
  return m;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<max>:
  endbr64
  mov    %edx,%eax
  cmp    %esi,%ecx
  jge    L6
  movslq %ecx,%r9
  movslq %edx,%rdx
  mov    (%rdi,%rdx,4),%edx
  cmp    %edx,(%rdi,%r9,4)
  cmovg  %ecx,%eax
L6:
  cmp    %r8d,%esi
  jle    L2
  movslq %r8d,%rcx
  movslq %eax,%rdx
  mov    (%rdi,%rdx,4),%edx
  cmp    %edx,(%rdi,%rcx,4)
  cmovg  %r8d,%eax
L2:
  ret
```

assistant: ```c
int max(int *a, int n, int i, int j, int k) {
  int m = i;
  if (j < n && a[j] > a[m])
    m = j;
  if (k < n && a[k] > a[m])
    m = k;
  return m;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<max>:
  endbr64
  mov    %edx,%eax
  cmp    %esi,%ecx
  jge    L10
  movslq %ecx,%r9
  movslq %edx,%rdx
  mov    (%rdi,%rdx,4),%edx
  cmp    %edx,(%rdi,%r9,4)
  cmovg  %ecx,%eax
L10:
  cmp    %r8d,%esi
  jle    L4
  movslq %eax,%rdx
  movslq %r8d,%rcx
  mov    (%rdi,%rdx,4),%edx
  cmp    %edx,(%rdi,%rcx,4)
  cmovg  %r8d,%eax
L4:
  ret
```

assistant: ```c
int max(int *a, int n, int i, int j, int k) {
  int m = i;
  if (j < n && a[j] > a[m])
    m = j;
  if (k < n && a[k] > a[m])
    m = k;
  return m;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<max>:
  endbr64
  mov    %edx,%eax
  cmp    %esi,%ecx
  jge    L18
  movslq %ecx,%r9
  movslq %edx,%rdx
  mov    (%rdi,%rdx,4),%edx
  cmp    %edx,(%rdi,%r9,4)
  cmovg  %ecx,%eax
L18:
  cmp    %r8d,%esi
  jle    L6
  movslq %eax,%rdx
  movslq %r8d,%rcx
  mov    (%rdi,%rdx,4),%edx
  cmp    %edx,(%rdi,%rcx,4)
  cmovg  %r8d,%eax
L6:
  ret
```

assistant: ```c
int max(int *a, int n, int i, int j, int k) {
  int m = i;
  if (j < n && a[j] > a[m])
    m = j;
  if (k < n && a[k] > a[m])
    m = k;
  return m;
}
```
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
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  je     L29
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jne    L18
  mov    $0x0,%eax
  jmp    L31
L29:
  nop
  jmp    L16
L18:
  nop
L16:
  mov    $0xffffffff,%eax
L31:
  leave
  ret
```

assistant: ```c
int main() {
  int a;
  int b;
  a = __VERIFIER_nondet_int();
  b = __VERIFIER_nondet_int();
  if (a == b) {
    goto ERROR;
  } else {
    a = b;
  }
  if (a == b) {
    return 0;
  } else {
  ERROR:
    return -1;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  cmp    %eax,%ebx
  sete   %al
  movzbl %al,%eax
  neg    %eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  int a;
  int b;
  a = __VERIFIER_nondet_int();
  b = __VERIFIER_nondet_int();
  if (a == b) {
    goto ERROR;
  } else {
    a = b;
  }
  if (a == b) {
    return 0;
  } else {
  ERROR:
    return -1;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <__VERIFIER_nondet_int@plt>
  cmp    %eax,%ebx
  pop    %rbx
  sete   %al
  movzbl %al,%eax
  neg    %eax
  ret
```

assistant: ```c
int main() {
  int a;
  int b;
  a = __VERIFIER_nondet_int();
  b = __VERIFIER_nondet_int();
  if (a == b) {
    goto ERROR;
  } else {
    a = b;
  }
  if (a == b) {
    return 0;
  } else {
  ERROR:
    return -1;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <__VERIFIER_nondet_int@plt>
  cmp    %eax,%ebx
  pop    %rbx
  sete   %al
  movzbl %al,%eax
  neg    %eax
  ret
```

assistant: ```c
int main() {
  int a;
  int b;
  a = __VERIFIER_nondet_int();
  b = __VERIFIER_nondet_int();
  if (a == b) {
    goto ERROR;
  } else {
    a = b;
  }
  if (a == b) {
    return 0;
  } else {
  ERROR:
    return -1;
  }
}
```