ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<set_number>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %rdx,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  lea    D9(%rip),%rdx
  mov    $0x20,%esi
  mov    %r13,%rdi
  call   <snprintf@plt>
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <set_value@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D9
data_type: string

tool: D9: %d

assistant: ```c
void set_number(const char *group, const char *name, const char *key, int value) {
  char buf[32];
  snprintf(buf, sizeof(buf), "%d", value);
  set_value(group, name, key, buf);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<set_number>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdx,%r13
  lea    D6(%rip),%rdx
  push   %r12
  mov    %rsi,%r12
  mov    $0x20,%esi
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rdi
  call   <snprintf@plt>
  mov    %r14,%rcx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <set_value@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x38,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D6
datatype: string

tool: D6: %d

assistant: ```c
void set_number(const char *group, const char *name, const char *key, int value) {
  char buf[32];
  snprintf(buf, sizeof(buf), "%d", value);
  set_value(group, name, key, buf);
}
```
user: What is the c source code of the assembly code below:

```asm
<set_number>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdx,%r13
  lea    D3(%rip),%rdx
  push   %r12
  mov    %rsi,%r12
  mov    $0x20,%esi
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rdi
  call   <snprintf@plt>
  mov    %r14,%rcx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <set_value@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x38,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D3
datatype: string

tool: D3: %d

assistant: ```c
void set_number(const char *group, const char *name, const char *key, int value) {
  char buf[32];
  snprintf(buf, sizeof(buf), "%d", value);
  set_value(group, name, key, buf);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<lio_listio>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x8(%rbp)
  mov    %rcx,-0x18(%rbp)
  mov    D6(%rip),%rax
  mov    (%rax),%edx
  mov    D15(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
int lio_listio(int mode, struct aiocb *const list[], int nent, struct sigevent *sig) {
  errno = ENOSYS;
  return -1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<lio_listio>:
  endbr64
  mov    D11(%rip),%rax
  mov    (%rax),%edx
  mov    D2(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int lio_listio(int mode, struct aiocb *const list[], int nent, struct sigevent *sig) {
  errno = ENOSYS;
  return -1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<lio_listio>:
  endbr64
  mov    D19(%rip),%rax
  mov    (%rax),%edx
  mov    D14(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int lio_listio(int mode, struct aiocb *const list[], int nent, struct sigevent *sig) {
  errno = ENOSYS;
  return -1;
}
```
user: What is the c source code of the assembly code below?

```asm
<lio_listio>:
  endbr64
  mov    D17(%rip),%rax
  mov    (%rax),%edx
  mov    D1(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int lio_listio(int mode, struct aiocb *const list[], int nent, struct sigevent *sig) {
  errno = ENOSYS;
  return -1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<mutate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %rdx,-0x30(%rbp)
  mov    $0x0,%eax
  call   <random@plt>
  cltd
  idivl  -0x24(%rbp)
  mov    %edx,-0x14(%rbp)
  mov    $0x0,%eax
  call   <random@plt>
  mov    -0x28(%rbp),%edx
  lea    -0x1(%rdx),%ecx
  cltd
  idiv   %ecx
  mov    %edx,-0x10(%rbp)
  mov    $0x0,%eax
  call   <random@plt>
  mov    -0x28(%rbp),%edx
  lea    -0x1(%rdx),%ecx
  cltd
  idiv   %ecx
  mov    %edx,%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  imul   -0x28(%rbp),%eax
  imul   -0x28(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%eax
  imul   -0x28(%rbp),%eax
  cltq
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  cltq
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  sub    $0x1,%eax
  mov    %eax,-0x18(%rbp)
  cmpl   $0x0,-0x18(%rbp)
  jns    L7
  movl   $0x0,-0x18(%rbp)
L7:
  mov    -0x8(%rbp),%rax
  mov    -0x18(%rbp),%edx
  mov    %edx,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void mutate(int mcount, int mwidth, int *individuum) {
  int letter = random() % mcount;
  int row = random() % (mwidth - 1);
  int col = 1 + random() % (mwidth - 1);
  int *pos = individuum + letter * mwidth * mwidth + row * mwidth + col;
  int new = *pos - 1;
  if (new < 0)
    new = 0;
  *pos = new;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<mutate>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r13d
  mov    %esi,%r12d
  mov    %rdx,%rbp
  mov    $0x0,%eax
  call   <random@plt>
  cltd
  idiv   %r13d
  mov    %edx,%ebx
  mov    $0x0,%eax
  call   <random@plt>
  lea    -0x1(%r12),%r14d
  cltd
  idiv   %r14d
  mov    %edx,%r13d
  mov    $0x0,%eax
  call   <random@plt>
  cltd
  idiv   %r14d
  lea    0x1(%rdx),%eax
  cltq
  imul   %r12d,%r13d
  movslq %r13d,%r13
  add    %r13,%rax
  imul   %r12d,%ebx
  imul   %r12d,%ebx
  movslq %ebx,%rbx
  lea    (%rax,%rbx,1),%rdx
  lea    0x0(%rbp,%rdx,4),%rax
  subl   $0x1,(%rax)
  mov    (%rax),%edx
  test   %edx,%edx
  mov    $0x0,%ecx
  cmovs  %ecx,%edx
  mov    %edx,(%rax)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void mutate(int mcount, int mwidth, int *individuum) {
  int letter = random() % mcount;
  int row = random() % (mwidth - 1);
  int col = 1 + random() % (mwidth - 1);
  int *pos = individuum + letter * mwidth * mwidth + row * mwidth + col;
  int new = *pos - 1;
  if (new < 0)
    new = 0;
  *pos = new;
}
```
user: What is the c source code of the assembly code below?

```asm
<mutate>:
  endbr64
  push   %r14
  xor    %eax,%eax
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  lea    -0x1(%rbp),%r14d
  call   <random@plt>
  cltd
  idiv   %r12d
  xor    %eax,%eax
  mov    %edx,%ebx
  call   <random@plt>
  imul   %ebp,%ebx
  cltd
  idiv   %r14d
  xor    %eax,%eax
  imul   %ebp,%ebx
  movslq %ebx,%rbx
  mov    %edx,%r12d
  call   <random@plt>
  imul   %ebp,%r12d
  xor    %ecx,%ecx
  cltd
  idiv   %r14d
  movslq %r12d,%r12
  lea    0x1(%rdx),%eax
  cltq
  add    %r12,%rax
  lea    (%rax,%rbx,1),%rdx
  lea    0x0(%r13,%rdx,4),%rax
  subl   $0x1,(%rax)
  mov    (%rax),%edx
  test   %edx,%edx
  cmovs  %ecx,%edx
  mov    %edx,(%rax)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void mutate(int mcount, int mwidth, int *individuum) {
  int letter = random() % mcount;
  int row = random() % (mwidth - 1);
  int col = 1 + random() % (mwidth - 1);
  int *pos = individuum + letter * mwidth * mwidth + row * mwidth + col;
  int new = *pos - 1;
  if (new < 0)
    new = 0;
  *pos = new;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<mutate>:
  endbr64
  push   %r14
  xor    %eax,%eax
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  lea    -0x1(%rbp),%r14d
  call   <random@plt>
  cltd
  idiv   %r12d
  xor    %eax,%eax
  mov    %edx,%ebx
  call   <random@plt>
  imul   %ebp,%ebx
  cltd
  idiv   %r14d
  xor    %eax,%eax
  imul   %ebp,%ebx
  movslq %ebx,%rbx
  mov    %edx,%r12d
  call   <random@plt>
  imul   %ebp,%r12d
  xor    %ecx,%ecx
  cltd
  idiv   %r14d
  movslq %r12d,%r12
  lea    0x1(%rdx),%eax
  cltq
  add    %r12,%rax
  lea    (%rax,%rbx,1),%rdx
  lea    0x0(%r13,%rdx,4),%rax
  subl   $0x1,(%rax)
  mov    (%rax),%edx
  test   %edx,%edx
  cmovs  %ecx,%edx
  mov    %edx,(%rax)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void mutate(int mcount, int mwidth, int *individuum) {
  int letter = random() % mcount;
  int row = random() % (mwidth - 1);
  int col = 1 + random() % (mwidth - 1);
  int *pos = individuum + letter * mwidth * mwidth + row * mwidth + col;
  int new = *pos - 1;
  if (new < 0)
    new = 0;
  *pos = new;
}
```
user: Translate the below assembly code into C programming language:

```asm
<AppendToLine>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L22
  mov    -0x38(%rbp),%rdx
  mov    -0x2c(%rbp),%ecx
  mov    -0x28(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <CreateLine@plt>
  jmp    L14
L22:
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    %ebx,%eax
  add    $0x1,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    -0x2c(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x28(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  mov    -0x14(%rbp),%edx
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,(%rbx)
  mov    (%rbx),%rax
  test   %rax,%rax
  je     L2
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    -0x38(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcat@plt>
  jmp    L14
L2:
  nop
L14:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void AppendToLine(char *lines[], int line, const char *text) {
  if (lines[line] == NULL)
    CreateLine(lines, line, text);
  else {
    int size = strlen(lines[line]) + strlen(text) + 1;
    if ((lines[line] = realloc(lines[line], size)) == NULL)
      return;
    strcat(lines[line], text);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<AppendToLine>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdx,%rbx
  movslq %esi,%rax
  lea    (%rdi,%rax,8),%rbp
  mov    0x0(%rbp),%rax
  test   %rax,%rax
  je     L19
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%r12d
  mov    %rbx,%rdi
  call   <strlen@plt>
  lea    0x1(%r12,%rax,1),%esi
  mov    0x0(%rbp),%rdi
  call   <realloc@plt>
  mov    %rax,%rdi
  mov    %rax,0x0(%rbp)
  test   %rax,%rax
  je     L28
  mov    %rbx,%rsi
  call   <strcat@plt>
L28:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L19:
  call   <CreateLine@plt>
  jmp    L28
```

assistant: ```c
void AppendToLine(char *lines[], int line, const char *text) {
  if (lines[line] == NULL)
    CreateLine(lines, line, text);
  else {
    int size = strlen(lines[line]) + strlen(text) + 1;
    if ((lines[line] = realloc(lines[line], size)) == NULL)
      return;
    strcat(lines[line], text);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<AppendToLine>:
  endbr64
  movslq %esi,%rax
  push   %r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  lea    (%rdi,%rax,8),%rbx
  mov    (%rbx),%r8
  test   %r8,%r8
  je     L18
  mov    %r8,%rdi
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%r12d
  call   <strlen@plt>
  mov    (%rbx),%rdi
  lea    0x1(%r12,%rax,1),%esi
  call   <realloc@plt>
  mov    %rax,(%rbx)
  mov    %rax,%rdi
  test   %rax,%rax
  je     L1
  pop    %rbx
  mov    %rbp,%rsi
  pop    %rbp
  pop    %r12
  jmp    10b0 <strcat@plt>
  nopl   0x0(%rax)
L1:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L18:
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1080 <CreateLine@plt>
```

assistant: ```c
void AppendToLine(char *lines[], int line, const char *text) {
  if (lines[line] == NULL)
    CreateLine(lines, line, text);
  else {
    int size = strlen(lines[line]) + strlen(text) + 1;
    if ((lines[line] = realloc(lines[line], size)) == NULL)
      return;
    strcat(lines[line], text);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<AppendToLine>:
  endbr64
  movslq %esi,%rax
  push   %r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  lea    (%rdi,%rax,8),%rbx
  mov    (%rbx),%r8
  test   %r8,%r8
  je     L13
  mov    %r8,%rdi
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%r12d
  call   <strlen@plt>
  mov    (%rbx),%rdi
  lea    0x1(%r12,%rax,1),%esi
  call   <realloc@plt>
  mov    %rax,(%rbx)
  mov    %rax,%rdi
  test   %rax,%rax
  je     L6
  pop    %rbx
  mov    %rbp,%rsi
  pop    %rbp
  pop    %r12
  jmp    10b0 <strcat@plt>
  nopl   0x0(%rax)
L6:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L13:
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1080 <CreateLine@plt>
```

assistant: ```c
void AppendToLine(char *lines[], int line, const char *text) {
  if (lines[line] == NULL)
    CreateLine(lines, line, text);
  else {
    int size = strlen(lines[line]) + strlen(text) + 1;
    if ((lines[line] = realloc(lines[line], size)) == NULL)
      return;
    strcat(lines[line], text);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<mat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jle    L5
  cmpl   $0x0,-0x18(%rbp)
  jg     L1
L5:
  mov    $0x0,%eax
  jmp    L8
L1:
  mov    -0x14(%rbp),%eax
  cltq
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  cltq
  imul   %edx,%eax
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L3
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <fatalerr@plt>
L3:
  mov    -0x8(%rbp),%rax
L8:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D6
datatype: string

tool: D6: matrix memory allocation error: n=%d,m=%d


assistant: ```c
extern double *mat(int n, int m) {
  double *p;
  if (n <= 0 || m <= 0) {
    return NULL;
  }
  if (!(p = (double *)malloc(sizeof(double) * n * m))) {
    fatalerr("matrix memory allocation error: n=%d,m=%d\n", n, m);
  }
  return p;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<mat>:
  endbr64
  test   %edi,%edi
  jle    L1
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  test   %esi,%esi
  jle    L26
  imul   %esi,%edi
  shl    $0x3,%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L23
L44:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L23:
  mov    %ebp,%edx
  mov    %ebx,%esi
  lea    D4(%rip),%rdi
  call   <fatalerr@plt>
  mov    $0x0,%eax
  jmp    L44
L26:
  mov    $0x0,%eax
  jmp    L44
L1:
  mov    $0x0,%eax
  ret
```

tool_calls: name: retrieve_data
arguments: label: D4
type_guessed: string

tool: D4: matrix memory allocation error: n=%d,m=%d


assistant: ```c
extern double *mat(int n, int m) {
  double *p;
  if (n <= 0 || m <= 0) {
    return NULL;
  }
  if (!(p = (double *)malloc(sizeof(double) * n * m))) {
    fatalerr("matrix memory allocation error: n=%d,m=%d\n", n, m);
  }
  return p;
}
```
user: What is the c source code of the assembly code below:

```asm
<mat>:
  endbr64
  test   %edi,%edi
  jle    L1
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  sub    $0x8,%rsp
  test   %esi,%esi
  jle    L18
  mov    %edi,%ebp
  imul   %esi,%edi
  shl    $0x3,%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L9
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
L9:
  mov    %r12d,%edx
  mov    %ebp,%esi
  lea    D4(%rip),%rdi
  call   <fatalerr@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L18:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  xor    %eax,%eax
  ret
```

tool_calls: name: parse_data
arguments: label: D4
type_guessed: string

tool: D4: matrix memory allocation error: n=%d,m=%d


assistant: ```c
extern double *mat(int n, int m) {
  double *p;
  if (n <= 0 || m <= 0) {
    return NULL;
  }
  if (!(p = (double *)malloc(sizeof(double) * n * m))) {
    fatalerr("matrix memory allocation error: n=%d,m=%d\n", n, m);
  }
  return p;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<mat>:
  endbr64
  test   %edi,%edi
  jle    L1
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  sub    $0x8,%rsp
  test   %esi,%esi
  jle    L18
  mov    %edi,%ebp
  imul   %esi,%edi
  shl    $0x3,%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L0
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
L0:
  mov    %r12d,%edx
  mov    %ebp,%esi
  lea    D8(%rip),%rdi
  call   <fatalerr@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L18:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  xor    %eax,%eax
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D8
datatype: string

tool: D8: matrix memory allocation error: n=%d,m=%d


assistant: ```c
extern double *mat(int n, int m) {
  double *p;
  if (n <= 0 || m <= 0) {
    return NULL;
  }
  if (!(p = (double *)malloc(sizeof(double) * n * m))) {
    fatalerr("matrix memory allocation error: n=%d,m=%d\n", n, m);
  }
  return p;
}
```
user: What is the c source code of the assembly code below?

```asm
<fBufferEmpty>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L32
  mov    $0x1,%eax
  jmp    L7
L3:
  addq   $0x1,-0x8(%rbp)
L32:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L10
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L3
L10:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L0
  mov    $0x1,%eax
  jmp    L7
L0:
  mov    $0x0,%eax
L7:
  leave
  ret
```

assistant: ```c
char fBufferEmpty(char *pBuffer) {
  if (pBuffer == NULL)
    return 1;
  for (; isspace(*pBuffer) && *pBuffer != '\0'; pBuffer++)
    ;
  if (*pBuffer == '\0')
    return 1;
  else
    return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<fBufferEmpty>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  test   %rdi,%rdi
  jne    L39
  mov    $0x1,%eax
  jmp    L21
L1:
  add    $0x1,%rbx
L39:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L16
  cmpb   $0x0,(%rbx)
  jne    L1
L16:
  cmpb   $0x0,(%rbx)
  sete   %al
L21:
  pop    %rbx
  ret
```

assistant: ```c
char fBufferEmpty(char *pBuffer) {
  if (pBuffer == NULL)
    return 1;
  for (; isspace(*pBuffer) && *pBuffer != '\0'; pBuffer++)
    ;
  if (*pBuffer == '\0')
    return 1;
  else
    return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<fBufferEmpty>:
  endbr64
  test   %rdi,%rdi
  je     L7
  push   %rbx
  mov    %rdi,%rbx
  jmp    L35
  nop
L20:
  test   %al,%al
  je     L37
  add    $0x1,%rbx
L35:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  movzbl (%rbx),%eax
  jne    L20
L37:
  test   %al,%al
  pop    %rbx
  sete   %al
  ret
  nop
L7:
  mov    $0x1,%eax
  ret
```

assistant: ```c
char fBufferEmpty(char *pBuffer) {
  if (pBuffer == NULL)
    return 1;
  for (; isspace(*pBuffer) && *pBuffer != '\0'; pBuffer++)
    ;
  if (*pBuffer == '\0')
    return 1;
  else
    return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<fBufferEmpty>:
  endbr64
  test   %rdi,%rdi
  je     L24
  push   %rbx
  mov    %rdi,%rbx
  jmp    L36
  nop
L21:
  test   %al,%al
  je     L17
  add    $0x1,%rbx
L36:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  movzbl (%rbx),%eax
  jne    L21
L17:
  test   %al,%al
  pop    %rbx
  sete   %al
  ret
  nop
L24:
  mov    $0x1,%eax
  ret
```

assistant: ```c
char fBufferEmpty(char *pBuffer) {
  if (pBuffer == NULL)
    return 1;
  for (; isspace(*pBuffer) && *pBuffer != '\0'; pBuffer++)
    ;
  if (*pBuffer == '\0')
    return 1;
  else
    return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<byteReverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
L8:
  mov    -0x18(%rbp),%rax
  add    $0x3,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  or     %edx,%eax
  shl    $0x10,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x8,%eax
  mov    %eax,%ecx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  or     %ecx,%eax
  or     %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    -0x4(%rbp),%edx
  mov    %edx,(%rax)
  addq   $0x4,-0x18(%rbp)
  subl   $0x1,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jne    L8
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void byteReverse(unsigned char *buf, unsigned longs) {
  unsigned int t;
  do {
    t = (unsigned int)((unsigned int)buf[3] << 8 | buf[2]) << 16 | ((unsigned)buf[1] << 8 | buf[0]);
    *(unsigned int *)buf = t;
    buf += 4;
  } while (--longs);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<byteReverse>:
  endbr64
  lea    -0x1(%rsi),%eax
  lea    0x4(%rdi,%rax,4),%rsi
L7:
  movzbl 0x3(%rdi),%eax
  shl    $0x8,%eax
  movzbl 0x2(%rdi),%edx
  or     %edx,%eax
  shl    $0x10,%eax
  movzbl 0x1(%rdi),%edx
  shl    $0x8,%edx
  movzbl (%rdi),%ecx
  or     %ecx,%edx
  or     %edx,%eax
  mov    %eax,(%rdi)
  add    $0x4,%rdi
  cmp    %rsi,%rdi
  jne    L7
  ret
```

assistant: ```c
void byteReverse(unsigned char *buf, unsigned longs) {
  unsigned int t;
  do {
    t = (unsigned int)((unsigned int)buf[3] << 8 | buf[2]) << 16 | ((unsigned)buf[1] << 8 | buf[0]);
    *(unsigned int *)buf = t;
    buf += 4;
  } while (--longs);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<byteReverse>:
  endbr64
  ret
```

assistant: ```c
void byteReverse(unsigned char *buf, unsigned longs) {
  unsigned int t;
  do {
    t = (unsigned int)((unsigned int)buf[3] << 8 | buf[2]) << 16 | ((unsigned)buf[1] << 8 | buf[0]);
    *(unsigned int *)buf = t;
    buf += 4;
  } while (--longs);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<byteReverse>:
  endbr64
  ret
```

assistant: ```c
void byteReverse(unsigned char *buf, unsigned longs) {
  unsigned int t;
  do {
    t = (unsigned int)((unsigned int)buf[3] << 8 | buf[2]) << 16 | ((unsigned)buf[1] << 8 | buf[0]);
    *(unsigned int *)buf = t;
    buf += 4;
  } while (--longs);
}
```