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
<read_pnm_header>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x118,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%r12
  mov    %rdx,%r13
  mov    %rcx,%r14
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    $0x100,%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
  mov    %rbx,%rdx
  mov    $0x100,%esi
  mov    %rbp,%rdi
  call   <fgets@plt>
  lea    D0(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  mov    $0x6,%r15d
  test   %rax,%rax
  jne    L6
L34:
  mov    %rsp,%rbp
L20:
  mov    %rbx,%rdx
  mov    $0x100,%esi
  mov    %rbp,%rdi
  call   <fgets@plt>
  cmpb   $0x23,(%rsp)
  je     L20
  mov    %rsp,%rdi
  mov    %r13,%rcx
  mov    %r12,%rdx
  lea    D22(%rip),%rsi
  mov    $0x0,%eax
  call   <sscanf@plt>
  cmpl   $0x0,(%r12)
  js     L12
  cmpl   $0x0,0x0(%r13)
  js     L31
  mov    %rsp,%rbp
  mov    %rbx,%rdx
  mov    $0x100,%esi
  mov    %rbp,%rdi
  call   <fgets@plt>
  mov    %r14,%rdx
  lea    D11(%rip),%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <sscanf@plt>
  cmpl   $0x100,(%r14)
  mov    $0x0,%eax
  cmovge %eax,%r15d
L36:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L71
  mov    %r15d,%eax
  add    $0x118,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L6:
  mov    %rsp,%rdi
  lea    D23(%rip),%rsi
  call   <strcmp@plt>
  mov    $0x0,%r15d
  test   %rax,%rax
  jne    L36
  mov    $0x5,%r15d
  jmp    L34
L12:
  mov    $0x0,%r15d
  jmp    L36
L31:
  mov    $0x0,%r15d
  jmp    L36
L71:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D0
type: string
name: parse_data
arguments: data_label: D22
type: string
name: parse_data
arguments: data_label: D11
type: string
name: parse_data
arguments: data_label: D23
type: string

tool: D0: P6


tool: D22: %d %d

tool: D11: %d

tool: D23: P5


assistant: ```c
int read_pnm_header(FILE *f, int *w, int *h, int *max) {
  char buf[256];
  memset(buf, 0, sizeof(buf));
  int ret;
  fgets(buf, 256, f);
  if (strcmp(buf, "P6\n") == 0) {
    ret = 6;
  } else if (strcmp(buf, "P5\n") == 0) {
    ret = 5;
  } else {
    return 0;
  }
  do {
    fgets(buf, 256, f);
  } while (buf[0] == '#');
  sscanf(buf, "%d %d", w, h);
  if (*w < 0 || *h < 0)
    return 0;
  fgets(buf, 256, f);
  sscanf(buf, "%d", max);
  if (*max > 255)
    return 0;
  return ret;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<read_pnm_header>:
  endbr64
  push   %r15
  mov    $0x6,%r15d
  push   %r14
  mov    %rcx,%r14
  push   %r13
  mov    %rdx,%r13
  mov    $0x100,%edx
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  xor    %esi,%esi
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <memset@plt>
  mov    %r12,%rdx
  mov    $0x100,%esi
  mov    %rbp,%rdi
  call   <fgets@plt>
  lea    D9(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L10
  nopl   0x0(%rax)
L24:
  mov    %r12,%rdx
  mov    $0x100,%esi
  mov    %rbp,%rdi
  call   <fgets@plt>
  cmpb   $0x23,(%rsp)
  je     L24
  mov    %rbx,%rdx
  mov    %r13,%rcx
  lea    D31(%rip),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <sscanf@plt>
  mov    (%rbx),%edx
  test   %edx,%edx
  js     L65
  mov    0x0(%r13),%eax
  test   %eax,%eax
  js     L65
  mov    %r12,%rdx
  mov    $0x100,%esi
  mov    %rbp,%rdi
  call   <fgets@plt>
  xor    %eax,%eax
  mov    %r14,%rdx
  mov    %rbp,%rdi
  lea    D26(%rip),%rsi
  call   <sscanf@plt>
  cmpl   $0xff,(%r14)
  jle    L22
L65:
  xor    %r15d,%r15d
L22:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L79
  add    $0x118,%rsp
  mov    %r15d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L10:
  lea    D27(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L65
  mov    $0x5,%r15d
  jmp    L24
L79:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D9
data_type: string
name: retrieve_data
arguments: label: D31
data_type: string
name: retrieve_data
arguments: label: D26
data_type: string
name: retrieve_data
arguments: label: D27
data_type: string

tool: D9: P6


tool: D31: %d %d

tool: D26: %d

tool: D27: P5


assistant: ```c
int read_pnm_header(FILE *f, int *w, int *h, int *max) {
  char buf[256];
  memset(buf, 0, sizeof(buf));
  int ret;
  fgets(buf, 256, f);
  if (strcmp(buf, "P6\n") == 0) {
    ret = 6;
  } else if (strcmp(buf, "P5\n") == 0) {
    ret = 5;
  } else {
    return 0;
  }
  do {
    fgets(buf, 256, f);
  } while (buf[0] == '#');
  sscanf(buf, "%d %d", w, h);
  if (*w < 0 || *h < 0)
    return 0;
  fgets(buf, 256, f);
  sscanf(buf, "%d", max);
  if (*max > 255)
    return 0;
  return ret;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<read_pnm_header>:
  endbr64
  push   %r15
  mov    $0x6,%r15d
  push   %r14
  mov    %rcx,%r14
  push   %r13
  mov    %rdx,%r13
  mov    $0x100,%edx
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  xor    %esi,%esi
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <memset@plt>
  mov    %r12,%rdx
  mov    $0x100,%esi
  mov    %rbp,%rdi
  call   <fgets@plt>
  lea    D8(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L57
  nopl   0x0(%rax)
L44:
  mov    %r12,%rdx
  mov    $0x100,%esi
  mov    %rbp,%rdi
  call   <fgets@plt>
  cmpb   $0x23,(%rsp)
  je     L44
  mov    %rbx,%rdx
  mov    %r13,%rcx
  lea    D39(%rip),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <sscanf@plt>
  mov    (%rbx),%edx
  test   %edx,%edx
  js     L45
  mov    0x0(%r13),%eax
  test   %eax,%eax
  js     L45
  mov    %r12,%rdx
  mov    $0x100,%esi
  mov    %rbp,%rdi
  call   <fgets@plt>
  xor    %eax,%eax
  mov    %r14,%rdx
  mov    %rbp,%rdi
  lea    D34(%rip),%rsi
  call   <sscanf@plt>
  cmpl   $0xff,(%r14)
  jle    L68
L45:
  xor    %r15d,%r15d
L68:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x118,%rsp
  mov    %r15d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L57:
  lea    D32(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L45
  mov    $0x5,%r15d
  jmp    L44
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D8
type_guessed: string
name: disassemble_data
arguments: data_tag: D39
type_guessed: string
name: disassemble_data
arguments: data_tag: D34
type_guessed: string
name: disassemble_data
arguments: data_tag: D32
type_guessed: string

tool: D8: P6


tool: D39: %d %d

tool: D34: %d

tool: D32: P5


assistant: ```c
int read_pnm_header(FILE *f, int *w, int *h, int *max) {
  char buf[256];
  memset(buf, 0, sizeof(buf));
  int ret;
  fgets(buf, 256, f);
  if (strcmp(buf, "P6\n") == 0) {
    ret = 6;
  } else if (strcmp(buf, "P5\n") == 0) {
    ret = 5;
  } else {
    return 0;
  }
  do {
    fgets(buf, 256, f);
  } while (buf[0] == '#');
  sscanf(buf, "%d %d", w, h);
  if (*w < 0 || *h < 0)
    return 0;
  fgets(buf, 256, f);
  sscanf(buf, "%d", max);
  if (*max > 255)
    return 0;
  return ret;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<hypre_CAlloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  imul   -0x18(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jle    L17
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L3
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <hypre_OutOfMemory@plt>
  jmp    L3
L17:
  movq   $0x0,-0x8(%rbp)
L3:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *hypre_CAlloc(int count, int elt_size) {
  char *ptr;
  int size = count * elt_size;
  if (size > 0) {
    ptr = calloc(count, elt_size);
    if (ptr == NULL) {
      hypre_OutOfMemory(size);
    }
  } else {
    ptr = NULL;
  }
  return ptr;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<hypre_CAlloc>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  imul   %esi,%ebp
  mov    $0x0,%ebx
  test   %ebp,%ebp
  jle    L6
  call   <calloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L24
L6:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L24:
  mov    %ebp,%edi
  call   <hypre_OutOfMemory@plt>
  jmp    L6
```

assistant: ```c
char *hypre_CAlloc(int count, int elt_size) {
  char *ptr;
  int size = count * elt_size;
  if (size > 0) {
    ptr = calloc(count, elt_size);
    if (ptr == NULL) {
      hypre_OutOfMemory(size);
    }
  } else {
    ptr = NULL;
  }
  return ptr;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<hypre_CAlloc>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %edi,%ebp
  imul   %esi,%ebp
  sub    $0x8,%rsp
  test   %ebp,%ebp
  jle    L2
  call   <calloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L20
L2:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nop
L20:
  mov    %ebp,%edi
  call   <hypre_OutOfMemory@plt>
  jmp    L2
```

assistant: ```c
char *hypre_CAlloc(int count, int elt_size) {
  char *ptr;
  int size = count * elt_size;
  if (size > 0) {
    ptr = calloc(count, elt_size);
    if (ptr == NULL) {
      hypre_OutOfMemory(size);
    }
  } else {
    ptr = NULL;
  }
  return ptr;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<hypre_CAlloc>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %edi,%ebp
  imul   %esi,%ebp
  sub    $0x8,%rsp
  test   %ebp,%ebp
  jle    L25
  call   <calloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L12
L25:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nop
L12:
  mov    %ebp,%edi
  call   <hypre_OutOfMemory@plt>
  jmp    L25
```

assistant: ```c
char *hypre_CAlloc(int count, int elt_size) {
  char *ptr;
  int size = count * elt_size;
  if (size > 0) {
    ptr = calloc(count, elt_size);
    if (ptr == NULL) {
      hypre_OutOfMemory(size);
    }
  } else {
    ptr = NULL;
  }
  return ptr;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x50(%rbp)
  jmp    L18
L35:
  movl   $0x0,-0x4c(%rbp)
  jmp    L56
L12:
  mov    -0x50(%rbp),%eax
  lea    (%rax,%rax,1),%ecx
  mov    -0x4c(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  lea    (%rcx,%rax,1),%edx
  mov    -0x4c(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x50(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rcx,%rax
  mov    %edx,-0x40(%rbp,%rax,4)
  addl   $0x1,-0x4c(%rbp)
L56:
  cmpl   $0x3,-0x4c(%rbp)
  jle    L12
  addl   $0x1,-0x50(%rbp)
L18:
  cmpl   $0x2,-0x50(%rbp)
  jle    L35
  movl   $0x0,-0x48(%rbp)
  jmp    L5
L55:
  movl   $0x0,-0x44(%rbp)
  jmp    L14
L84:
  mov    -0x44(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x48(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    -0x40(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x44(%rbp)
L14:
  cmpl   $0x3,-0x44(%rbp)
  jle    L84
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x48(%rbp)
L5:
  cmpl   $0x2,-0x48(%rbp)
  jle    L55
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L22
  call   <__stack_chk_fail@plt>
L22:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D3
data_type: string

tool: D3: %d	

assistant: ```c
int main() {
  int matris[3][4];
  int i, j, k, l;
  for (i = 0; i < 3; i++) {
    for (j = 0; j < 4; j++) {
      matris[i][j] = (2 * i) + (3 * j);
    }
  }
  for (k = 0; k < 3; k++) {
    for (l = 0; l < 4; l++) {
      printf("%d\t", matris[k][l]);
    }
    printf("\n");
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movl   $0x0,(%rsp)
  movl   $0x3,0x4(%rsp)
  movl   $0x6,0x8(%rsp)
  movl   $0x9,0xc(%rsp)
  movl   $0x2,0x10(%rsp)
  movl   $0x5,0x14(%rsp)
  movl   $0x8,0x18(%rsp)
  movl   $0xb,0x1c(%rsp)
  movl   $0x4,0x20(%rsp)
  movl   $0x7,0x24(%rsp)
  movl   $0xa,0x28(%rsp)
  movl   $0xd,0x2c(%rsp)
  lea    0x10(%rsp),%rbp
  lea    0x40(%rsp),%r13
  lea    D2(%rip),%r12
L9:
  lea    -0x10(%rbp),%rbx
L25:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %rbx,%rbp
  jne    L25
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x10,%rbp
  cmp    %rbp,%r13
  jne    L9
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  mov    $0x0,%eax
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D2
type_guessed: string

tool: D2: %d	

assistant: ```c
int main() {
  int matris[3][4];
  int i, j, k, l;
  for (i = 0; i < 3; i++) {
    for (j = 0; j < 4; j++) {
      matris[i][j] = (2 * i) + (3 * j);
    }
  }
  for (k = 0; k < 3; k++) {
    for (l = 0; l < 4; l++) {
      printf("%d\t", matris[k][l]);
    }
    printf("\n");
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  lea    D0(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  movabs $0x300000000,%rax
  lea    0x10(%rsp),%rbp
  lea    0x40(%rsp),%r13
  mov    %rax,(%rsp)
  movabs $0x900000006,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x500000002,%rax
  mov    %rax,0x10(%rsp)
  movabs $0xb00000008,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x700000004,%rax
  mov    %rax,0x20(%rsp)
  movabs $0xd0000000a,%rax
  mov    %rax,0x28(%rsp)
L17:
  lea    -0x10(%rbp),%rbx
L0:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,%rbp
  jne    L0
  mov    $0xa,%edi
  add    $0x10,%rbp
  call   <putchar@plt>
  cmp    %r13,%rbp
  jne    L17
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D0
data_type: string

tool: D0: %d	

assistant: ```c
int main() {
  int matris[3][4];
  int i, j, k, l;
  for (i = 0; i < 3; i++) {
    for (j = 0; j < 4; j++) {
      matris[i][j] = (2 * i) + (3 * j);
    }
  }
  for (k = 0; k < 3; k++) {
    for (l = 0; l < 4; l++) {
      printf("%d\t", matris[k][l]);
    }
    printf("\n");
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  lea    D11(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  movdqa D26(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  lea    0x40(%rsp),%r13
  movaps %xmm0,(%rsp)
  movdqa D9(%rip),%xmm0
  movaps %xmm0,0x10(%rsp)
  movdqa D23(%rip),%xmm0
  movaps %xmm0,0x20(%rsp)
L24:
  lea    -0x10(%rbp),%rbx
L25:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,%rbp
  jne    L25
  mov    $0xa,%edi
  add    $0x10,%rbp
  call   <putchar@plt>
  cmp    %r13,%rbp
  jne    L24
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D11
datatype: string
name: unpack_data
arguments: label: D26
datatype: byte[16]
name: unpack_data
arguments: label: D9
datatype: byte[16]
name: unpack_data
arguments: label: D23
datatype: byte[16]

tool: D11: %d	

tool: {"D26": [0x00,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x06,0x00,0x00,0x00,0x09,0x00,0x00,0x00]}

tool: {"D9": [0x02,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x08,0x00,0x00,0x00,0x0b,0x00,0x00,0x00]}

tool: {"D23": [0x04,0x00,0x00,0x00,0x07,0x00,0x00,0x00,0x0a,0x00,0x00,0x00,0x0d,0x00,0x00,0x00]}

assistant: ```c
int main() {
  int matris[3][4];
  int i, j, k, l;
  for (i = 0; i < 3; i++) {
    for (j = 0; j < 4; j++) {
      matris[i][j] = (2 * i) + (3 * j);
    }
  }
  for (k = 0; k < 3; k++) {
    for (l = 0; l < 4; l++) {
      printf("%d\t", matris[k][l]);
    }
    printf("\n");
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<isleap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  shr    $0x20,%rax
  sar    $0x5,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x64,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L16
  mov    -0x4(%rbp),%edx
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  shr    $0x20,%rax
  sar    $0x7,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x190,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L16
  mov    $0x1,%eax
  jmp    L8
L16:
  mov    -0x4(%rbp),%edx
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  shr    $0x20,%rax
  sar    $0x5,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x64,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  je     L11
  mov    -0x4(%rbp),%eax
  and    $0x3,%eax
  test   %eax,%eax
  jne    L11
  mov    $0x1,%eax
  jmp    L8
L11:
  mov    $0x0,%eax
L8:
  pop    %rbp
  ret
```

assistant: ```c
int isleap(int year) {
  if (year % 100 == 0 && year % 400 == 0)
    return 1;
  else if (year % 100 != 0 && year % 4 == 0)
    return 1;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<isleap>:
  endbr64
  movslq %edi,%rax
  imul   $0x51eb851f,%rax,%rax
  sar    $0x25,%rax
  mov    %edi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  imul   $0x64,%eax,%eax
  cmp    %eax,%edi
  jne    L5
  movslq %edi,%rax
  imul   $0x51eb851f,%rax,%rax
  sar    $0x27,%rax
  mov    %edi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  imul   $0x190,%eax,%eax
  cmp    %eax,%edi
  sete   %al
  movzbl %al,%eax
  ret
L5:
  test   $0x3,%dil
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int isleap(int year) {
  if (year % 100 == 0 && year % 400 == 0)
    return 1;
  else if (year % 100 != 0 && year % 4 == 0)
    return 1;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<isleap>:
  endbr64
  imul   $0xc28f5c29,%edi,%eax
  add    $0x51eb850,%eax
  mov    %eax,%edx
  ror    $0x2,%edx
  cmp    $0x28f5c28,%edx
  jbe    L3
  xor    %eax,%eax
  and    $0x3,%edi
  sete   %al
  ret
  nopl   (%rax)
L3:
  ror    $0x4,%eax
  cmp    $0xa3d70a,%eax
  setbe  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int isleap(int year) {
  if (year % 100 == 0 && year % 400 == 0)
    return 1;
  else if (year % 100 != 0 && year % 4 == 0)
    return 1;
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<isleap>:
  endbr64
  imul   $0xc28f5c29,%edi,%eax
  add    $0x51eb850,%eax
  mov    %eax,%edx
  ror    $0x2,%edx
  cmp    $0x28f5c28,%edx
  jbe    L2
  xor    %eax,%eax
  and    $0x3,%edi
  sete   %al
  ret
  nopl   (%rax)
L2:
  ror    $0x4,%eax
  cmp    $0xa3d70a,%eax
  setbe  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int isleap(int year) {
  if (year % 100 == 0 && year % 400 == 0)
    return 1;
  else if (year % 100 != 0 && year % 4 == 0)
    return 1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<Fact>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movq   $0x1,-0x8(%rbp)
  movl   $0x2,-0xc(%rbp)
  jmp    L7
L1:
  mov    -0xc(%rbp),%eax
  cltq
  mov    -0x8(%rbp),%rdx
  imul   %rdx,%rax
  mov    %rax,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L7:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L1
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
long long Fact(int k) {
  long long tmp = 1;
  int i;
  for (i = 2; i <= k; i++)
    tmp *= i;
  return tmp;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Fact>:
  endbr64
  cmp    $0x1,%edi
  jle    L25
  lea    0x1(%rdi),%ecx
  mov    $0x2,%eax
  mov    $0x1,%edx
L1:
  imul   %rax,%rdx
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L1
L16:
  mov    %rdx,%rax
  ret
L25:
  mov    $0x1,%edx
  jmp    L16
```

assistant: ```c
long long Fact(int k) {
  long long tmp = 1;
  int i;
  for (i = 2; i <= k; i++)
    tmp *= i;
  return tmp;
}
```
user: Translate the below assembly code into C programming language:

```asm
<Fact>:
  endbr64
  cmp    $0x1,%edi
  jle    L3
  lea    0x1(%rdi),%edx
  mov    $0x2,%eax
  mov    $0x1,%r8d
  nopw   0x0(%rax,%rax,1)
L18:
  imul   %rax,%r8
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L18
  mov    %r8,%rax
  ret
  nopl   0x0(%rax)
L3:
  mov    $0x1,%r8d
  mov    %r8,%rax
  ret
```

assistant: ```c
long long Fact(int k) {
  long long tmp = 1;
  int i;
  for (i = 2; i <= k; i++)
    tmp *= i;
  return tmp;
}
```
user: Translate the below assembly code into C programming language:

```asm
<Fact>:
  endbr64
  cmp    $0x1,%edi
  jle    L12
  lea    0x1(%rdi),%edx
  mov    $0x2,%eax
  mov    $0x1,%r8d
  nopw   0x0(%rax,%rax,1)
L15:
  imul   %rax,%r8
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L15
  mov    %r8,%rax
  ret
  nopl   0x0(%rax)
L12:
  mov    $0x1,%r8d
  mov    %r8,%rax
  ret
```

assistant: ```c
long long Fact(int k) {
  long long tmp = 1;
  int i;
  for (i = 2; i <= k; i++)
    tmp *= i;
  return tmp;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<open_weak_exclusive>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <open@plt>
  mov    %eax,-0xc(%rbp)
  cmpl   $0xffffffff,-0xc(%rbp)
  jne    L18
  mov    D25(%rip),%rax
  mov    (%rax),%eax
  and    -0x1c(%rbp),%eax
  test   %eax,%eax
  jne    L68
L18:
  mov    -0xc(%rbp),%eax
  jmp    L39
L68:
  mov    D29(%rip),%rax
  mov    (%rax),%eax
  add    $0xffffff80,%eax
  cmp    $0x7,%eax
  ja     L59
  mov    -0xc(%rbp),%eax
  jmp    L39
L59:
  lea    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <stat@plt>
  cmp    $0xffffffff,%eax
  je     L37
  mov    D29(%rip),%rax
  movl   $0x86,(%rax)
  mov    $0xffffffff,%eax
  jmp    L39
L37:
  mov    D25(%rip),%rax
  mov    (%rax),%eax
  not    %eax
  and    -0x1c(%rbp),%eax
  mov    %eax,%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <open@plt>
L39:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

assistant: ```c
int open_weak_exclusive(const char *pathname, int flags, mode_t mode) {
  int ret;
  struct stat filestat;
  ret = open(pathname, flags, mode);
  if (ret != -1 || !(flags & O_EXCL)) {
    return ret;
  }
  switch (errno) {
  case EACCES:
  case EEXIST:
  case EINTR:
  case EISDIR:
  case ELOOP:
  case ENAMETOOLONG:
  case ENOENT:
  case ENOTDIR:
    return ret;
  }
  if (stat(pathname, &filestat) != -1) {
    errno = EEXIST;
    return -1;
  } else {
    return open(pathname, flags & ~O_EXCL, mode);
  }
}
```