ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<uid_from_string>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x10000(%rsp),%r11
L109:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L109
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x10028(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x18(%rsp)
  test   %rdi,%rdi
  je     L76
  mov    %rdi,%rbx
  mov    %rsi,%r12
  lea    0x20(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    0x10(%rsp),%r8
  mov    $0x10000,%ecx
  call   <_getpwnam_r@plt>
  test   %rax,%rax
  jne    L16
  mov    0x10(%rsp),%rax
  test   %rax,%rax
  je     L16
  mov    (%rax),%rax
  mov    %rax,(%r12)
  mov    $0x0,%eax
  jmp    L157
L16:
  mov    D35(%rip),%r13
  movq   $0x0,0x0(%r13)
  lea    0x18(%rsp),%rsi
  mov    $0xa,%edx
  mov    %rbx,%rdi
  call   <strtol@plt>
  mov    %rax,%rbp
  mov    D8(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,0x0(%r13)
  je     L62
L4:
  mov    0x18(%rsp),%rax
  cmp    %rbx,%rax
  je     L93
  cmpb   $0x0,(%rax)
  jne    L95
  test   %rbp,%rbp
  js     L95
  mov    D14(%rip),%rax
  cmp    %rbp,(%rax)
  jl     L102
  lea    0x20(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    0x10(%rsp),%r8
  mov    $0x10000,%ecx
  mov    %rbp,%rdi
  call   <slurm_getpwuid_r@plt>
  test   %rax,%rax
  jne    L28
  mov    %rbp,(%r12)
L157:
  mov    0x10028(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L141
  add    $0x10038,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L62:
  mov    D21(%rip),%rax
  cmp    %rbp,(%rax)
  je     L150
  mov    D0(%rip),%rax
  cmp    %rbp,(%rax)
  jne    L4
  mov    $0xffffffff,%eax
  jmp    L157
L76:
  mov    $0xffffffff,%eax
  jmp    L157
L150:
  mov    $0xffffffff,%eax
  jmp    L157
L93:
  mov    $0xffffffff,%eax
  jmp    L157
L95:
  mov    $0xffffffff,%eax
  jmp    L157
L102:
  mov    $0xffffffff,%eax
  jmp    L157
L28:
  mov    $0xffffffff,%eax
  jmp    L157
L141:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int uid_from_string(char *name, uid_t *uidp) {
  struct passwd pwd, *result;
  char buffer[65536], *p = NULL;
  long l;
  if (!name)
    return -1;
  if ((_getpwnam_r(name, &pwd, buffer, 65536, &result) == 0) && result != NULL) {
    *uidp = result->pw_uid;
    return 0;
  }
  errno = 0;
  l = strtol(name, &p, 10);
  if (((errno == ERANGE) && ((l == LONG_MIN) || (l == LONG_MAX))) || (name == p) || (*p != '\0') || (l < 0) || (l > INT_MAX))
    return -1;
  if (slurm_getpwuid_r(l, &pwd, buffer, 65536, &result) != 0)
    return -1;
  *uidp = (uid_t)l;
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<uid_from_string>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x10000(%rsp),%r11
L54:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L54
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x10028(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x18(%rsp)
  test   %rdi,%rdi
  je     L116
  lea    0x10(%rsp),%r15
  lea    0x20(%rsp),%r14
  mov    %rsi,%r12
  mov    %rdi,%rbx
  lea    0x8(%rsp),%r13
  mov    %r15,%r8
  mov    $0x10000,%ecx
  mov    %r14,%rdx
  mov    %r13,%rsi
  call   <_getpwnam_r@plt>
  test   %rax,%rax
  jne    L123
  mov    0x10(%rsp),%rax
  test   %rax,%rax
  je     L123
  mov    (%rax),%rax
  mov    %rax,(%r12)
  xor    %eax,%eax
  jmp    L109
  nopw   0x0(%rax,%rax,1)
L123:
  mov    D0(%rip),%rax
  lea    0x18(%rsp),%rsi
  mov    $0xa,%edx
  mov    %rbx,%rdi
  movq   $0x0,(%rax)
  call   <strtol@plt>
  mov    D0(%rip),%rcx
  mov    %rax,%rbp
  mov    D15(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,(%rcx)
  je     L44
L43:
  mov    0x18(%rsp),%rax
  cmp    %rbx,%rax
  je     L116
  cmpb   $0x0,(%rax)
  jne    L116
  test   %rbp,%rbp
  js     L116
  mov    D20(%rip),%rax
  cmp    %rbp,(%rax)
  jl     L116
  mov    %r15,%r8
  mov    $0x10000,%ecx
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <slurm_getpwuid_r@plt>
  test   %rax,%rax
  jne    L116
  mov    %rbp,(%r12)
L109:
  mov    0x10028(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L55
  add    $0x10038,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L44:
  mov    D45(%rip),%rax
  cmp    %rbp,(%rax)
  je     L116
  mov    D26(%rip),%rax
  cmp    %rbp,(%rax)
  jne    L43
L116:
  mov    $0xffffffff,%eax
  jmp    L109
L55:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int uid_from_string(char *name, uid_t *uidp) {
  struct passwd pwd, *result;
  char buffer[65536], *p = NULL;
  long l;
  if (!name)
    return -1;
  if ((_getpwnam_r(name, &pwd, buffer, 65536, &result) == 0) && result != NULL) {
    *uidp = result->pw_uid;
    return 0;
  }
  errno = 0;
  l = strtol(name, &p, 10);
  if (((errno == ERANGE) && ((l == LONG_MIN) || (l == LONG_MAX))) || (name == p) || (*p != '\0') || (l < 0) || (l > INT_MAX))
    return -1;
  if (slurm_getpwuid_r(l, &pwd, buffer, 65536, &result) != 0)
    return -1;
  *uidp = (uid_t)l;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<uid_from_string>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x10000(%rsp),%r11
L90:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L90
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x10028(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x18(%rsp)
  test   %rdi,%rdi
  je     L86
  lea    0x10(%rsp),%r15
  lea    0x20(%rsp),%r14
  mov    %rsi,%r12
  mov    %rdi,%rbx
  lea    0x8(%rsp),%r13
  mov    %r15,%r8
  mov    $0x10000,%ecx
  mov    %r14,%rdx
  mov    %r13,%rsi
  call   <_getpwnam_r@plt>
  test   %rax,%rax
  jne    L55
  mov    0x10(%rsp),%rax
  test   %rax,%rax
  je     L55
  mov    (%rax),%rax
  mov    %rax,(%r12)
  xor    %eax,%eax
  jmp    L14
  nopw   0x0(%rax,%rax,1)
L55:
  mov    D8(%rip),%rax
  lea    0x18(%rsp),%rsi
  mov    $0xa,%edx
  mov    %rbx,%rdi
  movq   $0x0,(%rax)
  call   <strtol@plt>
  mov    D8(%rip),%rcx
  mov    %rax,%rbp
  mov    D15(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,(%rcx)
  je     L95
L85:
  mov    0x18(%rsp),%rax
  cmp    %rbx,%rax
  je     L86
  cmpb   $0x0,(%rax)
  jne    L86
  test   %rbp,%rbp
  js     L86
  mov    D36(%rip),%rax
  cmp    %rbp,(%rax)
  jl     L86
  mov    %r15,%r8
  mov    $0x10000,%ecx
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <slurm_getpwuid_r@plt>
  test   %rax,%rax
  jne    L86
  mov    %rbp,(%r12)
L14:
  mov    0x10028(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L121
  add    $0x10038,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L95:
  mov    D42(%rip),%rax
  cmp    %rbp,(%rax)
  je     L86
  mov    D32(%rip),%rax
  cmp    %rbp,(%rax)
  jne    L85
L86:
  mov    $0xffffffff,%eax
  jmp    L14
L121:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int uid_from_string(char *name, uid_t *uidp) {
  struct passwd pwd, *result;
  char buffer[65536], *p = NULL;
  long l;
  if (!name)
    return -1;
  if ((_getpwnam_r(name, &pwd, buffer, 65536, &result) == 0) && result != NULL) {
    *uidp = result->pw_uid;
    return 0;
  }
  errno = 0;
  l = strtol(name, &p, 10);
  if (((errno == ERANGE) && ((l == LONG_MIN) || (l == LONG_MAX))) || (name == p) || (*p != '\0') || (l < 0) || (l > INT_MAX))
    return -1;
  if (slurm_getpwuid_r(l, &pwd, buffer, 65536, &result) != 0)
    return -1;
  *uidp = (uid_t)l;
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<toto>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x9000(%rsp),%r11
L75:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L75
  sub    $0xdf0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x9dec(%rbp)
  jmp    L7
L8:
  movl   $0x1,-0x9de8(%rbp)
  jmp    L52
L29:
  movl   $0x1,-0x9de4(%rbp)
  jmp    L11
L88:
  mov    -0x9de8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x9dec(%rbp),%eax
  sub    $0x1,%eax
  movslq %eax,%rcx
  movslq %edx,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  add    %rdx,%rax
  shl    $0x2,%rax
  add    %rcx,%rax
  mov    -0x9c50(%rbp,%rax,4),%eax
  lea    0x2(%rax),%ecx
  mov    -0x9de4(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x9de8(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  add    %rdx,%rax
  shl    $0x2,%rax
  add    %rsi,%rax
  mov    %ecx,-0x9c50(%rbp,%rax,4)
  addl   $0x1,-0x9de4(%rbp)
L11:
  cmpl   $0x63,-0x9de4(%rbp)
  jle    L88
  mov    -0x9dec(%rbp),%eax
  add    %eax,-0x9de8(%rbp)
L52:
  cmpl   $0x63,-0x9de8(%rbp)
  jle    L29
  mov    -0x9dec(%rbp),%eax
  sub    $0x1,%eax
  cltq
  mov    -0x9de0(%rbp,%rax,4),%eax
  lea    0x2(%rax),%edx
  mov    -0x9dec(%rbp),%eax
  cltq
  mov    %edx,-0x9de0(%rbp,%rax,4)
  movl   $0x1,-0x9de8(%rbp)
  jmp    L61
L69:
  mov    -0x9de8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x9dec(%rbp),%eax
  sub    $0x1,%eax
  movslq %eax,%rcx
  movslq %edx,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  add    %rdx,%rax
  shl    $0x2,%rax
  add    %rcx,%rax
  mov    -0x9c50(%rbp,%rax,4),%eax
  lea    0x2(%rax),%ecx
  mov    -0x9dec(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x9de8(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  add    %rdx,%rax
  shl    $0x2,%rax
  add    %rsi,%rax
  mov    %ecx,-0x9c50(%rbp,%rax,4)
  addl   $0x1,-0x9de8(%rbp)
L61:
  cmpl   $0x63,-0x9de8(%rbp)
  jle    L69
  mov    -0x9dec(%rbp),%eax
  sub    $0x1,%eax
  cltq
  mov    -0x9de0(%rbp,%rax,4),%eax
  lea    0x2(%rax),%edx
  mov    -0x9dec(%rbp),%eax
  cltq
  mov    %edx,-0x9de0(%rbp,%rax,4)
  addl   $0x1,-0x9dec(%rbp)
L7:
  cmpl   $0x63,-0x9dec(%rbp)
  jle    L8
  mov    -0x978c(%rbp),%edx
  mov    -0x9ddc(%rbp),%eax
  add    %edx,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

assistant: ```c
int toto() {
  int i, j, k;
  int a[100][100];
  int b[100];
  for (i = 1; i < 100; i++) {
    for (j = 1; j < 100; j += i)
      for (k = 1; k < 100; k++)
        a[j][k] = a[j + 1][i - 1] + 2;
    b[i] = b[i - 1] + 2;
    for (j = 1; j < 100; j++)
      a[j][i] = a[j + 1][i - 1] + 2;
    b[i] = b[i - 1] + 2;
  }
  return a[3][5] + b[1];
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<toto>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x9000(%rsp),%r11
L20:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L20
  sub    $0xde0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x9dd8(%rsp)
  xor    %eax,%eax
  lea    0x190(%rsp),%r12
  lea    0x9f60(%rsp),%rdi
  mov    $0x190,%r9d
  mov    $0x1,%r11d
  mov    $0x0,%ebp
  jmp    L46
L18:
  movslq %ebx,%rax
  mov    (%rsp,%rax,4),%eax
  add    $0x2,%eax
  mov    %eax,(%rsp,%r11,4)
  lea    -0x9ab0(%rdi),%rax
  mov    %r11,%rcx
  sub    %rbp,%rcx
L65:
  mov    (%rax),%esi
  lea    0x2(%rsi),%edx
  mov    %edx,-0x190(%rax,%rcx,4)
  add    $0x190,%rax
  cmp    %rdi,%rax
  jne    L65
  movslq %ebx,%rbx
  mov    (%rsp,%rbx,4),%eax
  add    $0x2,%eax
  mov    %eax,(%rsp,%r11,4)
  add    $0x1,%rbp
  add    $0x1,%r11
  add    $0x4,%rdi
  add    $0x190,%r9
  cmp    $0x64,%r11
  je     L41
L46:
  mov    %r11d,%r10d
  lea    -0x9dcc(%rdi),%rsi
  lea    0x320(%r12),%rcx
  mov    $0x1,%r8d
  lea    -0x1(%r11),%ebx
L3:
  lea    -0x18c(%rcx),%rax
L79:
  mov    0x31c(%rsi),%edx
  add    $0x2,%edx
  mov    %edx,(%rax)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L79
  add    %r10d,%r8d
  add    %r9,%rsi
  add    %r9,%rcx
  cmp    $0x63,%r8d
  jle    L3
  jmp    L18
L41:
  mov    0x4(%rsp),%eax
  add    0x654(%rsp),%eax
  mov    0x9dd8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L21
  add    $0x9de0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L21:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int toto() {
  int i, j, k;
  int a[100][100];
  int b[100];
  for (i = 1; i < 100; i++) {
    for (j = 1; j < 100; j += i)
      for (k = 1; k < 100; k++)
        a[j][k] = a[j + 1][i - 1] + 2;
    b[i] = b[i - 1] + 2;
    for (j = 1; j < 100; j++)
      a[j][i] = a[j + 1][i - 1] + 2;
    b[i] = b[i - 1] + 2;
  }
  return a[3][5] + b[1];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<toto>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x9000(%rsp),%r11
L41:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L41
  sub    $0xde0,%rsp
  mov    $0x190,%r9d
  mov    $0x1,%ebx
  xor    %r11d,%r11d
  mov    %fs:0x28,%rax
  mov    %rax,0x9dd8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  lea    0x9f60(%rsp),%rdi
  lea    0x4b0(%rsp),%r12
L8:
  mov    %ebx,%r10d
  lea    -0x9dd0(%rdi),%rsi
  mov    %r12,%rcx
  mov    $0x1,%r8d
L26:
  lea    -0x18c(%rcx),%rax
  cs nopw 0x0(%rax,%rax,1)
L33:
  mov    0x320(%rsi),%edx
  add    $0x4,%rax
  add    $0x2,%edx
  mov    %edx,-0x4(%rax)
  cmp    %rcx,%rax
  jne    L33
  add    %r10d,%r8d
  add    %r9,%rsi
  lea    (%rax,%r9,1),%rcx
  cmp    $0x63,%r8d
  jle    L26
  mov    0x0(%rbp),%eax
  mov    %rbx,%rcx
  sub    %r11,%rcx
  add    $0x2,%eax
  mov    %eax,0x4(%rbp)
  lea    -0x9ab0(%rdi),%rax
  nopl   0x0(%rax,%rax,1)
L0:
  mov    (%rax),%esi
  lea    0x2(%rsi),%edx
  mov    %edx,-0x190(%rax,%rcx,4)
  add    $0x190,%rax
  cmp    %rdi,%rax
  jne    L0
  add    $0x1,%r11
  add    $0x1,%rbx
  lea    0x4(%rax),%rdi
  add    $0x4,%rbp
  add    $0x190,%r9
  cmp    $0x63,%r11
  jne    L8
  mov    0x4(%rsp),%eax
  add    0x654(%rsp),%eax
  mov    0x9dd8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L36
  add    $0x9de0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L36:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int toto() {
  int i, j, k;
  int a[100][100];
  int b[100];
  for (i = 1; i < 100; i++) {
    for (j = 1; j < 100; j += i)
      for (k = 1; k < 100; k++)
        a[j][k] = a[j + 1][i - 1] + 2;
    b[i] = b[i - 1] + 2;
    for (j = 1; j < 100; j++)
      a[j][i] = a[j + 1][i - 1] + 2;
    b[i] = b[i - 1] + 2;
  }
  return a[3][5] + b[1];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<toto>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x9000(%rsp),%r11
L16:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L16
  sub    $0xe18,%rsp
  mov    $0x190,%r8d
  mov    $0x1,%r14d
  mov    %fs:0x28,%rax
  mov    %rax,0x9e08(%rsp)
  xor    %eax,%eax
  lea    0x4d4(%rsp),%rax
  lea    0x30(%rsp),%r13
  movq   $0x0,0x8(%rsp)
  mov    %rax,0x10(%rsp)
  lea    0x354(%rsp),%rax
  mov    0x8(%rsp),%rbx
  lea    0x9f90(%rsp),%rdi
  mov    %rax,0x20(%rsp)
  lea    0x4e0(%rsp),%rax
  mov    %rax,0x28(%rsp)
L72:
  add    $0x1,%rbx
  mov    0x10(%rsp),%rdx
  mov    0x20(%rsp),%r11
  mov    %r14d,%ebp
  mov    0x28(%rsp),%rsi
  lea    0x0(,%rbx,4),%r12
  mov    %rbx,0x18(%rsp)
  lea    -0x9dd0(%rdi),%rax
  lea    0x188(%r12),%rbx
  mov    $0x194,%r9d
  mov    $0x1,%r10d
  add    $0x18c,%r12
L82:
  lea    0x18c(%r9),%r15
  lea    (%rbx,%r9,1),%rcx
  cmp    %rcx,%r15
  jle    L11
  lea    (%r12,%r9,1),%rcx
  cmp    %rcx,%r9
  jl     L81
L11:
  mov    0x320(%rax),%r15d
  mov    %r11,%rcx
  add    $0x2,%r15d
  movd   %r15d,%xmm0
  pshufd $0x0,%xmm0,%xmm0
  nopl   0x0(%rax)
L37:
  movups %xmm0,(%rcx)
  add    $0x10,%rcx
  cmp    %rdx,%rcx
  jne    L37
  mov    0x320(%rax),%ecx
  add    $0x2,%ecx
  movd   %ecx,%xmm0
  pshufd $0xe0,%xmm0,%xmm0
  movq   %xmm0,(%rdx)
  mov    0x320(%rax),%ecx
  add    $0x2,%ecx
  mov    %ecx,0x8(%rdx)
L40:
  add    %ebp,%r10d
  add    %r8,%r9
  add    %r8,%rax
  add    %r8,%rdx
  add    %r8,%r11
  add    %r8,%rsi
  cmp    $0x63,%r10d
  jle    L82
  mov    0x0(%r13),%eax
  mov    %r14,%rcx
  sub    0x8(%rsp),%rcx
  add    $0x2,%eax
  mov    %eax,0x4(%r13)
  lea    -0x9ab0(%rdi),%rax
  nopl   0x0(%rax)
L103:
  mov    (%rax),%ebx
  lea    0x2(%rbx),%edx
  mov    %edx,-0x190(%rax,%rcx,4)
  add    $0x190,%rax
  cmp    %rdi,%rax
  jne    L103
  mov    0x18(%rsp),%rax
  add    $0x1,%r14
  add    $0x4,%r13
  add    $0x4,%rdi
  add    $0x190,%r8
  cmp    $0x63,%rax
  je     L38
  mov    %rax,0x8(%rsp)
  mov    %rax,%rbx
  jmp    L72
L81:
  mov    %r11,%rcx
  nopl   (%rax)
L8:
  mov    0x320(%rax),%r15d
  add    $0x4,%rcx
  add    $0x2,%r15d
  mov    %r15d,-0x4(%rcx)
  cmp    %rsi,%rcx
  jne    L8
  jmp    L40
L38:
  mov    0x34(%rsp),%eax
  add    0x684(%rsp),%eax
  mov    0x9e08(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L88
  add    $0x9e18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L88:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int toto() {
  int i, j, k;
  int a[100][100];
  int b[100];
  for (i = 1; i < 100; i++) {
    for (j = 1; j < 100; j += i)
      for (k = 1; k < 100; k++)
        a[j][k] = a[j + 1][i - 1] + 2;
    b[i] = b[i - 1] + 2;
    for (j = 1; j < 100; j++)
      a[j][i] = a[j + 1][i - 1] + 2;
    b[i] = b[i - 1] + 2;
  }
  return a[3][5] + b[1];
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0xa,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    $0x64,%esi
  mov    %eax,%edi
  call   <random1@plt>
  mov    -0x4(%rbp),%eax
  mov    $0x64,%esi
  mov    %eax,%edi
  call   <random2@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int length = 10;
  random1(length, 100);
  random2(length, 100);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x64,%esi
  mov    $0xa,%edi
  call   <random1@plt>
  mov    $0x64,%esi
  mov    $0xa,%edi
  call   <random2@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int length = 10;
  random1(length, 100);
  random2(length, 100);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x64,%esi
  mov    $0xa,%edi
  call   <random1@plt>
  mov    $0x64,%esi
  mov    $0xa,%edi
  call   <random2@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int length = 10;
  random1(length, 100);
  random2(length, 100);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x64,%esi
  mov    $0xa,%edi
  call   <random1@plt>
  mov    $0x64,%esi
  mov    $0xa,%edi
  call   <random2@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int length = 10;
  random1(length, 100);
  random2(length, 100);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<append_printed_error>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xe0,%rsp
  mov    %edi,-0xd4(%rbp)
  mov    %rsi,-0xe0(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L0
  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)
L0:
  movl   $0x0,-0xc4(%rbp)
  movl   $0x1,-0xc0(%rbp)
  movl   $0x0,-0xbc(%rbp)
  mov    -0xe0(%rbp),%rdx
  mov    -0xb8(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xb8(%rbp),%r8d
  mov    -0xe0(%rbp),%rdi
  mov    -0xbc(%rbp),%ecx
  mov    -0xc0(%rbp),%edx
  mov    -0xc4(%rbp),%esi
  mov    -0xd4(%rbp),%eax
  mov    %r8d,%r9d
  mov    %rdi,%r8
  mov    %eax,%edi
  call   <print_string@plt>
  mov    %eax,-0xb4(%rbp)
  mov    -0xb8(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    -0xb4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int append_printed_error(bool all_processes, char const *restrict const fmt, ...) {
  bool const show_id = false;
  bool const error_mode = true;
  int const error_num = 0;
  va_list args;
  va_start(args, fmt);
  int const status = print_string(all_processes, show_id, error_mode, error_num, fmt, args);
  va_end(args);
  return status;
}
```
user: What is the c source code of the assembly code below?

```asm
<append_printed_error>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r12d
  mov    %rsi,%rbx
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%r9d
  mov    %rbx,%r8
  mov    $0x0,%ecx
  mov    $0x1,%edx
  mov    $0x0,%esi
  mov    %r12d,%edi
  call   <print_string@plt>
  mov    %eax,%ebx
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int append_printed_error(bool all_processes, char const *restrict const fmt, ...) {
  bool const show_id = false;
  bool const error_mode = true;
  int const error_num = 0;
  va_list args;
  va_start(args, fmt);
  int const status = print_string(all_processes, show_id, error_mode, error_num, fmt, args);
  va_end(args);
  return status;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<append_printed_error>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %ebp,%edi
  mov    %rsi,%rbx
  call   <va_start@plt>
  mov    %ebp,%r9d
  mov    %rbx,%r8
  mov    %r12d,%edi
  xor    %ecx,%ecx
  mov    $0x1,%edx
  xor    %esi,%esi
  call   <print_string@plt>
  mov    %ebp,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int append_printed_error(bool all_processes, char const *restrict const fmt, ...) {
  bool const show_id = false;
  bool const error_mode = true;
  int const error_num = 0;
  va_list args;
  va_start(args, fmt);
  int const status = print_string(all_processes, show_id, error_mode, error_num, fmt, args);
  va_end(args);
  return status;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<append_printed_error>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %ebp,%edi
  mov    %rsi,%rbx
  call   <va_start@plt>
  mov    %ebp,%r9d
  mov    %rbx,%r8
  mov    %r12d,%edi
  xor    %ecx,%ecx
  mov    $0x1,%edx
  xor    %esi,%esi
  call   <print_string@plt>
  mov    %ebp,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int append_printed_error(bool all_processes, char const *restrict const fmt, ...) {
  bool const show_id = false;
  bool const error_mode = true;
  int const error_num = 0;
  va_list args;
  va_start(args, fmt);
  int const status = print_string(all_processes, show_id, error_mode, error_num, fmt, args);
  va_end(args);
  return status;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x4,%edi
  call   <allouerchaine@plt>
  mov    %rax,-0x10(%rbp)
  mov    $0x8,%edi
  call   <allouerchaine@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D24
type_guessed: string
name: disassemble_data
arguments: data_tag: D19
type_guessed: string

tool: D24: %s

tool: D19: %s %s 


assistant: ```c
int main() {
  char *chaine1 = allouerchaine(4);
  char *chaine2 = allouerchaine(8);
  scanf("%s", chaine1);
  scanf("%s", chaine2);
  printf("%s %s \n", chaine1, chaine2);
  return 0;
}
```