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

```asm
<main>:
  endbr64
  push   %r14
  xor    %eax,%eax
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  call   <unknown1@plt>
  mov    %eax,%r14d
  jmp    L53
L4:
  add    $0x1,%r12d
  add    %r12d,%ebp
  add    %r12d,%r13d
  test   %r14d,%r14d
  lea    0x0(%rbp,%r12,1),%eax
  cmovne %eax,%ebp
L53:
  xor    %eax,%eax
  call   <unknown1@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jne    L4
  test   %r14d,%r14d
  je     L52
  add    %r13d,%r13d
  sub    %r13d,%ebp
  add    $0x2,%ebp
  js     L22
  nopw   0x0(%rax,%rax,1)
L0:
  xor    %eax,%eax
  call   <unknown2@plt>
  cmp    $0x1,%rax
  adc    $0x1,%ebx
  cmp    %ebp,%ebx
  jle    L0
L22:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L52:
  mov    $0x1,%ebp
  jmp    L0
```

assistant: ```c
int main() {
  int flag = unknown1();
  int t = 0;
  int s = 0;
  int a = 0;
  int b = 0;
  int x, y;
  while (unknown1()) {
    a++;
    b++;
    s += a;
    t += b;
    if (flag) {
      t += a;
    }
    t = t;
  }
  x = 1;
  if (flag) {
    x = t - 2 * s + 2;
  }
  y = 0;
  while (y <= x) {
    if (unknown2())
      y++;
    else
      y += 2;
    y = y;
  }
  if (y >= 5) {
    goto ERROR;
  ERROR:;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r12
  xor    %eax,%eax
  push   %rbp
  push   %rbx
  call   <unknown1@plt>
  test   %eax,%eax
  je     L56
  xor    %ebx,%ebx
  xor    %r12d,%r12d
  xor    %ebp,%ebp
  jmp    L30
  nopl   0x0(%rax)
L55:
  add    $0x1,%ebx
  add    %ebx,%r12d
  lea    0x0(%rbp,%rbx,2),%ebp
L30:
  xor    %eax,%eax
  call   <unknown1@plt>
  test   %eax,%eax
  jne    L55
  add    %r12d,%r12d
  sub    %r12d,%ebp
  add    $0x2,%ebp
  js     L52
L47:
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L17:
  xor    %eax,%eax
  call   <unknown2@plt>
  cmp    $0x1,%rax
  adc    $0x1,%ebx
  cmp    %ebp,%ebx
  jle    L17
L52:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L56:
  xor    %eax,%eax
  call   <unknown1@plt>
  test   %eax,%eax
  jne    L56
  mov    $0x1,%ebp
  jmp    L47
```

assistant: ```c
int main() {
  int flag = unknown1();
  int t = 0;
  int s = 0;
  int a = 0;
  int b = 0;
  int x, y;
  while (unknown1()) {
    a++;
    b++;
    s += a;
    t += b;
    if (flag) {
      t += a;
    }
    t = t;
  }
  x = 1;
  if (flag) {
    x = t - 2 * s + 2;
  }
  y = 0;
  while (y <= x) {
    if (unknown2())
      y++;
    else
      y += 2;
    y = y;
  }
  if (y >= 5) {
    goto ERROR;
  ERROR:;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<ft_iterative_factorial>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  cmpl   $0xc,-0x14(%rbp)
  jle    L37
  mov    $0x0,%eax
  jmp    L17
L37:
  cmpl   $0x0,-0x14(%rbp)
  jns    L55
  mov    $0x0,%eax
  jmp    L17
L55:
  movl   $0x1,-0x8(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L33
L23:
  mov    -0x8(%rbp),%eax
  imul   -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
L33:
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  cmp    %eax,-0x14(%rbp)
  jg     L23
  mov    -0x8(%rbp),%eax
L17:
  pop    %rbp
  ret
```

assistant: ```c
int ft_iterative_factorial(int nb) {
  int out;
  int i;
  if (nb > 12)
    return (0);
  if (nb < 0)
    return (0);
  out = 1;
  i = 1;
  while (i++ < nb)
    out *= i;
  return (out);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<ft_iterative_factorial>:
  endbr64
  mov    $0x0,%edx
  cmp    $0xc,%edi
  jbe    L31
L24:
  mov    %edx,%eax
  ret
L31:
  cmp    $0x1,%edi
  jle    L25
  add    $0x1,%edi
  mov    $0x2,%eax
  mov    $0x1,%edx
L29:
  imul   %eax,%edx
  add    $0x1,%eax
  cmp    %edi,%eax
  jne    L29
  jmp    L24
L25:
  mov    $0x1,%edx
  jmp    L24
```

assistant: ```c
int ft_iterative_factorial(int nb) {
  int out;
  int i;
  if (nb > 12)
    return (0);
  if (nb < 0)
    return (0);
  out = 1;
  i = 1;
  while (i++ < nb)
    out *= i;
  return (out);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ft_iterative_factorial>:
  endbr64
  xor    %r8d,%r8d
  cmp    $0xc,%edi
  jbe    L29
L8:
  mov    %r8d,%eax
  ret
L29:
  cmp    $0x1,%edi
  jle    L11
  add    $0x1,%edi
  mov    $0x2,%eax
  mov    $0x1,%r8d
  nopl   0x0(%rax,%rax,1)
L14:
  imul   %eax,%r8d
  add    $0x1,%eax
  cmp    %edi,%eax
  jne    L14
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L11:
  mov    $0x1,%r8d
  jmp    L8
```

assistant: ```c
int ft_iterative_factorial(int nb) {
  int out;
  int i;
  if (nb > 12)
    return (0);
  if (nb < 0)
    return (0);
  out = 1;
  i = 1;
  while (i++ < nb)
    out *= i;
  return (out);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ft_iterative_factorial>:
  endbr64
  xor    %eax,%eax
  cmp    $0xc,%edi
  jbe    L65
  ret
  nopl   0x0(%rax)
L65:
  cmp    $0x1,%edi
  jle    L38
  cmp    $0x2,%edi
  je     L43
  cmp    $0x3,%edi
  je     L16
  cmp    $0x4,%edi
  je     L48
  cmp    $0x5,%edi
  je     L44
  cmp    $0x6,%edi
  je     L100
  cmp    $0x7,%edi
  je     L21
  cmp    $0x8,%edi
  je     L66
  cmp    $0x9,%edi
  je     L32
  cmp    $0xa,%edi
  je     L99
  cmp    $0xc,%edi
  mov    $0x2611500,%edx
  mov    $0x1c8cfc00,%eax
  cmovne %edx,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L38:
  mov    $0x1,%eax
  ret
L66:
  mov    $0x9d80,%eax
  ret
  nopl   0x0(%rax)
L43:
  mov    $0x2,%eax
  ret
L16:
  mov    $0x6,%eax
  ret
L48:
  mov    $0x18,%eax
  ret
L44:
  mov    $0x78,%eax
  ret
L100:
  mov    $0x2d0,%eax
  ret
L21:
  mov    $0x13b0,%eax
  ret
L32:
  mov    $0x58980,%eax
  ret
L99:
  mov    $0x375f00,%eax
  ret
```

assistant: ```c
int ft_iterative_factorial(int nb) {
  int out;
  int i;
  if (nb > 12)
    return (0);
  if (nb < 0)
    return (0);
  out = 1;
  i = 1;
  while (i++ < nb)
    out *= i;
  return (out);
}
```
user: What is the c source code of the assembly code below?

```asm
<copy_fwd_rest_bytes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  cmpq   $0x7,-0x18(%rbp)
  ja     L1
  mov    -0x18(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  lea    D6(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D6(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    -0x10(%rbp),%rdx
  lea    0x4(%rdx),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x4(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  mov    (%rdx),%edx
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rdx
  lea    0x4(%rdx),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x4(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  mov    (%rdx),%edx
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rdx
  lea    0x4(%rdx),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x4(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  mov    (%rdx),%edx
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rdx
  lea    0x4(%rdx),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x4(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  mov    (%rdx),%edx
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rdx
  lea    0x4(%rdx),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x4(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  mov    (%rdx),%edx
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rdx
  lea    0x4(%rdx),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x4(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  mov    (%rdx),%edx
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
L1:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void copy_fwd_rest_bytes(uint8_t *d, const uint8_t *s, size_t len) {
  switch (len) {
  case 7:
    *d++ = *s++;
  case 6:
    *d++ = *s++;
  case 5:
    *d++ = *s++;
  case 4:
    *d++ = *s++;
  case 3:
    *d++ = *s++;
  case 2:
    *d++ = *s++;
  case 1:
    *d = *s;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<copy_fwd_rest_bytes>:
  endbr64
  mov    %rdi,%rax
  mov    %rsi,%rcx
  cmp    $0x7,%rdx
  ja     L1
  lea    D5(%rip),%rsi
  movslq (%rsi,%rdx,4),%rdx
  add    %rsi,%rdx
  notrack jmp *%rdx
  mov    (%rcx),%edx
  mov    %edx,(%rdi)
  lea    0x4(%rcx),%rcx
  lea    0x4(%rdi),%rax
  mov    (%rcx),%edx
  mov    %edx,(%rax)
  lea    0x4(%rcx),%rcx
  lea    0x4(%rax),%rax
  lea    0x4(%rcx),%rsi
  lea    0x4(%rax),%rdx
  mov    (%rcx),%ecx
  mov    %ecx,(%rax)
L5:
  lea    0x4(%rsi),%rcx
  lea    0x4(%rdx),%rax
  mov    (%rsi),%esi
  mov    %esi,(%rdx)
  lea    0x4(%rcx),%rsi
  lea    0x4(%rax),%rdx
  mov    (%rcx),%ecx
  mov    %ecx,(%rax)
L15:
  lea    0x4(%rsi),%rcx
  lea    0x4(%rdx),%rax
  mov    (%rsi),%esi
  mov    %esi,(%rdx)
  mov    (%rcx),%edx
  mov    %edx,(%rax)
L1:
  ret
  mov    %rcx,%rsi
  mov    %rdi,%rdx
  jmp    L5
  mov    %rcx,%rsi
  mov    %rdi,%rdx
  jmp    L15
```

assistant: ```c
void copy_fwd_rest_bytes(uint8_t *d, const uint8_t *s, size_t len) {
  switch (len) {
  case 7:
    *d++ = *s++;
  case 6:
    *d++ = *s++;
  case 5:
    *d++ = *s++;
  case 4:
    *d++ = *s++;
  case 3:
    *d++ = *s++;
  case 2:
    *d++ = *s++;
  case 1:
    *d = *s;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<copy_fwd_rest_bytes>:
  endbr64
  mov    %rdi,%rax
  mov    %rsi,%rcx
  cmp    $0x7,%rdx
  ja     L0
  lea    D1(%rip),%rsi
  movslq (%rsi,%rdx,4),%rdx
  add    %rsi,%rdx
  notrack jmp *%rdx
  nopl   0x0(%rax)
  mov    (%rcx),%edx
  add    $0x4,%rax
  add    $0x4,%rcx
  mov    %edx,(%rdi)
  mov    (%rcx),%edx
  add    $0x4,%rax
  add    $0x4,%rcx
  mov    %edx,-0x4(%rax)
  lea    0x4(%rcx),%rsi
  mov    (%rcx),%ecx
  lea    0x4(%rax),%rdx
  mov    %ecx,(%rax)
L29:
  lea    0x4(%rsi),%rcx
  mov    (%rsi),%esi
  lea    0x4(%rdx),%rax
  mov    %esi,(%rdx)
  lea    0x4(%rcx),%rsi
  mov    (%rcx),%ecx
  lea    0x4(%rax),%rdx
  mov    %ecx,(%rax)
L20:
  lea    0x4(%rsi),%rcx
  mov    (%rsi),%esi
  lea    0x4(%rdx),%rax
  mov    %esi,(%rdx)
  mov    (%rcx),%edx
  mov    %edx,(%rax)
L0:
  ret
  cs nopw 0x0(%rax,%rax,1)
  mov    %rcx,%rsi
  mov    %rdi,%rdx
  jmp    L29
  nopl   0x0(%rax,%rax,1)
  mov    %rcx,%rsi
  mov    %rdi,%rdx
  jmp    L20
```

assistant: ```c
void copy_fwd_rest_bytes(uint8_t *d, const uint8_t *s, size_t len) {
  switch (len) {
  case 7:
    *d++ = *s++;
  case 6:
    *d++ = *s++;
  case 5:
    *d++ = *s++;
  case 4:
    *d++ = *s++;
  case 3:
    *d++ = *s++;
  case 2:
    *d++ = *s++;
  case 1:
    *d = *s;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<copy_fwd_rest_bytes>:
  endbr64
  mov    %rdi,%rax
  mov    %rsi,%rcx
  cmp    $0x7,%rdx
  ja     L15
  lea    D6(%rip),%rsi
  movslq (%rsi,%rdx,4),%rdx
  add    %rsi,%rdx
  notrack jmp *%rdx
  nopl   0x0(%rax)
  mov    (%rcx),%edx
  add    $0x4,%rax
  add    $0x4,%rcx
  mov    %edx,(%rdi)
  mov    (%rcx),%edx
  add    $0x4,%rax
  add    $0x4,%rcx
  mov    %edx,-0x4(%rax)
  lea    0x4(%rcx),%rsi
  mov    (%rcx),%ecx
  lea    0x4(%rax),%rdx
  mov    %ecx,(%rax)
L6:
  lea    0x4(%rsi),%rcx
  mov    (%rsi),%esi
  lea    0x4(%rdx),%rax
  mov    %esi,(%rdx)
  lea    0x4(%rcx),%rsi
  mov    (%rcx),%ecx
  lea    0x4(%rax),%rdx
  mov    %ecx,(%rax)
L20:
  lea    0x4(%rsi),%rcx
  mov    (%rsi),%esi
  lea    0x4(%rdx),%rax
  mov    %esi,(%rdx)
  mov    (%rcx),%edx
  mov    %edx,(%rax)
L15:
  ret
  cs nopw 0x0(%rax,%rax,1)
  mov    %rcx,%rsi
  mov    %rdi,%rdx
  jmp    L6
  nopl   0x0(%rax,%rax,1)
  mov    %rcx,%rsi
  mov    %rdi,%rdx
  jmp    L20
```

assistant: ```c
void copy_fwd_rest_bytes(uint8_t *d, const uint8_t *s, size_t len) {
  switch (len) {
  case 7:
    *d++ = *s++;
  case 6:
    *d++ = *s++;
  case 5:
    *d++ = *s++;
  case 4:
    *d++ = *s++;
  case 3:
    *d++ = *s++;
  case 2:
    *d++ = *s++;
  case 1:
    *d = *s;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xa8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  lea    D41(%rip),%rdi
  call   <puts@plt>
  movq   $0x0,0x8(%rsp)
  mov    $0x5,%r13d
  lea    D11(%rip),%r15
  lea    D45(%rip),%r14
L51:
  lea    -0x4(%r13),%ebx
  lea    0x10(%rsp),%r12
  mov    0x8(%rsp),%rax
  lea    (%r12,%rax,4),%rbp
L34:
  mov    %ebx,%esi
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  add    $0x1,%ebx
  add    $0x4,%rbp
  cmp    %r13d,%ebx
  jne    L34
  add    $0x4,%r13d
  addq   $0x4,0x8(%rsp)
  cmp    $0x15,%r13d
  jne    L51
  lea    0x50(%rsp),%r13
  lea    0x40(%r12),%rcx
  mov    %r13,%rax
L20:
  mov    (%r12),%edx
  mov    %edx,(%rax)
  mov    0x4(%r12),%edx
  mov    %edx,0x10(%rax)
  mov    0x8(%r12),%edx
  mov    %edx,0x20(%rax)
  mov    0xc(%r12),%edx
  mov    %edx,0x30(%rax)
  add    $0x10,%r12
  add    $0x4,%rax
  cmp    %rcx,%r12
  jne    L20
  lea    D37(%rip),%rdi
  call   <puts@plt>
  lea    0x10(%r13),%rbp
  add    $0x50,%r13
  lea    D12(%rip),%r12
L2:
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    -0x10(%rbp),%rbx
L23:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L23
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x10,%rbp
  cmp    %r13,%rbp
  jne    L2
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  mov    $0x0,%eax
  add    $0xa8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D41
type_infered: string
name: disassemble_data
arguments: tag: D11
type_infered: string
name: disassemble_data
arguments: tag: D45
type_infered: string
name: disassemble_data
arguments: tag: D37
type_infered: string
name: disassemble_data
arguments: tag: D12
type_infered: string

tool: D41: 
	 Enter Numbers: 

tool: D11: 
 Cell %d: 

tool: D45: %d

tool: D37: 
	 The Transpose is: 

tool: D12: 	 %d 	

assistant: ```c
int main() {
  int a[4][4], b[4][4], m, n, i = 1;
  printf("\n\t Enter Numbers: \n");
  for (m = 0; m < 4; m++) {
    for (n = 0; n < 4; n++) {
      printf("\n Cell %d: ", i);
      scanf("%d", &a[m][n]);
      i++;
    }
  }
  for (m = 0; m < 4; m++) {
    for (n = 0; n < 4; n++) {
      b[n][m] = a[m][n];
    }
  }
  printf("\n\t The Transpose is: \n");
  for (m = 0; m < 4; m++) {
    printf("\n");
    for (n = 0; n < 4; n++) {
      printf("\t %d \t", b[m][n]);
    }
    printf("\n");
  }
  printf("\n");
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D38(%rip),%rdi
  push   %r14
  push   %r13
  lea    D1(%rip),%r13
  push   %r12
  lea    D16(%rip),%r12
  push   %rbp
  mov    $0x5,%ebp
  push   %rbx
  sub    $0xa8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x10(%rsp),%rax
  movq   $0x0,(%rsp)
  mov    %rax,0x8(%rsp)
L55:
  mov    0x8(%rsp),%rax
  mov    (%rsp),%rcx
  lea    -0x4(%rbp),%r14d
  mov    %rax,%r15
  lea    (%rax,%rcx,4),%rbx
L40:
  mov    %r14d,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%r14d
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %r12,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    %ebp,%r14d
  jne    L40
  addq   $0x4,(%rsp)
  lea    0x4(%r14),%ebp
  cmp    $0x11,%r14d
  jne    L55
  mov    0x8(%rsp),%r14
  lea    0x50(%rsp),%r13
  mov    %r13,%rax
  add    $0x40,%r14
L45:
  mov    (%r15),%edx
  add    $0x10,%r15
  add    $0x4,%rax
  mov    %edx,-0x4(%rax)
  mov    -0xc(%r15),%edx
  mov    %edx,0xc(%rax)
  mov    -0x8(%r15),%edx
  mov    %edx,0x1c(%rax)
  mov    -0x4(%r15),%edx
  mov    %edx,0x2c(%rax)
  cmp    %r14,%r15
  jne    L45
  lea    D28(%rip),%rdi
  lea    0x10(%r13),%rbp
  add    $0x50,%r13
  call   <puts@plt>
  lea    D12(%rip),%r12
L15:
  mov    $0xa,%edi
  lea    -0x10(%rbp),%rbx
  call   <putchar@plt>
L2:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  jne    L2
  mov    $0xa,%edi
  lea    0x10(%rbx),%rbp
  call   <putchar@plt>
  cmp    %r13,%rbp
  jne    L15
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L50
  add    $0xa8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L50:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D38
data_type: string
name: retrieve_data
arguments: tag: D1
data_type: string
name: retrieve_data
arguments: tag: D16
data_type: string
name: retrieve_data
arguments: tag: D28
data_type: string
name: retrieve_data
arguments: tag: D12
data_type: string

tool: D38: 
	 Enter Numbers: 

tool: D1: 
 Cell %d: 

tool: D16: %d

tool: D28: 
	 The Transpose is: 

tool: D12: 	 %d 	

assistant: ```c
int main() {
  int a[4][4], b[4][4], m, n, i = 1;
  printf("\n\t Enter Numbers: \n");
  for (m = 0; m < 4; m++) {
    for (n = 0; n < 4; n++) {
      printf("\n Cell %d: ", i);
      scanf("%d", &a[m][n]);
      i++;
    }
  }
  for (m = 0; m < 4; m++) {
    for (n = 0; n < 4; n++) {
      b[n][m] = a[m][n];
    }
  }
  printf("\n\t The Transpose is: \n");
  for (m = 0; m < 4; m++) {
    printf("\n");
    for (n = 0; n < 4; n++) {
      printf("\t %d \t", b[m][n]);
    }
    printf("\n");
  }
  printf("\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D6(%rip),%rdi
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  lea    D29(%rip),%r13
  push   %r12
  lea    D28(%rip),%r12
  push   %rbp
  mov    $0x5,%ebp
  push   %rbx
  sub    $0xa8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x10(%rsp),%rax
  mov    %rax,0x8(%rsp)
L31:
  mov    0x8(%rsp),%rax
  lea    -0x4(%rbp),%r15d
  lea    (%rax,%r14,4),%rbx
L11:
  mov    %r15d,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%r15d
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %r12,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    %ebp,%r15d
  jne    L11
  lea    0x4(%r15),%ebp
  add    $0x4,%r14
  cmp    $0x11,%r15d
  jne    L31
  movdqa 0x10(%rsp),%xmm0
  movdqa 0x30(%rsp),%xmm1
  lea    D41(%rip),%rdi
  lea    0x60(%rsp),%rbp
  lea    0xa0(%rsp),%r13
  lea    D33(%rip),%r12
  movdqa %xmm0,%xmm2
  shufps $0x88,0x20(%rsp),%xmm2
  movdqa %xmm2,%xmm4
  shufps $0xdd,0x20(%rsp),%xmm0
  movdqa %xmm1,%xmm3
  shufps $0x88,0x40(%rsp),%xmm3
  shufps $0x88,%xmm3,%xmm4
  movaps %xmm4,0x50(%rsp)
  movdqa %xmm0,%xmm4
  shufps $0xdd,0x40(%rsp),%xmm1
  shufps $0xdd,%xmm3,%xmm2
  shufps $0xdd,%xmm1,%xmm0
  shufps $0x88,%xmm1,%xmm4
  movaps %xmm2,0x70(%rsp)
  movaps %xmm4,0x60(%rsp)
  movaps %xmm0,0x80(%rsp)
  call   <puts@plt>
L33:
  mov    $0xa,%edi
  lea    -0x10(%rbp),%rbx
  call   <putchar@plt>
L1:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  jne    L1
  mov    $0xa,%edi
  lea    0x10(%rbx),%rbp
  call   <putchar@plt>
  cmp    %rbp,%r13
  jne    L33
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0xa8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D6
type_infered: string
name: extract_symbolic_data
arguments: label: D29
type_infered: string
name: extract_symbolic_data
arguments: label: D28
type_infered: string
name: extract_symbolic_data
arguments: label: D41
type_infered: string
name: extract_symbolic_data
arguments: label: D33
type_infered: string

tool: D6: 
	 Enter Numbers: 

tool: D29: 
 Cell %d: 

tool: D28: %d

tool: D41: 
	 The Transpose is: 

tool: D33: 	 %d 	

assistant: ```c
int main() {
  int a[4][4], b[4][4], m, n, i = 1;
  printf("\n\t Enter Numbers: \n");
  for (m = 0; m < 4; m++) {
    for (n = 0; n < 4; n++) {
      printf("\n Cell %d: ", i);
      scanf("%d", &a[m][n]);
      i++;
    }
  }
  for (m = 0; m < 4; m++) {
    for (n = 0; n < 4; n++) {
      b[n][m] = a[m][n];
    }
  }
  printf("\n\t The Transpose is: \n");
  for (m = 0; m < 4; m++) {
    printf("\n");
    for (n = 0; n < 4; n++) {
      printf("\t %d \t", b[m][n]);
    }
    printf("\n");
  }
  printf("\n");
  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
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x1,-0x4(%rbp)
  jmp    L6
L0:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <check@plt>
  addl   $0x1,-0x4(%rbp)
L6:
  cmpl   $0xa,-0x4(%rbp)
  jle    L0
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(void) {
  int i;
  for (i = 1; i < 11; i = i + 1) {
    check(i);
  }
  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
<main>:
  endbr64
  push   %rbx
  mov    $0x1,%ebx
L4:
  mov    %ebx,%edi
  call   <check@plt>
  add    $0x1,%ebx
  cmp    $0xb,%ebx
  jne    L4
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(void) {
  int i;
  for (i = 1; i < 11; i = i + 1) {
    check(i);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x1,%ebx
  nopw   0x0(%rax,%rax,1)
L5:
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <check@plt>
  cmp    $0xb,%ebx
  jne    L5
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(void) {
  int i;
  for (i = 1; i < 11; i = i + 1) {
    check(i);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x1,%ebx
  nopw   0x0(%rax,%rax,1)
L1:
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <check@plt>
  cmp    $0xb,%ebx
  jne    L1
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(void) {
  int i;
  for (i = 1; i < 11; i = i + 1) {
    check(i);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<pastShield>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L15
L8:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  lea    0x1(%rdx),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  subl   $0x1,-0x4(%rbp)
L15:
  cmpl   $0x0,-0x4(%rbp)
  jns    L8
  mov    -0x18(%rbp),%rax
  movb   $0x2f,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void pastShield(char *sendbuff) {
  for (int i = strlen(sendbuff) - 1; i >= 0; i--)
    sendbuff[i + 1] = sendbuff[i];
  sendbuff[0] = '/';
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<pastShield>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %eax,%edx
  sub    $0x1,%edx
  js     L15
  movslq %edx,%rdx
  add    %rbx,%rdx
  movslq %eax,%rcx
  lea    -0x2(%rbx,%rcx,1),%rcx
  lea    -0x1(%rax),%eax
  sub    %rax,%rcx
L3:
  movzbl (%rdx),%eax
  mov    %al,0x1(%rdx)
  sub    $0x1,%rdx
  cmp    %rcx,%rdx
  jne    L3
L15:
  movb   $0x2f,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void pastShield(char *sendbuff) {
  for (int i = strlen(sendbuff) - 1; i >= 0; i--)
    sendbuff[i + 1] = sendbuff[i];
  sendbuff[0] = '/';
  return;
}
```
user: What is the c source code of the assembly code below:

```asm
<pastShield>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %eax,%edx
  sub    $0x1,%edx
  js     L7
  movslq %eax,%rdi
  mov    %edx,%ecx
  movslq %edx,%rsi
  mov    %rdi,%rax
  lea    0x1(%rcx),%r8
  sub    %rcx,%rsi
  sub    %rcx,%rax
  lea    (%rbx,%rsi,1),%rsi
  mov    %r8,%rdx
  lea    (%rbx,%rax,1),%rdi
  call   <memmove@plt>
L7:
  movb   $0x2f,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void pastShield(char *sendbuff) {
  for (int i = strlen(sendbuff) - 1; i >= 0; i--)
    sendbuff[i + 1] = sendbuff[i];
  sendbuff[0] = '/';
  return;
}
```