ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<r_combination>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    %edx,%ebx
  call   <r_reflection@plt>
  mov    %ebx,%eax
  sub    $0x1,%eax
  js     L40
  cltq
  lea    (%r12,%rax,8),%rsi
  mov    $0x1,%r9d
  mov    %ebx,%r8d
  jmp    L26
L52:
  movslq %r9d,%rax
  lea    -0x8(%rbp,%rax,8),%rdi
  mov    $0x0,%eax
L30:
  mov    (%rsi),%rdx
  movzbl (%rdx,%rax,1),%ecx
  mov    (%rdi),%rdx
  mov    %cl,(%rdx,%rax,1)
  add    $0x1,%rax
  cmp    %r8,%rax
  jne    L30
L12:
  sub    $0x8,%rsi
  lea    0x1(%r9),%eax
  cmp    %ebx,%r9d
  je     L40
  mov    %eax,%r9d
L26:
  test   %ebx,%ebx
  jg     L52
  jmp    L12
L40:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void r_combination(char **arr, char **arr2, int n) {
  int i, j;
  r_reflection(arr, arr2, n);
  for (i = n - 1; i >= 0; i--)
    for (j = 0; j < n; j++)
      arr2[n - i - 1][j] = arr[i][j];
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<r_combination>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %edx,%ebx
  call   <r_reflection@plt>
  mov    %ebx,%r8d
  sub    $0x1,%r8d
  js     L16
  movslq %r8d,%rax
  mov    $0x1,%r9d
  mov    %r8d,%r8d
  lea    (%r12,%rax,8),%rsi
  xchg   %ax,%ax
L40:
  test   %ebx,%ebx
  jle    L41
  mov    %r9d,%eax
  lea    -0x8(%rbp,%rax,8),%rdi
  xor    %eax,%eax
  xchg   %ax,%ax
L38:
  mov    (%rsi),%rdx
  movzbl (%rdx,%rax,1),%ecx
  mov    (%rdi),%rdx
  mov    %cl,(%rdx,%rax,1)
  mov    %rax,%rdx
  add    $0x1,%rax
  cmp    %rdx,%r8
  jne    L38
L41:
  sub    $0x8,%rsi
  lea    0x1(%r9),%eax
  cmp    %ebx,%r9d
  je     L16
  mov    %eax,%r9d
  jmp    L40
  nopl   0x0(%rax,%rax,1)
L16:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void r_combination(char **arr, char **arr2, int n) {
  int i, j;
  r_reflection(arr, arr2, n);
  for (i = n - 1; i >= 0; i--)
    for (j = 0; j < n; j++)
      arr2[n - i - 1][j] = arr[i][j];
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<r_combination>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %rsi,%rbx
  call   <r_reflection@plt>
  test   %ebp,%ebp
  jle    L5
  lea    -0x1(%rbp),%eax
  mov    %rbx,%rsi
  mov    %ebp,%r8d
  movslq %eax,%rdx
  mov    %eax,%eax
  lea    (%r12,%rdx,8),%rdi
  lea    0x8(%rbx,%rax,8),%r9
L14:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L2:
  mov    (%rdi),%rdx
  movzbl (%rdx,%rax,1),%ecx
  mov    (%rsi),%rdx
  mov    %cl,(%rdx,%rax,1)
  add    $0x1,%rax
  cmp    %r8,%rax
  jne    L2
  add    $0x8,%rsi
  sub    $0x8,%rdi
  cmp    %r9,%rsi
  jne    L14
L5:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void r_combination(char **arr, char **arr2, int n) {
  int i, j;
  r_reflection(arr, arr2, n);
  for (i = n - 1; i >= 0; i--)
    for (j = 0; j < n; j++)
      arr2[n - i - 1][j] = arr[i][j];
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<Catalan>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  movl   $0x1,(%rax)
  movl   $0x1,-0x8(%rbp)
  jmp    L32
L34:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  movl   $0x0,-0x4(%rbp)
  jmp    L23
L11:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%ecx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%eax
  sub    $0x1,%eax
  sub    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rsi
  mov    -0x20(%rbp),%rax
  add    %rsi,%rax
  mov    (%rax),%eax
  imul   %eax,%edx
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rsi
  mov    -0x20(%rbp),%rax
  add    %rsi,%rax
  add    %ecx,%edx
  mov    %edx,(%rax)
  addl   $0x1,-0x4(%rbp)
L23:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jg     L11
  addl   $0x1,-0x8(%rbp)
L32:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L34
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void Catalan(int n, int *d) {
  int i, j;
  d[0] = 1;
  for (i = 1; i <= n; i++) {
    d[i] = 0;
    for (j = 0; j <= i - 1; j++)
      d[i] += d[j] * d[i - 1 - j];
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<Catalan>:
  endbr64
  movl   $0x1,(%rsi)
  test   %edi,%edi
  jle    L9
  lea    0x4(%rsi),%r9
  mov    $0x0,%r10d
L11:
  mov    %r9,%r8
  movl   $0x0,(%r9)
  add    $0x1,%r10d
  test   %r10d,%r10d
  jle    L19
  mov    %rsi,%rax
  mov    %r9,%rcx
L29:
  mov    (%rax),%edx
  imul   -0x4(%rcx),%edx
  add    %edx,(%r8)
  add    $0x4,%rax
  sub    $0x4,%rcx
  cmp    %r9,%rax
  jne    L29
L19:
  add    $0x4,%r9
  cmp    %edi,%r10d
  jne    L11
L9:
  ret
```

assistant: ```c
void Catalan(int n, int *d) {
  int i, j;
  d[0] = 1;
  for (i = 1; i <= n; i++) {
    d[i] = 0;
    for (j = 0; j <= i - 1; j++)
      d[i] += d[j] * d[i - 1 - j];
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<Catalan>:
  endbr64
  movl   $0x1,(%rsi)
  test   %edi,%edi
  jle    L23
  movslq %edi,%r10
  lea    0x4(%rsi),%r8
  mov    $0x4,%edi
  xor    %r9d,%r9d
  nopl   (%rax)
L12:
  movl   $0x0,(%r8)
  mov    %rsi,%rax
  xor    %ecx,%ecx
  nopl   0x0(%rax)
L0:
  mov    %rdi,%rdx
  sub    %rax,%rdx
  add    $0x4,%rax
  mov    -0x4(%rdx,%rsi,2),%edx
  imul   -0x4(%rax),%edx
  add    %edx,%ecx
  mov    %ecx,(%r8)
  cmp    %rax,%r8
  jne    L0
  add    $0x1,%r9
  add    $0x4,%r8
  add    $0x4,%rdi
  cmp    %r10,%r9
  jne    L12
L23:
  ret
```

assistant: ```c
void Catalan(int n, int *d) {
  int i, j;
  d[0] = 1;
  for (i = 1; i <= n; i++) {
    d[i] = 0;
    for (j = 0; j <= i - 1; j++)
      d[i] += d[j] * d[i - 1 - j];
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Catalan>:
  endbr64
  movl   $0x1,(%rsi)
  test   %edi,%edi
  jle    L8
  movslq %edi,%r10
  lea    0x4(%rsi),%r8
  mov    $0x4,%edi
  xor    %r9d,%r9d
  nopl   (%rax)
L28:
  movl   $0x0,(%r8)
  mov    %rsi,%rax
  xor    %ecx,%ecx
  nopl   0x0(%rax)
L10:
  mov    %rdi,%rdx
  sub    %rax,%rdx
  add    $0x4,%rax
  mov    -0x4(%rdx,%rsi,2),%edx
  imul   -0x4(%rax),%edx
  add    %edx,%ecx
  mov    %ecx,(%r8)
  cmp    %rax,%r8
  jne    L10
  add    $0x1,%r9
  add    $0x4,%r8
  add    $0x4,%rdi
  cmp    %r10,%r9
  jne    L28
L8:
  ret
```

assistant: ```c
void Catalan(int n, int *d) {
  int i, j;
  d[0] = 1;
  for (i = 1; i <= n; i++) {
    d[i] = 0;
    for (j = 0; j <= i - 1; j++)
      d[i] += d[j] * d[i - 1 - j];
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<utilIpCmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L50
  cmpq   $0x0,-0x20(%rbp)
  jne    L50
  mov    $0x0,%eax
  jmp    L131
L50:
  cmpq   $0x0,-0x18(%rbp)
  jne    L1
  cmpq   $0x0,-0x20(%rbp)
  je     L1
  mov    $0xffffffffffffffff,%rax
  jmp    L131
L1:
  cmpq   $0x0,-0x18(%rbp)
  je     L146
  cmpq   $0x0,-0x20(%rbp)
  jne    L146
  mov    $0x1,%eax
  jmp    L131
L146:
  movb   $0x0,-0x1(%rbp)
  jmp    L126
L106:
  movzbl -0x1(%rbp),%edx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  movzbl -0x1(%rbp),%ecx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jbe    L104
  mov    $0x1,%eax
  jmp    L131
L104:
  movzbl -0x1(%rbp),%edx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  movzbl -0x1(%rbp),%ecx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jae    L132
  mov    $0xffffffffffffffff,%rax
  jmp    L131
L132:
  movzbl -0x1(%rbp),%eax
  add    $0x1,%eax
  mov    %al,-0x1(%rbp)
L126:
  cmpb   $0x3,-0x1(%rbp)
  jbe    L106
  mov    $0x0,%eax
L131:
  pop    %rbp
  ret
```

assistant: ```c
long utilIpCmp(unsigned char *byIp1, unsigned char *byIp2) {
  unsigned char iCnt;
  if (byIp1 == (0) && byIp2 == (0))
    return 0;
  if (byIp1 == (0) && byIp2 != (0))
    return -1;
  if (byIp1 != (0) && byIp2 == (0))
    return 1;
  for (iCnt = 0; iCnt < 4; iCnt++) {
    if (byIp1[iCnt] > byIp2[iCnt])
      return 1;
    if (byIp1[iCnt] < byIp2[iCnt])
      return -1;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<utilIpCmp>:
  endbr64
  mov    $0x0,%eax
  mov    %rdi,%rdx
  or     %rsi,%rdx
  je     L25
  test   %rdi,%rdi
  jne    L13
  mov    $0xffffffffffffffff,%rax
  test   %rsi,%rsi
  je     L13
L25:
  ret
L13:
  test   %rsi,%rsi
  jne    L80
  mov    $0x1,%eax
  test   %rdi,%rdi
  jne    L25
L80:
  mov    $0x0,%eax
L6:
  movzbl (%rdi,%rax,1),%ecx
  movzbl (%rsi,%rax,1),%edx
  cmp    %dl,%cl
  ja     L83
  jb     L77
  add    $0x1,%rax
  cmp    $0x4,%rax
  jne    L6
  mov    $0x0,%eax
  ret
L83:
  mov    $0x1,%eax
  ret
L77:
  mov    $0xffffffffffffffff,%rax
  jmp    L25
```

assistant: ```c
long utilIpCmp(unsigned char *byIp1, unsigned char *byIp2) {
  unsigned char iCnt;
  if (byIp1 == (0) && byIp2 == (0))
    return 0;
  if (byIp1 == (0) && byIp2 != (0))
    return -1;
  if (byIp1 != (0) && byIp2 == (0))
    return 1;
  for (iCnt = 0; iCnt < 4; iCnt++) {
    if (byIp1[iCnt] > byIp2[iCnt])
      return 1;
    if (byIp1[iCnt] < byIp2[iCnt])
      return -1;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<utilIpCmp>:
  endbr64
  mov    %rdi,%rdx
  xor    %eax,%eax
  or     %rsi,%rdx
  je     L6
  test   %rdi,%rdi
  jne    L39
  mov    $0xffffffffffffffff,%rax
  test   %rsi,%rsi
  jne    L6
L77:
  mov    $0x1,%eax
  test   %rdi,%rdi
  je     L76
L6:
  ret
  nopw   0x0(%rax,%rax,1)
L39:
  test   %rsi,%rsi
  je     L77
L76:
  xor    %eax,%eax
L45:
  movzbl (%rsi,%rax,1),%ecx
  cmp    %cl,(%rdi,%rax,1)
  ja     L53
  jb     L13
  add    $0x1,%rax
  cmp    $0x4,%rax
  jne    L45
  xor    %eax,%eax
  ret
  nop
L53:
  mov    $0x1,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L13:
  mov    $0xffffffffffffffff,%rax
  ret
```

assistant: ```c
long utilIpCmp(unsigned char *byIp1, unsigned char *byIp2) {
  unsigned char iCnt;
  if (byIp1 == (0) && byIp2 == (0))
    return 0;
  if (byIp1 == (0) && byIp2 != (0))
    return -1;
  if (byIp1 != (0) && byIp2 == (0))
    return 1;
  for (iCnt = 0; iCnt < 4; iCnt++) {
    if (byIp1[iCnt] > byIp2[iCnt])
      return 1;
    if (byIp1[iCnt] < byIp2[iCnt])
      return -1;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<utilIpCmp>:
  endbr64
  mov    %rdi,%rdx
  xor    %eax,%eax
  or     %rsi,%rdx
  je     L73
  test   %rdi,%rdi
  jne    L43
  mov    $0xffffffffffffffff,%rax
  test   %rsi,%rsi
  jne    L73
L53:
  mov    $0x1,%eax
  test   %rdi,%rdi
  je     L29
L73:
  ret
  nopw   0x0(%rax,%rax,1)
L43:
  test   %rsi,%rsi
  je     L53
L29:
  movzbl (%rsi),%eax
  cmp    %al,(%rdi)
  ja     L28
  jb     L90
  movzbl 0x1(%rsi),%eax
  cmp    %al,0x1(%rdi)
  ja     L28
  jb     L90
  movzbl 0x2(%rsi),%eax
  cmp    %al,0x2(%rdi)
  ja     L28
  jb     L90
  movzbl 0x3(%rsi),%eax
  cmp    %al,0x3(%rdi)
  ja     L28
  sbb    %rax,%rax
  ret
  nopl   0x0(%rax)
L28:
  mov    $0x1,%eax
  ret
  xchg   %ax,%ax
L90:
  mov    $0xffffffffffffffff,%rax
  ret
```

assistant: ```c
long utilIpCmp(unsigned char *byIp1, unsigned char *byIp2) {
  unsigned char iCnt;
  if (byIp1 == (0) && byIp2 == (0))
    return 0;
  if (byIp1 == (0) && byIp2 != (0))
    return -1;
  if (byIp1 != (0) && byIp2 == (0))
    return 1;
  for (iCnt = 0; iCnt < 4; iCnt++) {
    if (byIp1[iCnt] > byIp2[iCnt])
      return 1;
    if (byIp1[iCnt] < byIp2[iCnt])
      return -1;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<perl_split_on_nothing>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x38(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%rax
  shl    $0x3,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  movl   $0x0,-0x24(%rbp)
  jmp    L10
L0:
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    $0x2,%edi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%rax
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  add    $0x1,%rax
  movb   $0x0,(%rax)
  addl   $0x1,-0x24(%rbp)
L10:
  mov    -0x24(%rbp),%eax
  movslq %eax,%rbx
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %rax,%rbx
  jb     L0
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movq   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char **perl_split_on_nothing(buf)
char *buf;
{
  char **return_buf;
  int cntr;
  char *temp2;
  temp2 = (char *)malloc(sizeof(char) * (1 + strlen(buf)));
  strcpy(temp2, buf);
  return_buf = (char **)malloc(sizeof(char *) * (strlen(buf) + 1));
  for (cntr = 0; cntr < strlen(buf); ++cntr) {
    return_buf[cntr] = (char *)malloc(sizeof(char) * 2);
    return_buf[cntr][0] = temp2[cntr];
    return_buf[cntr][1] = '\0';
  }
  return_buf[cntr] = 0;
  return (return_buf);
}
```
user: What is the c source code of the assembly code below:

```asm
<perl_split_on_nothing>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%r13
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    0x8(,%rax,8),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    $0x0,%ebx
  jmp    L12
L4:
  mov    $0x2,%edi
  call   <malloc@plt>
  mov    %rax,(%r12,%rbx,8)
  movzbl 0x0(%r13,%rbx,1),%edx
  mov    %dl,(%rax)
  movb   $0x0,0x1(%rax)
  add    $0x1,%rbx
L12:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %rbx,%rax
  ja     L4
  movq   $0x0,(%r12,%rbx,8)
  mov    %r12,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char **perl_split_on_nothing(buf)
char *buf;
{
  char **return_buf;
  int cntr;
  char *temp2;
  temp2 = (char *)malloc(sizeof(char) * (1 + strlen(buf)));
  strcpy(temp2, buf);
  return_buf = (char **)malloc(sizeof(char *) * (strlen(buf) + 1));
  for (cntr = 0; cntr < strlen(buf); ++cntr) {
    return_buf[cntr] = (char *)malloc(sizeof(char) * 2);
    return_buf[cntr][0] = temp2[cntr];
    return_buf[cntr][1] = '\0';
  }
  return_buf[cntr] = 0;
  return (return_buf);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<perl_split_on_nothing>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strlen@plt>
  lea    0x1(%rax),%rbx
  mov    %rax,%r12
  mov    %rbx,%rdi
  call   <malloc@plt>
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r14
  call   <memcpy@plt>
  lea    0x0(,%rbx,8),%rdi
  xor    %ebx,%ebx
  call   <malloc@plt>
  mov    %rax,%r13
  mov    %rax,%rbp
  jmp    L6
  nopl   0x0(%rax,%rax,1)
L11:
  mov    $0x2,%edi
  add    $0x8,%rbp
  call   <malloc@plt>
  movzbl (%r14,%rbx,1),%edx
  add    $0x1,%rbx
  mov    %rax,-0x8(%rbp)
  mov    %dl,(%rax)
  movb   $0x0,0x1(%rax)
L6:
  cmp    %rbx,%r12
  ja     L11
  movq   $0x0,0x0(%rbp)
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char **perl_split_on_nothing(buf)
char *buf;
{
  char **return_buf;
  int cntr;
  char *temp2;
  temp2 = (char *)malloc(sizeof(char) * (1 + strlen(buf)));
  strcpy(temp2, buf);
  return_buf = (char **)malloc(sizeof(char *) * (strlen(buf) + 1));
  for (cntr = 0; cntr < strlen(buf); ++cntr) {
    return_buf[cntr] = (char *)malloc(sizeof(char) * 2);
    return_buf[cntr][0] = temp2[cntr];
    return_buf[cntr][1] = '\0';
  }
  return_buf[cntr] = 0;
  return (return_buf);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<perl_split_on_nothing>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strlen@plt>
  lea    0x1(%rax),%rbx
  mov    %rax,%r12
  mov    %rbx,%rdi
  call   <malloc@plt>
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r14
  call   <memcpy@plt>
  lea    0x0(,%rbx,8),%rdi
  xor    %ebx,%ebx
  call   <malloc@plt>
  mov    %rax,%r13
  mov    %rax,%rbp
  jmp    L11
  nopl   0x0(%rax,%rax,1)
L7:
  mov    $0x2,%edi
  add    $0x8,%rbp
  call   <malloc@plt>
  movzbl (%r14,%rbx,1),%edx
  add    $0x1,%rbx
  mov    %rax,-0x8(%rbp)
  mov    %dl,(%rax)
  movb   $0x0,0x1(%rax)
L11:
  cmp    %rbx,%r12
  ja     L7
  movq   $0x0,0x0(%rbp)
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char **perl_split_on_nothing(buf)
char *buf;
{
  char **return_buf;
  int cntr;
  char *temp2;
  temp2 = (char *)malloc(sizeof(char) * (1 + strlen(buf)));
  strcpy(temp2, buf);
  return_buf = (char **)malloc(sizeof(char *) * (strlen(buf) + 1));
  for (cntr = 0; cntr < strlen(buf); ++cntr) {
    return_buf[cntr] = (char *)malloc(sizeof(char) * 2);
    return_buf[cntr][0] = temp2[cntr];
    return_buf[cntr][1] = '\0';
  }
  return_buf[cntr] = 0;
  return (return_buf);
}
```
user: Translate the below assembly code into C programming language:

```asm
<supprimerSuite>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,%eax
  mov    %al,-0x14(%rbp)
  nop
L17:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x4(%rbp)
  movsbl -0x14(%rbp),%eax
  cmp    %eax,-0x4(%rbp)
  je     L2
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jne    L17
L2:
  nop
  leave
  ret
```

assistant: ```c
void supprimerSuite(char caractereRecherche) {
  int caractere;
  while ((caractere = getchar()) != caractereRecherche && caractere != EOF) {
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<supprimerSuite>:
  endbr64
  push   %rbx
  movsbl %dil,%ebx
L5:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    %eax,%ebx
  je     L3
  mov    D5(%rip),%rdx
  cmp    %eax,(%rdx)
  jne    L5
L3:
  pop    %rbx
  ret
```

assistant: ```c
void supprimerSuite(char caractereRecherche) {
  int caractere;
  while ((caractere = getchar()) != caractereRecherche && caractere != EOF) {
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<supprimerSuite>:
  endbr64
  push   %rbx
  movsbl %dil,%ebx
  jmp    L11
  nopl   0x0(%rax,%rax,1)
L19:
  mov    D4(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L12
L11:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    %ebx,%eax
  jne    L19
L12:
  pop    %rbx
  ret
```

assistant: ```c
void supprimerSuite(char caractereRecherche) {
  int caractere;
  while ((caractere = getchar()) != caractereRecherche && caractere != EOF) {
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<supprimerSuite>:
  endbr64
  push   %rbx
  movsbl %dil,%ebx
  jmp    L6
  nopl   0x0(%rax,%rax,1)
L4:
  mov    D4(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L3
L6:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    %ebx,%eax
  jne    L4
L3:
  pop    %rbx
  ret
```

assistant: ```c
void supprimerSuite(char caractereRecherche) {
  int caractere;
  while ((caractere = getchar()) != caractereRecherche && caractere != EOF) {
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<Cyc_Sexp_parse_short>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x0,-0x18(%rbp)
L2:
  cmpl   $0x3,-0x18(%rbp)
  ja     L16
  nop
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <Cyc_Sexp_pgetc@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <Cyc_Sexp_hex2value@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x1c(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  or     -0x4(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  addl   $0x1,-0x18(%rbp)
  jmp    L2
L16:
  nop
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %ax,-0x1e(%rbp)
  movzwl -0x1e(%rbp),%eax
  leave
  ret
```

assistant: ```c
unsigned short Cyc_Sexp_parse_short(struct Cyc_Sexp_Parser *p) {
  unsigned _T0;
  unsigned _T1;
  unsigned _T2;
  unsigned short _T3;
  unsigned res = 0U;
  {
    unsigned i = 0U;
  _TL6B:
    if (i < 4U)
      goto _TL69;
    else {
      goto _TL6A;
    }
  _TL69 : {
    int ch = Cyc_Sexp_pgetc(p);
    unsigned v = Cyc_Sexp_hex2value(p, ch);
    _T0 = res << 4;
    _T1 = v;
    res = _T0 | _T1;
  }
    i = i + 1;
    goto _TL6B;
  _TL6A:;
  }
  _T2 = res;
  _T3 = (unsigned short)_T2;
  return _T3;
}
```
user: What is the c source code of the assembly code below?

```asm
<Cyc_Sexp_parse_short>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    $0x4,%ebp
  mov    $0x0,%ebx
L8:
  mov    %r12,%rdi
  call   <Cyc_Sexp_pgetc@plt>
  mov    %eax,%esi
  mov    %r12,%rdi
  call   <Cyc_Sexp_hex2value@plt>
  shl    $0x4,%ebx
  or     %eax,%ebx
  sub    $0x1,%ebp
  jne    L8
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
unsigned short Cyc_Sexp_parse_short(struct Cyc_Sexp_Parser *p) {
  unsigned _T0;
  unsigned _T1;
  unsigned _T2;
  unsigned short _T3;
  unsigned res = 0U;
  {
    unsigned i = 0U;
  _TL6B:
    if (i < 4U)
      goto _TL69;
    else {
      goto _TL6A;
    }
  _TL69 : {
    int ch = Cyc_Sexp_pgetc(p);
    unsigned v = Cyc_Sexp_hex2value(p, ch);
    _T0 = res << 4;
    _T1 = v;
    res = _T0 | _T1;
  }
    i = i + 1;
    goto _TL6B;
  _TL6A:;
  }
  _T2 = res;
  _T3 = (unsigned short)_T2;
  return _T3;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Cyc_Sexp_parse_short>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    $0x4,%ebp
  push   %rbx
  xor    %ebx,%ebx
L7:
  mov    %r12,%rdi
  shl    $0x4,%ebx
  call   <Cyc_Sexp_pgetc@plt>
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <Cyc_Sexp_hex2value@plt>
  or     %eax,%ebx
  sub    $0x1,%ebp
  jne    L7
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
unsigned short Cyc_Sexp_parse_short(struct Cyc_Sexp_Parser *p) {
  unsigned _T0;
  unsigned _T1;
  unsigned _T2;
  unsigned short _T3;
  unsigned res = 0U;
  {
    unsigned i = 0U;
  _TL6B:
    if (i < 4U)
      goto _TL69;
    else {
      goto _TL6A;
    }
  _TL69 : {
    int ch = Cyc_Sexp_pgetc(p);
    unsigned v = Cyc_Sexp_hex2value(p, ch);
    _T0 = res << 4;
    _T1 = v;
    res = _T0 | _T1;
  }
    i = i + 1;
    goto _TL6B;
  _TL6A:;
  }
  _T2 = res;
  _T3 = (unsigned short)_T2;
  return _T3;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<Cyc_Sexp_parse_short>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    $0x4,%ebp
  push   %rbx
  xor    %ebx,%ebx
L1:
  mov    %r12,%rdi
  shl    $0x4,%ebx
  call   <Cyc_Sexp_pgetc@plt>
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <Cyc_Sexp_hex2value@plt>
  or     %eax,%ebx
  sub    $0x1,%ebp
  jne    L1
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
unsigned short Cyc_Sexp_parse_short(struct Cyc_Sexp_Parser *p) {
  unsigned _T0;
  unsigned _T1;
  unsigned _T2;
  unsigned short _T3;
  unsigned res = 0U;
  {
    unsigned i = 0U;
  _TL6B:
    if (i < 4U)
      goto _TL69;
    else {
      goto _TL6A;
    }
  _TL69 : {
    int ch = Cyc_Sexp_pgetc(p);
    unsigned v = Cyc_Sexp_hex2value(p, ch);
    _T0 = res << 4;
    _T1 = v;
    res = _T0 | _T1;
  }
    i = i + 1;
    goto _TL6B;
  _TL6A:;
  }
  _T2 = res;
  _T3 = (unsigned short)_T2;
  return _T3;
}
```