ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<findMedianSortedArrays>:
  endbr64
  push   %r15
  mov    %rdx,%r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  lea    (%rsi,%rcx,1),%r13d
  push   %r12
  lea    0x0(,%r13,4),%edi
  mov    %ecx,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %r13d,%ecx
  shr    $0x1f,%ecx
  mov    %rax,%rsi
  add    %r13d,%ecx
  sar    %ecx
  cmp    $0xffffffff,%r13d
  jl     L29
  xor    %eax,%eax
  xor    %r9d,%r9d
  xor    %edx,%edx
  jmp    L80
  nop
L2:
  movslq %edx,%rdi
  movslq %r9d,%r8
  mov    (%r14,%rdi,4),%edi
  mov    (%r15,%r8,4),%ebx
  cmp    %ebx,%edi
  jge    L32
  add    $0x1,%edx
  mov    %edi,%ebx
L27:
  mov    %ebx,(%rsi,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%ecx
  jl     L47
L80:
  cmp    %edx,%ebp
  jle    L66
  cmp    %r9d,%r12d
  jg     L2
  cmp    %edx,%ebp
  jle    L66
  movslq %edx,%rdi
  add    $0x1,%edx
  mov    (%r14,%rdi,4),%ebx
  mov    %ebx,(%rsi,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%ecx
  jge    L80
L47:
  xor    %eax,%eax
  test   %ecx,%ecx
  cmovns %ecx,%eax
  add    $0x1,%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
L113:
  and    $0x1,%r13d
  mov    (%rsi,%rdx,1),%edx
  je     L114
  add    $0x8,%rsp
  pxor   %xmm0,%xmm0
  pop    %rbx
  cvtsi2sd %edx,%xmm0
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L66:
  cmp    %r9d,%r12d
  jle    L27
  movslq %r9d,%rdi
  add    $0x1,%r9d
  mov    (%r15,%rdi,4),%ebx
  jmp    L27
  nopw   0x0(%rax,%rax,1)
L32:
  add    $0x1,%r9d
  jmp    L27
  nopl   0x0(%rax)
L114:
  add    -0x8(%rsi,%rax,1),%edx
  pxor   %xmm0,%xmm0
  cvtsi2sd %edx,%xmm0
  mulsd  D7(%rip),%xmm0
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L29:
  mov    $0xfffffffffffffffc,%rdx
  xor    %eax,%eax
  jmp    L113
```

tool_calls: name: retrieve_data
arguments: data_tag: D7
data_type: byte[8]

tool: {"D7": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

assistant: ```c
double findMedianSortedArrays(int *nums1, int nums1Size, int *nums2, int nums2Size) {
  int sum = nums1Size + nums2Size;
  int *nums = malloc(sizeof(int) * sum);
  int i = 0, j = 0, k = 0;
  int half = sum / 2 + 1;
  while (k < half) {
    int n;
    if (i < nums1Size && j < nums2Size) {
      n = (nums1[i] < nums2[j]) ? nums1[i++] : nums2[j++];
    } else if (i < nums1Size) {
      n = nums1[i++];
    } else if (j < nums2Size) {
      n = nums2[j++];
    }
    nums[k++] = n;
  }
  if (sum % 2) {
    return nums[k - 1];
  } else {
    return (nums[k - 1] + nums[k - 2]) / 2.0;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<findMedianSortedArrays>:
  endbr64
  push   %r15
  mov    %rdx,%r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  lea    (%rsi,%rcx,1),%r13d
  push   %r12
  lea    0x0(,%r13,4),%edi
  mov    %ecx,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %r13d,%ecx
  shr    $0x1f,%ecx
  mov    %rax,%rsi
  add    %r13d,%ecx
  sar    %ecx
  cmp    $0xffffffff,%r13d
  jl     L74
  xor    %eax,%eax
  xor    %r9d,%r9d
  xor    %edx,%edx
  jmp    L2
  nop
L17:
  movslq %edx,%rdi
  movslq %r9d,%r8
  mov    (%r14,%rdi,4),%edi
  mov    (%r15,%r8,4),%ebx
  cmp    %ebx,%edi
  jge    L105
  add    $0x1,%edx
  mov    %edi,%ebx
L114:
  mov    %ebx,(%rsi,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%ecx
  jl     L75
L2:
  cmp    %edx,%ebp
  jle    L1
  cmp    %r9d,%r12d
  jg     L17
  cmp    %edx,%ebp
  jle    L1
  movslq %edx,%rdi
  add    $0x1,%edx
  mov    (%r14,%rdi,4),%ebx
  mov    %ebx,(%rsi,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%ecx
  jge    L2
L75:
  xor    %eax,%eax
  test   %ecx,%ecx
  cmovns %ecx,%eax
  add    $0x1,%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
L99:
  and    $0x1,%r13d
  mov    (%rsi,%rdx,1),%edx
  je     L57
  add    $0x8,%rsp
  pxor   %xmm0,%xmm0
  pop    %rbx
  cvtsi2sd %edx,%xmm0
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L1:
  cmp    %r9d,%r12d
  jle    L114
  movslq %r9d,%rdi
  add    $0x1,%r9d
  mov    (%r15,%rdi,4),%ebx
  jmp    L114
  nopw   0x0(%rax,%rax,1)
L105:
  add    $0x1,%r9d
  jmp    L114
  nopl   0x0(%rax)
L57:
  add    -0x8(%rsi,%rax,1),%edx
  pxor   %xmm0,%xmm0
  cvtsi2sd %edx,%xmm0
  mulsd  D7(%rip),%xmm0
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L74:
  mov    $0xfffffffffffffffc,%rdx
  xor    %eax,%eax
  jmp    L99
```

tool_calls: name: disassemble_data
arguments: tag: D7
datatype: byte[8]

tool: {"D7": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

assistant: ```c
double findMedianSortedArrays(int *nums1, int nums1Size, int *nums2, int nums2Size) {
  int sum = nums1Size + nums2Size;
  int *nums = malloc(sizeof(int) * sum);
  int i = 0, j = 0, k = 0;
  int half = sum / 2 + 1;
  while (k < half) {
    int n;
    if (i < nums1Size && j < nums2Size) {
      n = (nums1[i] < nums2[j]) ? nums1[i++] : nums2[j++];
    } else if (i < nums1Size) {
      n = nums1[i++];
    } else if (j < nums2Size) {
      n = nums2[j++];
    }
    nums[k++] = n;
  }
  if (sum % 2) {
    return nums[k - 1];
  } else {
    return (nums[k - 1] + nums[k - 2]) / 2.0;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x420,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  jmp    L8
L14:
  mov    -0x414(%rbp),%edx
  lea    -0x410(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <write@plt>
L8:
  lea    -0x410(%rbp),%rax
  mov    $0x400,%edx
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <read@plt>
  mov    %eax,-0x414(%rbp)
  cmpl   $0x0,-0x414(%rbp)
  jg     L14
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L22
  call   <__stack_chk_fail@plt>
L22:
  leave
  ret
```

assistant: ```c
int main() {
  char buf[1024];
  int n;
  while ((n = read(0, buf, 1024)) > 0)
    write(1, buf, n);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  jmp    L25
L14:
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <write@plt>
L25:
  mov    $0x400,%edx
  mov    %rbx,%rsi
  mov    $0x0,%edi
  call   <read@plt>
  mov    %eax,%edx
  test   %eax,%eax
  jg     L14
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  mov    $0x0,%eax
  add    $0x410,%rsp
  pop    %rbx
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  char buf[1024];
  int n;
  while ((n = read(0, buf, 1024)) > 0)
    write(1, buf, n);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  jmp    L17
  nopl   0x0(%rax)
L6:
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <write@plt>
L17:
  mov    $0x400,%edx
  mov    %rbx,%rsi
  xor    %edi,%edi
  call   <read@plt>
  mov    %eax,%edx
  test   %eax,%eax
  jg     L6
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x410,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L20:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  char buf[1024];
  int n;
  while ((n = read(0, buf, 1024)) > 0)
    write(1, buf, n);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  jmp    L16
  nopl   0x0(%rax)
L7:
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <write@plt>
L16:
  mov    $0x400,%edx
  mov    %rbx,%rsi
  xor    %edi,%edi
  call   <read@plt>
  mov    %eax,%edx
  test   %eax,%eax
  jg     L7
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x410,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  char buf[1024];
  int n;
  while ((n = read(0, buf, 1024)) > 0)
    write(1, buf, n);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RSA_padding_add_SSLv23>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    $0xa,%eax
  cmp    %eax,-0x20(%rbp)
  jl     L7
  mov    D13(%rip),%rax
  mov    (%rax),%edx
  mov    D9(%rip),%rax
  mov    (%rax),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <RSAerr@plt>
  mov    $0x0,%eax
  jmp    L44
L7:
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x2,(%rax)
  mov    -0x1c(%rbp),%eax
  sub    $0xb,%eax
  sub    -0x20(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <RAND_bytes@plt>
  test   %rax,%rax
  jg     L32
  mov    $0x0,%eax
  jmp    L44
L32:
  movl   $0x0,-0x10(%rbp)
  jmp    L74
L47:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L26
L94:
  mov    -0x8(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <RAND_bytes@plt>
  test   %rax,%rax
  jg     L14
  mov    $0x0,%eax
  jmp    L44
L14:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L94
L26:
  addq   $0x1,-0x8(%rbp)
  addl   $0x1,-0x10(%rbp)
L74:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L47
  mov    -0x8(%rbp),%rax
  mov    $0x8,%edx
  mov    $0x3,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  addq   $0x8,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x0,(%rax)
  mov    -0x20(%rbp),%edx
  mov    -0x28(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    $0x1,%eax
L44:
  leave
  ret
```

assistant: ```c
int RSA_padding_add_SSLv23(unsigned char *to, int tlen, const unsigned char *from, int flen) {
  int i, j;
  unsigned char *p;
  if (flen > (tlen - 11)) {
    RSAerr(RSA_F_RSA_PADDING_ADD_SSLV23, RSA_R_DATA_TOO_LARGE_FOR_KEY_SIZE);
    return (0);
  }
  p = (unsigned char *)to;
  *(p++) = 0;
  *(p++) = 2;
  j = tlen - 3 - 8 - flen;
  if (RAND_bytes(p, j) <= 0)
    return (0);
  for (i = 0; i < j; i++) {
    if (*p == '\0')
      do {
        if (RAND_bytes(p, 1) <= 0)
          return (0);
      } while (*p == '\0');
    p++;
  }
  memset(p, 3, 8);
  p += 8;
  *(p++) = '\0';
  memcpy(p, from, (unsigned int)flen);
  return (1);
}
```
user: What is the c source code of the assembly code below?

```asm
<RSA_padding_add_SSLv23>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0xa(%rsi),%eax
  cmp    %ecx,%eax
  jle    L22
  mov    %rdi,%rbp
  mov    %rdx,%r13
  mov    %ecx,%r12d
  movb   $0x0,(%rdi)
  lea    0x2(%rdi),%rbx
  movb   $0x2,0x1(%rdi)
  lea    -0xb(%rsi),%r14d
  sub    %ecx,%r14d
  mov    %r14d,%esi
  mov    %rbx,%rdi
  call   <RAND_bytes@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  jle    L90
  test   %r14d,%r14d
  jle    L28
  lea    0x2(%r14),%eax
  add    %rax,%rbp
  jmp    L50
L22:
  mov    D13(%rip),%rax
  mov    (%rax),%esi
  mov    D5(%rip),%rax
  mov    (%rax),%edi
  call   <RSAerr@plt>
  mov    $0x0,%eax
  jmp    L90
L53:
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  je     L44
L50:
  cmpb   $0x0,(%rbx)
  jne    L53
L52:
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <RAND_bytes@plt>
  test   %rax,%rax
  jle    L89
  cmpb   $0x0,(%rbx)
  je     L52
  jmp    L53
L28:
  mov    %rbx,%rbp
L44:
  mov    $0x8,%edx
  mov    $0x3,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
  movb   $0x0,0x8(%rbp)
  lea    0x9(%rbp),%rdi
  mov    %r12d,%edx
  mov    %r13,%rsi
  call   <memcpy@plt>
  mov    $0x1,%eax
  jmp    L90
L89:
  mov    $0x0,%eax
L90:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int RSA_padding_add_SSLv23(unsigned char *to, int tlen, const unsigned char *from, int flen) {
  int i, j;
  unsigned char *p;
  if (flen > (tlen - 11)) {
    RSAerr(RSA_F_RSA_PADDING_ADD_SSLV23, RSA_R_DATA_TOO_LARGE_FOR_KEY_SIZE);
    return (0);
  }
  p = (unsigned char *)to;
  *(p++) = 0;
  *(p++) = 2;
  j = tlen - 3 - 8 - flen;
  if (RAND_bytes(p, j) <= 0)
    return (0);
  for (i = 0; i < j; i++) {
    if (*p == '\0')
      do {
        if (RAND_bytes(p, 1) <= 0)
          return (0);
      } while (*p == '\0');
    p++;
  }
  memset(p, 3, 8);
  p += 8;
  *(p++) = '\0';
  memcpy(p, from, (unsigned int)flen);
  return (1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<RSA_padding_add_SSLv23>:
  endbr64
  lea    -0xa(%rsi),%eax
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  cmp    %ecx,%eax
  jle    L71
  mov    $0x200,%eax
  lea    -0xb(%rsi),%ebp
  lea    0x2(%rdi),%rbx
  mov    %rdi,%r14
  mov    %ax,(%rdi)
  sub    %ecx,%ebp
  mov    %rbx,%rdi
  mov    %rdx,%r13
  mov    %ebp,%esi
  mov    %ecx,%r12d
  call   <RAND_bytes@plt>
  test   %rax,%rax
  jle    L73
  test   %ebp,%ebp
  jle    L64
  lea    0x2(%rbp),%eax
  add    %rax,%r14
  jmp    L55
L92:
  add    $0x1,%rbx
  cmp    %rbx,%r14
  je     L36
  nopl   0x0(%rax)
L55:
  cmpb   $0x0,(%rbx)
  je     L53
  add    $0x1,%rbx
  cmp    %rbx,%r14
  jne    L55
L36:
  mov    %r14,%rdi
  mov    $0x8,%edx
  mov    $0x3,%esi
  call   <memset@plt>
  movb   $0x0,0x8(%r14)
  mov    %r12d,%edx
  mov    %r13,%rsi
  lea    0x9(%r14),%rdi
  call   <memcpy@plt>
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L65:
  cmpb   $0x0,(%rbx)
  jne    L92
L53:
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <RAND_bytes@plt>
  test   %rax,%rax
  jg     L65
L73:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L71:
  mov    D18(%rip),%rax
  mov    (%rax),%esi
  mov    D15(%rip),%rax
  mov    (%rax),%edi
  call   <RSAerr@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L64:
  mov    %rbx,%r14
  jmp    L36
```

assistant: ```c
int RSA_padding_add_SSLv23(unsigned char *to, int tlen, const unsigned char *from, int flen) {
  int i, j;
  unsigned char *p;
  if (flen > (tlen - 11)) {
    RSAerr(RSA_F_RSA_PADDING_ADD_SSLV23, RSA_R_DATA_TOO_LARGE_FOR_KEY_SIZE);
    return (0);
  }
  p = (unsigned char *)to;
  *(p++) = 0;
  *(p++) = 2;
  j = tlen - 3 - 8 - flen;
  if (RAND_bytes(p, j) <= 0)
    return (0);
  for (i = 0; i < j; i++) {
    if (*p == '\0')
      do {
        if (RAND_bytes(p, 1) <= 0)
          return (0);
      } while (*p == '\0');
    p++;
  }
  memset(p, 3, 8);
  p += 8;
  *(p++) = '\0';
  memcpy(p, from, (unsigned int)flen);
  return (1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RSA_padding_add_SSLv23>:
  endbr64
  lea    -0xa(%rsi),%eax
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  cmp    %ecx,%eax
  jle    L48
  mov    $0x200,%eax
  lea    -0xb(%rsi),%ebp
  lea    0x2(%rdi),%rbx
  mov    %rdi,%r14
  mov    %ax,(%rdi)
  sub    %ecx,%ebp
  mov    %rbx,%rdi
  mov    %rdx,%r13
  mov    %ebp,%esi
  mov    %ecx,%r12d
  call   <RAND_bytes@plt>
  test   %rax,%rax
  jle    L22
  test   %ebp,%ebp
  jle    L82
  lea    0x2(%rbp),%eax
  add    %rax,%r14
  jmp    L98
L37:
  add    $0x1,%rbx
  cmp    %rbx,%r14
  je     L87
  nopl   0x0(%rax)
L98:
  cmpb   $0x0,(%rbx)
  je     L75
  add    $0x1,%rbx
  cmp    %rbx,%r14
  jne    L98
L87:
  mov    %r14,%rdi
  mov    $0x8,%edx
  mov    $0x3,%esi
  call   <memset@plt>
  movb   $0x0,0x8(%r14)
  mov    %r12d,%edx
  mov    %r13,%rsi
  lea    0x9(%r14),%rdi
  call   <memcpy@plt>
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L4:
  cmpb   $0x0,(%rbx)
  jne    L37
L75:
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <RAND_bytes@plt>
  test   %rax,%rax
  jg     L4
L22:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L48:
  mov    D13(%rip),%rax
  mov    (%rax),%esi
  mov    D14(%rip),%rax
  mov    (%rax),%edi
  call   <RSAerr@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L82:
  mov    %rbx,%r14
  jmp    L87
```

assistant: ```c
int RSA_padding_add_SSLv23(unsigned char *to, int tlen, const unsigned char *from, int flen) {
  int i, j;
  unsigned char *p;
  if (flen > (tlen - 11)) {
    RSAerr(RSA_F_RSA_PADDING_ADD_SSLV23, RSA_R_DATA_TOO_LARGE_FOR_KEY_SIZE);
    return (0);
  }
  p = (unsigned char *)to;
  *(p++) = 0;
  *(p++) = 2;
  j = tlen - 3 - 8 - flen;
  if (RAND_bytes(p, j) <= 0)
    return (0);
  for (i = 0; i < j; i++) {
    if (*p == '\0')
      do {
        if (RAND_bytes(p, 1) <= 0)
          return (0);
      } while (*p == '\0');
    p++;
  }
  memset(p, 3, 8);
  p += 8;
  *(p++) = '\0';
  memcpy(p, from, (unsigned int)flen);
  return (1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<print_hr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  jmp    L10
L14:
  mov    $0x2d,%edi
  call   <putchar@plt>
  addl   $0x1,-0x4(%rbp)
L10:
  cmpl   $0x4,-0x4(%rbp)
  jle    L14
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

assistant: ```c
void print_hr() {
  for (int i = 0; i < 5; ++i)
    putchar('-');
  putchar('\n');
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<print_hr>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2d,%edi
  call   <putchar@plt>
  mov    $0x2d,%edi
  call   <putchar@plt>
  mov    $0x2d,%edi
  call   <putchar@plt>
  mov    $0x2d,%edi
  call   <putchar@plt>
  mov    $0x2d,%edi
  call   <putchar@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void print_hr() {
  for (int i = 0; i < 5; ++i)
    putchar('-');
  putchar('\n');
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<print_hr>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2d,%edi
  call   <putchar@plt>
  mov    $0x2d,%edi
  call   <putchar@plt>
  mov    $0x2d,%edi
  call   <putchar@plt>
  mov    $0x2d,%edi
  call   <putchar@plt>
  mov    $0x2d,%edi
  call   <putchar@plt>
  mov    $0xa,%edi
  add    $0x8,%rsp
  jmp    1050 <putchar@plt>
```

assistant: ```c
void print_hr() {
  for (int i = 0; i < 5; ++i)
    putchar('-');
  putchar('\n');
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<print_hr>:
  endbr64
  push   %rbx
  mov    $0x5,%ebx
L2:
  mov    $0x2d,%edi
  call   <putchar@plt>
  sub    $0x1,%ebx
  jne    L2
  mov    $0xa,%edi
  pop    %rbx
  jmp    1050 <putchar@plt>
```

assistant: ```c
void print_hr() {
  for (int i = 0; i < 5; ++i)
    putchar('-');
  putchar('\n');
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L11
L14:
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  movl   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  addl   $0x1,-0x10(%rbp)
L11:
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  cmp    %eax,-0x10(%rbp)
  jl     L14
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int i, a;
  int *t;
  for (i = 0; i < atoi(argv[1]); i++) {
    t = (int *)malloc(sizeof(int));
    a = *t;
    *t = 0;
    free(t);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbp
  mov    $0x0,%ebx
  jmp    L13
L10:
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  movl   $0x0,(%rax)
  call   <free@plt>
  add    $0x1,%ebx
L13:
  mov    0x8(%rbp),%rdi
  call   <atoi@plt>
  cmp    %ebx,%eax
  jg     L10
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int i, a;
  int *t;
  for (i = 0; i < atoi(argv[1]); i++) {
    t = (int *)malloc(sizeof(int));
    a = *t;
    *t = 0;
    free(t);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  jmp    L16
  nopl   0x0(%rax)
L11:
  mov    $0x4,%edi
  add    $0x1,%ebx
  call   <malloc@plt>
  movl   $0x0,(%rax)
  mov    %rax,%rdi
  call   <free@plt>
L16:
  mov    0x8(%rbp),%rdi
  call   <atoi@plt>
  cmp    %ebx,%eax
  jg     L11
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int i, a;
  int *t;
  for (i = 0; i < atoi(argv[1]); i++) {
    t = (int *)malloc(sizeof(int));
    a = *t;
    *t = 0;
    free(t);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  jmp    L0
  nopl   0x0(%rax)
L13:
  mov    $0x4,%edi
  add    $0x1,%ebx
  call   <malloc@plt>
  movl   $0x0,(%rax)
  mov    %rax,%rdi
  call   <free@plt>
L0:
  mov    0x8(%rbp),%rdi
  call   <atoi@plt>
  cmp    %ebx,%eax
  jg     L13
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int i, a;
  int *t;
  for (i = 0; i < atoi(argv[1]); i++) {
    t = (int *)malloc(sizeof(int));
    a = *t;
    *t = 0;
    free(t);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<gen_DANGER>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%edi
  call   <handle_DANGER@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int gen_DANGER() {
  handle_DANGER(0);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<gen_DANGER>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <handle_DANGER@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int gen_DANGER() {
  handle_DANGER(0);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<gen_DANGER>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <handle_DANGER@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int gen_DANGER() {
  handle_DANGER(0);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<gen_DANGER>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <handle_DANGER@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int gen_DANGER() {
  handle_DANGER(0);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<close_or_die>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <close@plt>
  test   %rax,%rax
  je     L1
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  lea    D0(%rip),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <err@plt>
L1:
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D0
type_guessed: string

tool: D0: close(2) failed: d=%d

assistant: ```c
void close_or_die(int d) {
  if (close(d) != 0) {
    err(1, "close(2) failed: d=%d", d);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<close_or_die>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  call   <close@plt>
  test   %rax,%rax
  jne    L5
L11:
  pop    %rbx
  ret
L5:
  mov    %ebx,%edx
  lea    D6(%rip),%rsi
  mov    $0x1,%edi
  call   <err@plt>
  jmp    L11
```

tool_calls: name: parse_data
arguments: data_label: D6
type_guessed: string

tool: D6: close(2) failed: d=%d

assistant: ```c
void close_or_die(int d) {
  if (close(d) != 0) {
    err(1, "close(2) failed: d=%d", d);
  }
}
```