ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的：

```asm
<hlines>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    %r8d,-0x28(%rbp)
  movss  %xmm0,-0x2c(%rbp)
  mov    %r9d,-0x30(%rbp)
  movb   $0x0,-0xa(%rbp)
  movb   $0xff,-0x9(%rbp)
  cmpl   $0x0,-0x30(%rbp)
  je     L32
  movl   $0x0,-0x8(%rbp)
  jmp    L34
L8:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl -0xa(%rbp),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x8(%rbp)
L34:
  mov    -0x1c(%rbp),%eax
  imul   -0x20(%rbp),%eax
  cmp    %eax,-0x8(%rbp)
  jl     L8
L32:
  cmpl   $0x0,-0x24(%rbp)
  jg     L13
  movl   $0x1,-0x24(%rbp)
L13:
  cmpl   $0x0,-0x28(%rbp)
  jg     L54
  movl   $0x1,-0x28(%rbp)
L54:
  mov    -0x20(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cltd
  idivl  -0x24(%rbp)
  mov    %edx,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L15
L41:
  movzbl -0x9(%rbp),%ecx
  mov    -0x28(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  neg    %eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  lea    (%rdx,%rax,1),%r8d
  mov    -0x1c(%rbp),%edi
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%esi
  mov    -0x18(%rbp),%rax
  push   %rcx
  mov    -0x28(%rbp),%ecx
  push   %rcx
  mov    %edi,%r9d
  mov    $0x0,%ecx
  mov    %rax,%rdi
  call   <draw_rectangle@plt>
  add    $0x10,%rsp
  mov    -0x24(%rbp),%eax
  add    %eax,-0x8(%rbp)
L15:
  mov    -0x8(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L41
  nop
  nop
  leave
  ret
```

assistant: ```c
void hlines(unsigned char *sl, int w, int h, int size1, int size2, float ar, int clr) {
  int i, iz;
  unsigned char black, white;
  black = 0;
  white = 255;
  if (clr != 0)
    for (i = 0; i < (w * h); i++)
      sl[i] = black;
  if (size1 < 1)
    size1 = 1;
  if (size2 < 1)
    size2 = 1;
  iz = h / 2 - size1 * ((h / 2) / size1);
  for (i = iz; i < h; i = i + size1)
    draw_rectangle(sl, w, h, 0, i - size2 / 2, w, size2, white);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<hlines>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r15
  mov    %esi,%r12d
  mov    %edx,%ebp
  test   %r9d,%r9d
  je     L15
  mov    %esi,%edx
  imul   %ebp,%edx
  test   %edx,%edx
  jle    L15
  mov    %rdi,%rax
  lea    -0x1(%rdx),%edx
  lea    0x1(%rdi,%rdx,1),%rdx
L13:
  movb   $0x0,(%rax)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L13
L15:
  test   %ecx,%ecx
  mov    $0x1,%eax
  cmovg  %ecx,%eax
  mov    %eax,%esi
  mov    %eax,0xc(%rsp)
  test   %r8d,%r8d
  mov    $0x1,%eax
  cmovg  %r8d,%eax
  mov    %eax,%r13d
  mov    %ebp,%eax
  shr    $0x1f,%eax
  add    %ebp,%eax
  sar    %eax
  cltd
  idiv   %esi
  mov    %edx,%ebx
  cmp    %edx,%ebp
  jle    L47
  mov    %r13d,%r14d
  shr    $0x1f,%r14d
  add    %r13d,%r14d
  sar    %r14d
  neg    %r14d
L35:
  push   $0xff
  push   %r13
  mov    %r12d,%r9d
  lea    (%r14,%rbx,1),%r8d
  mov    $0x0,%ecx
  mov    %ebp,%edx
  mov    %r12d,%esi
  mov    %r15,%rdi
  call   <draw_rectangle@plt>
  add    0x1c(%rsp),%ebx
  add    $0x10,%rsp
  cmp    %ebx,%ebp
  jg     L35
L47:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void hlines(unsigned char *sl, int w, int h, int size1, int size2, float ar, int clr) {
  int i, iz;
  unsigned char black, white;
  black = 0;
  white = 255;
  if (clr != 0)
    for (i = 0; i < (w * h); i++)
      sl[i] = black;
  if (size1 < 1)
    size1 = 1;
  if (size2 < 1)
    size2 = 1;
  iz = h / 2 - size1 * ((h / 2) / size1);
  for (i = iz; i < h; i = i + size1)
    draw_rectangle(sl, w, h, 0, i - size2 / 2, w, size2, white);
}
```
user: What is the c source code of the assembly code below?

```asm
<hlines>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %r8d,%r13d
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  sub    $0x18,%rsp
  test   %r9d,%r9d
  je     L32
  mov    %esi,%edx
  imul   %ebp,%edx
  test   %edx,%edx
  jle    L32
  movslq %edx,%rdx
  xor    %esi,%esi
  call   <memset@plt>
  nopl   0x0(%rax,%rax,1)
L32:
  mov    $0x1,%eax
  test   %r12d,%r12d
  cmovle %eax,%r12d
  test   %r13d,%r13d
  cmovle %eax,%r13d
  mov    %ebp,%eax
  shr    $0x1f,%eax
  add    %ebp,%eax
  sar    %eax
  cltd
  idiv   %r12d
  mov    %edx,%ebx
  cmp    %edx,%ebp
  jle    L20
  mov    %r13d,%eax
  shr    $0x1f,%eax
  add    %r13d,%eax
  sar    %eax
  neg    %eax
  mov    %eax,0xc(%rsp)
  nopw   0x0(%rax,%rax,1)
L29:
  push   $0xff
  mov    %ebp,%edx
  mov    %r14d,%r9d
  xor    %ecx,%ecx
  push   %r13
  mov    0x1c(%rsp),%eax
  mov    %r14d,%esi
  mov    %r15,%rdi
  lea    (%rbx,%rax,1),%r8d
  add    %r12d,%ebx
  call   <draw_rectangle@plt>
  pop    %rax
  pop    %rdx
  cmp    %ebx,%ebp
  jg     L29
L20:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void hlines(unsigned char *sl, int w, int h, int size1, int size2, float ar, int clr) {
  int i, iz;
  unsigned char black, white;
  black = 0;
  white = 255;
  if (clr != 0)
    for (i = 0; i < (w * h); i++)
      sl[i] = black;
  if (size1 < 1)
    size1 = 1;
  if (size2 < 1)
    size2 = 1;
  iz = h / 2 - size1 * ((h / 2) / size1);
  for (i = iz; i < h; i = i + size1)
    draw_rectangle(sl, w, h, 0, i - size2 / 2, w, size2, white);
}
```
user: What is the c source code of the assembly code below?

```asm
<hlines>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %r8d,%r13d
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  sub    $0x18,%rsp
  test   %r9d,%r9d
  je     L28
  mov    %esi,%edx
  imul   %ebp,%edx
  test   %edx,%edx
  jle    L28
  movslq %edx,%rdx
  xor    %esi,%esi
  call   <memset@plt>
  nopl   0x0(%rax,%rax,1)
L28:
  mov    $0x1,%eax
  test   %r12d,%r12d
  cmovle %eax,%r12d
  test   %r13d,%r13d
  cmovle %eax,%r13d
  mov    %ebp,%eax
  shr    $0x1f,%eax
  add    %ebp,%eax
  sar    %eax
  cltd
  idiv   %r12d
  mov    %edx,%ebx
  cmp    %edx,%ebp
  jle    L33
  mov    %r13d,%eax
  shr    $0x1f,%eax
  add    %r13d,%eax
  sar    %eax
  neg    %eax
  mov    %eax,0xc(%rsp)
  nopw   0x0(%rax,%rax,1)
L32:
  push   $0xff
  mov    %ebp,%edx
  mov    %r14d,%r9d
  xor    %ecx,%ecx
  push   %r13
  mov    0x1c(%rsp),%eax
  mov    %r14d,%esi
  mov    %r15,%rdi
  lea    (%rbx,%rax,1),%r8d
  add    %r12d,%ebx
  call   <draw_rectangle@plt>
  pop    %rax
  pop    %rdx
  cmp    %ebx,%ebp
  jg     L32
L33:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void hlines(unsigned char *sl, int w, int h, int size1, int size2, float ar, int clr) {
  int i, iz;
  unsigned char black, white;
  black = 0;
  white = 255;
  if (clr != 0)
    for (i = 0; i < (w * h); i++)
      sl[i] = black;
  if (size1 < 1)
    size1 = 1;
  if (size2 < 1)
    size2 = 1;
  iz = h / 2 - size1 * ((h / 2) / size1);
  for (i = iz; i < h; i = i + size1)
    draw_rectangle(sl, w, h, 0, i - size2 / 2, w, size2, white);
}
```
user: What is the c source code of the assembly code below:

```asm
<my_aes_encrypt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    %rcx,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x2c(%rbp)
  movl   $0x0,-0x2c(%rbp)
  jmp    L69
L65:
  mov    -0x2c(%rbp),%eax
  mov    -0x50(%rbp),%rdx
  shr    $0x4,%rdx
  cmp    %rdx,%rax
  jne    L73
  mov    -0x50(%rbp),%rax
  and    $0xf,%eax
  mov    %eax,-0x24(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  je     L61
  lea    -0x20(%rbp),%rax
  mov    $0x10,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  jmp    L36
L73:
  movl   $0x10,-0x24(%rbp)
L36:
  mov    -0x24(%rbp),%edx
  mov    -0x2c(%rbp),%eax
  shl    $0x4,%rax
  mov    %rax,%rcx
  mov    -0x40(%rbp),%rax
  add    %rax,%rcx
  lea    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movl   $0x0,-0x28(%rbp)
  jmp    L60
L16:
  mov    -0x2c(%rbp),%eax
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%eax
  add    %rax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x28(%rbp),%ecx
  mov    -0x38(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  xor    %eax,%edx
  mov    -0x28(%rbp),%eax
  mov    %dl,-0x20(%rbp,%rax,1)
  addl   $0x1,-0x28(%rbp)
L60:
  mov    -0x28(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jb     L16
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <encrypt@plt>
  lea    -0x20(%rbp),%rcx
  mov    -0x38(%rbp),%rax
  mov    $0x10,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x2c(%rbp),%eax
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x48(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  lea    -0x20(%rbp),%rax
  mov    $0x10,%edx
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcpy@plt>
  addl   $0x1,-0x2c(%rbp)
L69:
  mov    -0x2c(%rbp),%eax
  mov    -0x50(%rbp),%rdx
  shr    $0x4,%rdx
  cmp    %rdx,%rax
  jbe    L65
  jmp    L53
L61:
  nop
L53:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L26
  call   <__stack_chk_fail@plt>
L26:
  leave
  ret
```

assistant: ```c
void my_aes_encrypt(unsigned char *iv, unsigned char *inbuf, unsigned char *outbuf, unsigned long long len) {
  unsigned char block[16];
  unsigned int blockno = 0, i;
  for (blockno = 0; blockno <= (len / sizeof(block)); blockno++) {
    unsigned int fraction;
    if (blockno == (len / sizeof(block))) {
      fraction = len % sizeof(block);
      if (fraction == 0)
        break;
      memset(block, 0, sizeof(block));
    } else
      fraction = 16;
    memcpy(block, inbuf + blockno * sizeof(block), fraction);
    for (i = 0; i < fraction; i++)
      block[i] = inbuf[blockno * sizeof(block) + i] ^ iv[i];
    encrypt((char *)block);
    memcpy(iv, block, sizeof(block));
    memcpy(outbuf + blockno * sizeof(block), block, sizeof(block));
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<my_aes_encrypt>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rdi,%r12
  mov    %rsi,0x10(%rsp)
  mov    %rdx,0x8(%rsp)
  mov    %rcx,0x18(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rcx,%r15
  shr    $0x4,%r15
  mov    $0x0,%r13d
  mov    $0x0,%ebp
  mov    $0x10,%r14d
  jmp    L48
L62:
  mov    %r14d,%ebx
L50:
  shl    $0x4,%rbp
  mov    %rbp,%rax
  mov    %rbp,(%rsp)
  mov    0x10(%rsp),%rbp
  add    %rax,%rbp
  lea    0x20(%rsp),%rdi
  mov    %ebx,%edx
  mov    %rbp,%rsi
  call   <memcpy@plt>
  mov    $0x0,%eax
L10:
  movzbl 0x0(%rbp,%rax,1),%edx
  xor    (%r12,%rax,1),%dl
  mov    %dl,0x20(%rsp,%rax,1)
  add    $0x1,%rax
  cmp    %eax,%ebx
  ja     L10
  lea    0x20(%rsp),%rbx
  mov    %rbx,%rdi
  call   <encrypt@plt>
  mov    %r14d,%edx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
  mov    (%rsp),%rdi
  add    0x8(%rsp),%rdi
  mov    %r14d,%edx
  mov    %rbx,%rsi
  call   <memcpy@plt>
  add    $0x1,%r13d
  mov    %r13d,%ebp
  cmp    %r15,%rbp
  ja     L9
L48:
  cmp    %rbp,%r15
  jne    L62
  mov    0x18(%rsp),%ebx
  and    $0xf,%ebx
  je     L9
  lea    0x20(%rsp),%rdi
  mov    %r14d,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  jmp    L50
L9:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L53
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L53:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void my_aes_encrypt(unsigned char *iv, unsigned char *inbuf, unsigned char *outbuf, unsigned long long len) {
  unsigned char block[16];
  unsigned int blockno = 0, i;
  for (blockno = 0; blockno <= (len / sizeof(block)); blockno++) {
    unsigned int fraction;
    if (blockno == (len / sizeof(block))) {
      fraction = len % sizeof(block);
      if (fraction == 0)
        break;
      memset(block, 0, sizeof(block));
    } else
      fraction = 16;
    memcpy(block, inbuf + blockno * sizeof(block), fraction);
    for (i = 0; i < fraction; i++)
      block[i] = inbuf[blockno * sizeof(block) + i] ^ iv[i];
    encrypt((char *)block);
    memcpy(iv, block, sizeof(block));
    memcpy(outbuf + blockno * sizeof(block), block, sizeof(block));
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<my_aes_encrypt>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rsi,0x8(%rsp)
  lea    0x20(%rsp),%r15
  mov    %rdx,0x10(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  mov    %rcx,%rax
  shr    $0x4,%rax
  mov    %rax,(%rsp)
  mov    %ecx,%eax
  and    $0xf,%eax
  mov    %eax,0x1c(%rsp)
  nopl   0x0(%rax,%rax,1)
L0:
  cmp    (%rsp),%rbp
  jne    L21
  mov    0x1c(%rsp),%r13d
  mov    %r13d,%ebx
  test   %r13d,%r13d
  je     L54
  mov    $0x10,%edx
  xor    %esi,%esi
  mov    %r15,%rdi
  call   <memset@plt>
  mov    %r13d,%edx
L48:
  mov    0x8(%rsp),%rax
  shl    $0x4,%rbp
  mov    %r15,%rdi
  lea    (%rax,%rbp,1),%r13
  mov    %r13,%rsi
  call   <memcpy@plt>
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L23:
  movzbl 0x0(%r13,%rax,1),%edx
  xor    (%r14,%rax,1),%dl
  mov    %dl,(%r15,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rbx
  jne    L23
  mov    %r15,%rdi
  call   <encrypt@plt>
  mov    $0x10,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <memcpy@plt>
  mov    0x10(%rsp),%rax
  mov    $0x10,%edx
  mov    %r15,%rsi
  lea    (%rax,%rbp,1),%rdi
  lea    0x1(%r12),%ebp
  call   <memcpy@plt>
  mov    %rbp,%r12
  cmp    (%rsp),%rbp
  jbe    L0
L54:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L53
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L21:
  mov    $0x10,%edx
  mov    $0x10,%ebx
  jmp    L48
L53:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void my_aes_encrypt(unsigned char *iv, unsigned char *inbuf, unsigned char *outbuf, unsigned long long len) {
  unsigned char block[16];
  unsigned int blockno = 0, i;
  for (blockno = 0; blockno <= (len / sizeof(block)); blockno++) {
    unsigned int fraction;
    if (blockno == (len / sizeof(block))) {
      fraction = len % sizeof(block);
      if (fraction == 0)
        break;
      memset(block, 0, sizeof(block));
    } else
      fraction = 16;
    memcpy(block, inbuf + blockno * sizeof(block), fraction);
    for (i = 0; i < fraction; i++)
      block[i] = inbuf[blockno * sizeof(block) + i] ^ iv[i];
    encrypt((char *)block);
    memcpy(iv, block, sizeof(block));
    memcpy(outbuf + blockno * sizeof(block), block, sizeof(block));
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<tcc_fileextension>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <tcc_basename@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    $0x2e,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L17
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  jmp    L4
L17:
  mov    -0x8(%rbp),%rax
L4:
  leave
  ret
```

assistant: ```c
char *tcc_fileextension(const char *name) {
  char *b = tcc_basename(name);
  char *e = strrchr(b, '.');
  return e ? e : strchr(b, 0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<tcc_fileextension>:
  endbr64
  push   %rbx
  call   <tcc_basename@plt>
  mov    %rax,%rbx
  mov    $0x2e,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  test   %rax,%rax
  je     L2
L19:
  pop    %rbx
  ret
L2:
  mov    %rbx,%rdi
  call   <strlen@plt>
  add    %rbx,%rax
  jmp    L19
```

assistant: ```c
char *tcc_fileextension(const char *name) {
  char *b = tcc_basename(name);
  char *e = strrchr(b, '.');
  return e ? e : strchr(b, 0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<tcc_fileextension>:
  endbr64
  push   %rbx
  call   <tcc_basename@plt>
  mov    $0x2e,%esi
  mov    %rax,%rdi
  mov    %rax,%rbx
  call   <strrchr@plt>
  test   %rax,%rax
  je     L0
  pop    %rbx
  ret
  nopl   0x0(%rax)
L0:
  mov    %rbx,%rdi
  call   <strlen@plt>
  add    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *tcc_fileextension(const char *name) {
  char *b = tcc_basename(name);
  char *e = strrchr(b, '.');
  return e ? e : strchr(b, 0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<tcc_fileextension>:
  endbr64
  push   %rbx
  call   <tcc_basename@plt>
  mov    $0x2e,%esi
  mov    %rax,%rdi
  mov    %rax,%rbx
  call   <strrchr@plt>
  test   %rax,%rax
  je     L5
  pop    %rbx
  ret
  nopl   0x0(%rax)
L5:
  mov    %rbx,%rdi
  call   <strlen@plt>
  add    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *tcc_fileextension(const char *name) {
  char *b = tcc_basename(name);
  char *e = strrchr(b, '.');
  return e ? e : strchr(b, 0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<MaxSubSequenceSum2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L1
L42:
  movl   $0x0,-0x10(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L4
L17:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jle    L32
  mov    -0x10(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L32:
  addl   $0x1,-0x4(%rbp)
L4:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L17
  addl   $0x1,-0x8(%rbp)
L1:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L42
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int MaxSubSequenceSum2(const int A[], int N) {
  int ThisSum, MaxSum, i, j;
  MaxSum = 0;
  for (i = 0; i < N; i++) {
    ThisSum = 0;
    for (j = i; j < N; j++) {
      ThisSum += A[j];
      if (ThisSum > MaxSum)
        MaxSum = ThisSum;
    }
  }
  return MaxSum;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<MaxSubSequenceSum2>:
  endbr64
  test   %esi,%esi
  jle    L6
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x1(%rsi),%r12d
  mov    %esi,%ebx
  mov    %rdi,%r10
  mov    $0x0,%r9d
  mov    $0x0,%ecx
  lea    0x4(%rdi),%r11
  mov    $0x0,%ebp
  jmp    L1
L45:
  mov    %r12d,%eax
  sub    %r9d,%eax
  add    %r9,%rax
  lea    (%r11,%rax,4),%r8
  mov    %r10,%rax
  mov    %ebp,%edi
L20:
  add    (%rax),%edi
  cmp    %edi,%ecx
  cmovl  %edi,%ecx
  add    $0x4,%rax
  cmp    %r8,%rax
  jne    L20
L25:
  add    $0x1,%r9
  add    $0x4,%r10
  cmp    %rbx,%r9
  je     L38
L1:
  cmp    %r9d,%esi
  jg     L45
  jmp    L25
L6:
  mov    $0x0,%ecx
  mov    %ecx,%eax
  ret
L38:
  mov    %ecx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int MaxSubSequenceSum2(const int A[], int N) {
  int ThisSum, MaxSum, i, j;
  MaxSum = 0;
  for (i = 0; i < N; i++) {
    ThisSum = 0;
    for (j = i; j < N; j++) {
      ThisSum += A[j];
      if (ThisSum > MaxSum)
        MaxSum = ThisSum;
    }
  }
  return MaxSum;
}
```
user: Translate the below assembly code into C programming language:

```asm
<MaxSubSequenceSum2>:
  endbr64
  test   %esi,%esi
  jle    L14
  movslq %esi,%r9
  xor    %ecx,%ecx
  xor    %r8d,%r8d
L13:
  mov    %rcx,%rdx
  xor    %eax,%eax
  nopl   (%rax)
L16:
  add    (%rdi,%rdx,4),%eax
  cmp    %eax,%r8d
  cmovl  %eax,%r8d
  add    $0x1,%rdx
  cmp    %edx,%esi
  jg     L16
  add    $0x1,%rcx
  cmp    %rcx,%r9
  jne    L13
  mov    %r8d,%eax
  ret
L14:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int MaxSubSequenceSum2(const int A[], int N) {
  int ThisSum, MaxSum, i, j;
  MaxSum = 0;
  for (i = 0; i < N; i++) {
    ThisSum = 0;
    for (j = i; j < N; j++) {
      ThisSum += A[j];
      if (ThisSum > MaxSum)
        MaxSum = ThisSum;
    }
  }
  return MaxSum;
}
```
user: What is the c source code of the assembly code below?

```asm
<MaxSubSequenceSum2>:
  endbr64
  test   %esi,%esi
  jle    L20
  movslq %esi,%r9
  xor    %ecx,%ecx
  xor    %r8d,%r8d
L26:
  mov    %rcx,%rdx
  xor    %eax,%eax
  nopl   (%rax)
L12:
  add    (%rdi,%rdx,4),%eax
  cmp    %eax,%r8d
  cmovl  %eax,%r8d
  add    $0x1,%rdx
  cmp    %edx,%esi
  jg     L12
  add    $0x1,%rcx
  cmp    %rcx,%r9
  jne    L26
  mov    %r8d,%eax
  ret
L20:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int MaxSubSequenceSum2(const int A[], int N) {
  int ThisSum, MaxSum, i, j;
  MaxSum = 0;
  for (i = 0; i < N; i++) {
    ThisSum = 0;
    for (j = i; j < N; j++) {
      ThisSum += A[j];
      if (ThisSum > MaxSum)
        MaxSum = ThisSum;
    }
  }
  return MaxSum;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<copy_chanpressure>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%eax
  mov    %al,-0x9(%rbp)
  lea    -0x9(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  mov    $0x1,%ecx
  mov    %eax,%esi
  mov    $0xd0,%edi
  call   <mf_write_midi_event@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
void copy_chanpressure(int chan, int c1) {
  char data[1];
  data[0] = (char)c1;
  mf_write_midi_event(0xd0, chan, data, 1);
}
```
user: What is the c source code of the assembly code below?

```asm
<copy_chanpressure>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %sil,0x7(%rsp)
  lea    0x7(%rsp),%rdx
  mov    $0x1,%ecx
  mov    %edi,%esi
  mov    $0xd0,%edi
  call   <mf_write_midi_event@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void copy_chanpressure(int chan, int c1) {
  char data[1];
  data[0] = (char)c1;
  mf_write_midi_event(0xd0, chan, data, 1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<copy_chanpressure>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x1,%ecx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%rdx
  mov    %sil,0x7(%rsp)
  mov    %edi,%esi
  mov    $0xd0,%edi
  call   <mf_write_midi_event@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void copy_chanpressure(int chan, int c1) {
  char data[1];
  data[0] = (char)c1;
  mf_write_midi_event(0xd0, chan, data, 1);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<copy_chanpressure>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x1,%ecx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%rdx
  mov    %sil,0x7(%rsp)
  mov    %edi,%esi
  mov    $0xd0,%edi
  call   <mf_write_midi_event@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void copy_chanpressure(int chan, int c1) {
  char data[1];
  data[0] = (char)c1;
  mf_write_midi_event(0xd0, chan, data, 1);
}
```