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

```asm
<escape>:
  endbr64
  movzbl (%rsi),%eax
  mov    D9(%rip),%rdx
  cmp    (%rdx),%al
  je     L64
  mov    $0x1,%ecx
  mov    $0x0,%edx
  mov    D9(%rip),%r8
  jmp    L26
L59:
  movslq %edx,%rax
  movb   $0x5c,(%rdi,%rax,1)
  lea    0x1(%rdx),%eax
  cltq
  movb   $0x74,(%rdi,%rax,1)
  add    $0x2,%edx
  mov    %ecx,%eax
L69:
  cltq
  movzbl (%rsi,%rax,1),%eax
  add    $0x1,%ecx
  cmp    (%r8),%al
  je     L64
L26:
  cmp    $0x9,%al
  je     L59
  cmp    $0xa,%al
  je     L7
  movslq %edx,%r9
  mov    %al,(%rdi,%r9,1)
  mov    %ecx,%eax
  add    $0x1,%edx
  jmp    L69
L7:
  movslq %edx,%rax
  movb   $0x5c,(%rdi,%rax,1)
  lea    0x1(%rdx),%eax
  cltq
  movb   $0x6e,(%rdi,%rax,1)
  add    $0x2,%edx
  mov    %ecx,%eax
  jmp    L69
L64:
  ret
```

assistant: ```c
void escape(char to[200], char from[100]) {
  int i = 0;
  int j = 0;
  while (from[i] != EOF) {
    switch (from[i]) {
    case '\t':
      to[j] = '\\';
      ++j;
      to[j] = 't';
      ++j;
      ++i;
      break;
    case '\n':
      to[j] = '\\';
      ++j;
      to[j] = 'n';
      ++j;
      ++i;
      break;
    default:
      to[j] = from[i];
      ++i;
      ++j;
      break;
    }
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<escape>:
  endbr64
  mov    D9(%rip),%r9
  movzbl (%rsi),%eax
  xor    %ecx,%ecx
  add    $0x1,%rsi
  cmp    (%r9),%al
  jne    L47
  jmp    L41
  nopl   0x0(%rax,%rax,1)
L14:
  mov    %al,(%rdx)
  mov    %r8d,%ecx
L25:
  movzbl (%rsi),%eax
  add    $0x1,%rsi
  cmp    (%r9),%al
  je     L41
L47:
  movslq %ecx,%rdx
  lea    0x1(%rcx),%r8d
  add    %rdi,%rdx
  cmp    $0x9,%al
  je     L42
  cmp    $0xa,%al
  jne    L14
  movslq %r8d,%r8
  movb   $0x5c,(%rdx)
  add    $0x2,%ecx
  add    $0x1,%rsi
  movb   $0x6e,(%rdi,%r8,1)
  movzbl -0x1(%rsi),%eax
  cmp    (%r9),%al
  jne    L47
L41:
  ret
  nop
L42:
  movslq %r8d,%r8
  movb   $0x5c,(%rdx)
  add    $0x2,%ecx
  movb   $0x74,(%rdi,%r8,1)
  jmp    L25
```

assistant: ```c
void escape(char to[200], char from[100]) {
  int i = 0;
  int j = 0;
  while (from[i] != EOF) {
    switch (from[i]) {
    case '\t':
      to[j] = '\\';
      ++j;
      to[j] = 't';
      ++j;
      ++i;
      break;
    case '\n':
      to[j] = '\\';
      ++j;
      to[j] = 'n';
      ++j;
      ++i;
      break;
    default:
      to[j] = from[i];
      ++i;
      ++j;
      break;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<escape>:
  endbr64
  mov    D4(%rip),%r9
  movzbl (%rsi),%eax
  xor    %ecx,%ecx
  add    $0x1,%rsi
  cmp    (%r9),%al
  jne    L54
  jmp    L29
  nopl   0x0(%rax,%rax,1)
L33:
  mov    %al,(%rdx)
  mov    %r8d,%ecx
L60:
  movzbl (%rsi),%eax
  add    $0x1,%rsi
  cmp    (%r9),%al
  je     L29
L54:
  movslq %ecx,%rdx
  lea    0x1(%rcx),%r8d
  add    %rdi,%rdx
  cmp    $0x9,%al
  je     L37
  cmp    $0xa,%al
  jne    L33
  movslq %r8d,%r8
  movb   $0x5c,(%rdx)
  add    $0x2,%ecx
  add    $0x1,%rsi
  movb   $0x6e,(%rdi,%r8,1)
  movzbl -0x1(%rsi),%eax
  cmp    (%r9),%al
  jne    L54
L29:
  ret
  nop
L37:
  movslq %r8d,%r8
  movb   $0x5c,(%rdx)
  add    $0x2,%ecx
  movb   $0x74,(%rdi,%r8,1)
  jmp    L60
```

assistant: ```c
void escape(char to[200], char from[100]) {
  int i = 0;
  int j = 0;
  while (from[i] != EOF) {
    switch (from[i]) {
    case '\t':
      to[j] = '\\';
      ++j;
      to[j] = 't';
      ++j;
      ++i;
      break;
    case '\n':
      to[j] = '\\';
      ++j;
      to[j] = 'n';
      ++j;
      ++i;
      break;
    default:
      to[j] = from[i];
      ++i;
      ++j;
      break;
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<answer2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x1,-0x8(%rbp)
  jmp    L0
L7:
  mov    -0x8(%rbp),%eax
  imul   %eax,%eax
  imul   -0x8(%rbp),%eax
  add    %eax,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L0:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L7
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int answer2(int n) {
  int i;
  int answer2 = 0;
  for (i = 1; i <= n; i++)
    answer2 += i * i * i;
  return answer2;
}
```
user: What is the c source code of the assembly code below:

```asm
<answer2>:
  endbr64
  test   %edi,%edi
  jle    L6
  add    $0x1,%edi
  mov    $0x0,%ecx
  mov    $0x1,%eax
L5:
  mov    %eax,%edx
  imul   %eax,%edx
  imul   %eax,%edx
  add    %edx,%ecx
  add    $0x1,%eax
  cmp    %edi,%eax
  jne    L5
L28:
  mov    %ecx,%eax
  ret
L6:
  mov    $0x0,%ecx
  jmp    L28
```

assistant: ```c
int answer2(int n) {
  int i;
  int answer2 = 0;
  for (i = 1; i <= n; i++)
    answer2 += i * i * i;
  return answer2;
}
```
user: What is the c source code of the assembly code below:

```asm
<answer2>:
  endbr64
  test   %edi,%edi
  jle    L7
  add    $0x1,%edi
  xor    %r8d,%r8d
  mov    $0x1,%eax
  nopl   0x0(%rax,%rax,1)
L8:
  mov    %eax,%edx
  imul   %eax,%edx
  imul   %eax,%edx
  add    $0x1,%eax
  add    %edx,%r8d
  cmp    %edi,%eax
  jne    L8
  mov    %r8d,%eax
  ret
  xchg   %ax,%ax
L7:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int answer2(int n) {
  int i;
  int answer2 = 0;
  for (i = 1; i <= n; i++)
    answer2 += i * i * i;
  return answer2;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<answer2>:
  endbr64
  test   %edi,%edi
  jle    L3
  add    $0x1,%edi
  xor    %r8d,%r8d
  mov    $0x1,%eax
  nopl   0x0(%rax,%rax,1)
L15:
  mov    %eax,%edx
  imul   %eax,%edx
  imul   %eax,%edx
  add    $0x1,%eax
  add    %edx,%r8d
  cmp    %eax,%edi
  jne    L15
  mov    %r8d,%eax
  ret
  xchg   %ax,%ax
L3:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int answer2(int n) {
  int i;
  int answer2 = 0;
  for (i = 1; i <= n; i++)
    answer2 += i * i * i;
  return answer2;
}
```
user: Translate the below assembly code into C programming language:

```asm
<copy_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *copy_string(char *s, int l) {
  char *result = (char *)malloc(l + 1);
  strncpy(result, s, l);
  result[l] = 0;
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<copy_string>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebx
  lea    0x1(%rsi),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %ebx,%edx
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  movslq %ebx,%rbx
  movb   $0x0,0x0(%rbp,%rbx,1)
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *copy_string(char *s, int l) {
  char *result = (char *)malloc(l + 1);
  strncpy(result, s, l);
  result[l] = 0;
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<copy_string>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movslq %esi,%rbx
  lea    0x1(%rbx),%edi
  call   <malloc@plt>
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strncpy@plt>
  movb   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *copy_string(char *s, int l) {
  char *result = (char *)malloc(l + 1);
  strncpy(result, s, l);
  result[l] = 0;
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<pipedata>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x138,%rsp
  mov    %edi,%r12d
  mov    %esi,%r14d
  mov    %fs:0x28,%rax
  mov    %rax,0x128(%rsp)
  xor    %eax,%eax
  lea    0x18(%rsp),%rdi
  call   <FD_ZERO@plt>
  lea    0x1c(%rsp),%rdi
  call   <FD_ZERO@plt>
  lea    0x1(%r14),%r15d
  lea    0x1(%r12),%eax
  cmp    %r14d,%r12d
  cmovg  %eax,%r15d
  mov    $0x0,%ebx
  lea    0x18(%rsp),%rbp
  lea    0x1c(%rsp),%r13
L30:
  lea    0x20(%rsp),%rax
  mov    %rax,0x8(%rsp)
  jmp    L122
L110:
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <FD_SET@plt>
  jmp    L123
L118:
  mov    %r13,%rsi
  mov    %r14d,%edi
  call   <FD_SET@plt>
  jmp    L90
L129:
  mov    %r13,%rsi
  mov    %r14d,%edi
  call   <FD_ISSET@plt>
  test   %rax,%rax
  jne    L114
L122:
  cmp    $0xff,%ebx
  jle    L110
L123:
  test   %ebx,%ebx
  jne    L118
L90:
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %r15d,%edi
  call   <select@plt>
  test   %eax,%eax
  js     L124
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <FD_ISSET@plt>
  test   %rax,%rax
  je     L129
  mov    $0x100,%edx
  sub    %ebx,%edx
  movslq %ebx,%rsi
  add    0x8(%rsp),%rsi
  mov    %r12d,%edi
  call   <read@plt>
  test   %eax,%eax
  jle    L124
  add    %eax,%ebx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <FD_CLR@plt>
  jmp    L129
L114:
  lea    0x20(%rsp),%rsi
  mov    %ebx,%edx
  mov    %r14d,%edi
  call   <write@plt>
  test   %eax,%eax
  jle    L124
  sub    %eax,%ebx
  lea    0x20(%rsp),%rdi
  cltq
  lea    (%rdi,%rax,1),%rsi
  mov    %ebx,%edx
  call   <memmove@plt>
  mov    %r13,%rsi
  mov    %r14d,%edi
  call   <FD_CLR@plt>
  jmp    L30
L124:
  mov    0x128(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  mov    $0x0,%eax
  add    $0x138,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L34:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int pipedata(int fdr, int fdw) {
  char ch;
  char buf[256];
  int i = 0, retv, n;
  fd_set fsr, fsw;
  FD_ZERO(&fsr);
  FD_ZERO(&fsw);
  while (1) {
    if (i < 256)
      FD_SET(fdr, &fsr);
    if (i)
      FD_SET(fdw, &fsw);
    retv = select((fdr > fdw) ? (1 + fdr) : (1 + fdw), &fsr, &fsw, NULL, NULL);
    if (retv < 0)
      return;
    if (FD_ISSET(fdr, &fsr)) {
      n = read(fdr, buf + i, sizeof(buf) - i);
      if (n <= 0)
        return;
      i += n;
      FD_CLR(fdr, &fsr);
    }
    if (FD_ISSET(fdw, &fsw)) {
      n = write(fdw, buf, i);
      if (n <= 0)
        return;
      i -= n;
      memmove(buf, buf + n, i);
      FD_CLR(fdw, &fsw);
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<pipedata>:
  endbr64
  push   %r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  lea    0x1(%r14),%r15d
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  push   %rbx
  sub    $0x138,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x128(%rsp)
  xor    %eax,%eax
  lea    0x18(%rsp),%rbp
  lea    0x1c(%rsp),%r13
  mov    %rbp,%rdi
  call   <FD_ZERO@plt>
  mov    %r13,%rdi
  call   <FD_ZERO@plt>
  cmp    %r14d,%r12d
  lea    0x1(%r12),%eax
  cmovg  %eax,%r15d
  xor    %ebx,%ebx
  jmp    L12
  nopl   0x0(%rax)
L53:
  mov    %r13,%rsi
  mov    %r14d,%edi
  call   <FD_SET@plt>
L33:
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %r15d,%edi
  call   <select@plt>
  test   %eax,%eax
  js     L49
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <FD_ISSET@plt>
  test   %rax,%rax
  je     L44
  mov    $0x100,%edx
  movslq %ebx,%rax
  mov    %r12d,%edi
  sub    %ebx,%edx
  lea    0x20(%rsp,%rax,1),%rsi
  call   <read@plt>
  test   %eax,%eax
  jle    L49
  mov    %rbp,%rsi
  mov    %r12d,%edi
  add    %eax,%ebx
  call   <FD_CLR@plt>
L44:
  mov    %r13,%rsi
  mov    %r14d,%edi
  call   <FD_ISSET@plt>
  test   %rax,%rax
  jne    L40
L12:
  cmp    $0xff,%ebx
  jg     L53
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <FD_SET@plt>
  test   %ebx,%ebx
  je     L33
  jmp    L53
  cs nopw 0x0(%rax,%rax,1)
L49:
  mov    0x128(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L70
  add    $0x138,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L40:
  lea    0x20(%rsp),%r8
  mov    %ebx,%edx
  mov    %r14d,%edi
  mov    %r8,%rsi
  mov    %r8,0x8(%rsp)
  call   <write@plt>
  test   %eax,%eax
  jle    L49
  mov    0x8(%rsp),%r8
  sub    %eax,%ebx
  cltq
  mov    %ebx,%edx
  lea    (%r8,%rax,1),%rsi
  mov    %r8,%rdi
  call   <memmove@plt>
  mov    %r13,%rsi
  mov    %r14d,%edi
  call   <FD_CLR@plt>
  jmp    L12
L70:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int pipedata(int fdr, int fdw) {
  char ch;
  char buf[256];
  int i = 0, retv, n;
  fd_set fsr, fsw;
  FD_ZERO(&fsr);
  FD_ZERO(&fsw);
  while (1) {
    if (i < 256)
      FD_SET(fdr, &fsr);
    if (i)
      FD_SET(fdw, &fsw);
    retv = select((fdr > fdw) ? (1 + fdr) : (1 + fdw), &fsr, &fsw, NULL, NULL);
    if (retv < 0)
      return;
    if (FD_ISSET(fdr, &fsr)) {
      n = read(fdr, buf + i, sizeof(buf) - i);
      if (n <= 0)
        return;
      i += n;
      FD_CLR(fdr, &fsr);
    }
    if (FD_ISSET(fdw, &fsw)) {
      n = write(fdw, buf, i);
      if (n <= 0)
        return;
      i -= n;
      memmove(buf, buf + n, i);
      FD_CLR(fdw, &fsw);
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<pipedata>:
  endbr64
  push   %r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  lea    0x1(%r14),%r15d
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  push   %rbx
  sub    $0x138,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x128(%rsp)
  xor    %eax,%eax
  lea    0x18(%rsp),%rbp
  lea    0x1c(%rsp),%r13
  mov    %rbp,%rdi
  call   <FD_ZERO@plt>
  mov    %r13,%rdi
  call   <FD_ZERO@plt>
  cmp    %r14d,%r12d
  lea    0x1(%r12),%eax
  cmovg  %eax,%r15d
  xor    %ebx,%ebx
  jmp    L89
  nopl   0x0(%rax)
L18:
  mov    %r13,%rsi
  mov    %r14d,%edi
  call   <FD_SET@plt>
L101:
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %r15d,%edi
  call   <select@plt>
  test   %eax,%eax
  js     L105
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <FD_ISSET@plt>
  test   %rax,%rax
  je     L100
  mov    $0x100,%edx
  movslq %ebx,%rax
  mov    %r12d,%edi
  sub    %ebx,%edx
  lea    0x20(%rsp,%rax,1),%rsi
  call   <read@plt>
  test   %eax,%eax
  jle    L105
  mov    %rbp,%rsi
  mov    %r12d,%edi
  add    %eax,%ebx
  call   <FD_CLR@plt>
L100:
  mov    %r13,%rsi
  mov    %r14d,%edi
  call   <FD_ISSET@plt>
  test   %rax,%rax
  jne    L87
L89:
  cmp    $0xff,%ebx
  jg     L18
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <FD_SET@plt>
  test   %ebx,%ebx
  je     L101
  jmp    L18
  cs nopw 0x0(%rax,%rax,1)
L105:
  mov    0x128(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L53
  add    $0x138,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L87:
  lea    0x20(%rsp),%r8
  mov    %ebx,%edx
  mov    %r14d,%edi
  mov    %r8,%rsi
  mov    %r8,0x8(%rsp)
  call   <write@plt>
  test   %eax,%eax
  jle    L105
  mov    0x8(%rsp),%r8
  sub    %eax,%ebx
  cltq
  mov    %ebx,%edx
  lea    (%r8,%rax,1),%rsi
  mov    %r8,%rdi
  call   <memmove@plt>
  mov    %r13,%rsi
  mov    %r14d,%edi
  call   <FD_CLR@plt>
  jmp    L89
L53:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int pipedata(int fdr, int fdw) {
  char ch;
  char buf[256];
  int i = 0, retv, n;
  fd_set fsr, fsw;
  FD_ZERO(&fsr);
  FD_ZERO(&fsw);
  while (1) {
    if (i < 256)
      FD_SET(fdr, &fsr);
    if (i)
      FD_SET(fdw, &fsw);
    retv = select((fdr > fdw) ? (1 + fdr) : (1 + fdw), &fsr, &fsw, NULL, NULL);
    if (retv < 0)
      return;
    if (FD_ISSET(fdr, &fsr)) {
      n = read(fdr, buf + i, sizeof(buf) - i);
      if (n <= 0)
        return;
      i += n;
      FD_CLR(fdr, &fsr);
    }
    if (FD_ISSET(fdw, &fsw)) {
      n = write(fdw, buf, i);
      if (n <= 0)
        return;
      i -= n;
      memmove(buf, buf + n, i);
      FD_CLR(fdw, &fsw);
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x10(%rbp)
  movss  D2(%rip),%xmm0
  movss  %xmm0,-0xc(%rbp)
  movss  D24(%rip),%xmm0
  movss  %xmm0,-0x8(%rbp)
  movss  -0x10(%rbp),%xmm0
  movss  %xmm0,-0x14(%rbp)
  jmp    L17
L12:
  pxor   %xmm0,%xmm0
  cvtss2sd -0x14(%rbp),%xmm0
  movsd  D39(%rip),%xmm2
  movapd %xmm0,%xmm1
  subsd  %xmm2,%xmm1
  movsd  D18(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,-0x4(%rbp)
  movss  -0x4(%rbp),%xmm0
  mov    -0x14(%rbp),%eax
  movaps %xmm0,%xmm1
  movd   %eax,%xmm0
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movss  -0x14(%rbp),%xmm0
  addss  -0x8(%rbp),%xmm0
  movss  %xmm0,-0x14(%rbp)
L17:
  movss  -0xc(%rbp),%xmm0
  comiss -0x14(%rbp),%xmm0
  jae    L12
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D2
type: byte[4]
name: unpack_data
arguments: data_label: D24
type: byte[4]
name: unpack_data
arguments: data_label: D39
type: f64
name: unpack_data
arguments: data_label: D18
type: byte[8]
name: unpack_data
arguments: data_label: D15
type: string

tool: {"D2": [0x00,0x00,0x96,0x43]}

tool: {"D24": [0x00,0x00,0xa0,0x41]}

tool: D39: 32

tool: {"D18": [0x72,0x1c,0xc7,0x71,0x1c,0xc7,0xe1,0x3f]}

tool: D15: %3.0f %6.1f


assistant: ```c
int main(void) {
  float fahr, celsius;
  float lower, upper, step;
  lower = 0;
  upper = 300;
  step = 20;
  fahr = lower;
  while (fahr <= upper) {
    celsius = (5.0 / 9.0) * (fahr - 32.0);
    printf("%3.0f %6.1f\n", fahr, celsius);
    fahr = fahr + step;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    $0x10,%ebx
  pxor   %xmm4,%xmm4
  movss  %xmm4,0xc(%rsp)
  lea    D10(%rip),%rbp
L6:
  movss  0xc(%rsp),%xmm2
  pxor   %xmm1,%xmm1
  cvtss2sd %xmm2,%xmm1
  subsd  D26(%rip),%xmm1
  mulsd  D0(%rip),%xmm1
  cvtsd2ss %xmm1,%xmm1
  movaps %xmm2,%xmm0
  mov    %rbp,%rdi
  call   <printf@plt>
  movss  D1(%rip),%xmm3
  addss  0xc(%rsp),%xmm3
  movss  %xmm3,0xc(%rsp)
  sub    $0x1,%ebx
  jne    L6
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_label: D10
datatype: string
name: read_data
arguments: data_label: D26
datatype: f64
name: read_data
arguments: data_label: D0
datatype: byte[8]
name: read_data
arguments: data_label: D1
datatype: byte[4]

tool: D10: %3.0f %6.1f


tool: D26: 32

tool: {"D0": [0x72,0x1c,0xc7,0x71,0x1c,0xc7,0xe1,0x3f]}

tool: {"D1": [0x00,0x00,0xa0,0x41]}

assistant: ```c
int main(void) {
  float fahr, celsius;
  float lower, upper, step;
  lower = 0;
  upper = 300;
  step = 20;
  fahr = lower;
  while (fahr <= upper) {
    celsius = (5.0 / 9.0) * (fahr - 32.0);
    printf("%3.0f %6.1f\n", fahr, celsius);
    fahr = fahr + step;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  pxor   %xmm0,%xmm0
  lea    D14(%rip),%rbp
  push   %rbx
  mov    $0x10,%ebx
  sub    $0x18,%rsp
  nopw   0x0(%rax,%rax,1)
L4:
  pxor   %xmm1,%xmm1
  mov    %rbp,%rdi
  movss  %xmm0,0xc(%rsp)
  cvtss2sd %xmm0,%xmm1
  subsd  D38(%rip),%xmm1
  mulsd  D22(%rip),%xmm1
  cvtsd2ss %xmm1,%xmm1
  call   <printf@plt>
  movss  0xc(%rsp),%xmm0
  addss  D2(%rip),%xmm0
  sub    $0x1,%ebx
  jne    L4
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D14
datatype: string
name: retrieve_data
arguments: tag: D38
datatype: f64
name: retrieve_data
arguments: tag: D22
datatype: byte[8]
name: retrieve_data
arguments: tag: D2
datatype: byte[4]

tool: D14: %3.0f %6.1f


tool: D38: 32

tool: {"D22": [0x72,0x1c,0xc7,0x71,0x1c,0xc7,0xe1,0x3f]}

tool: {"D2": [0x00,0x00,0xa0,0x41]}

assistant: ```c
int main(void) {
  float fahr, celsius;
  float lower, upper, step;
  lower = 0;
  upper = 300;
  step = 20;
  fahr = lower;
  while (fahr <= upper) {
    celsius = (5.0 / 9.0) * (fahr - 32.0);
    printf("%3.0f %6.1f\n", fahr, celsius);
    fahr = fahr + step;
  }
  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
  pxor   %xmm0,%xmm0
  lea    D1(%rip),%rbp
  push   %rbx
  mov    $0x10,%ebx
  sub    $0x18,%rsp
  nopw   0x0(%rax,%rax,1)
L5:
  pxor   %xmm1,%xmm1
  mov    %rbp,%rdi
  movss  %xmm0,0xc(%rsp)
  cvtss2sd %xmm0,%xmm1
  subsd  D39(%rip),%xmm1
  mulsd  D16(%rip),%xmm1
  cvtsd2ss %xmm1,%xmm1
  call   <printf@plt>
  movss  0xc(%rsp),%xmm0
  addss  D23(%rip),%xmm0
  sub    $0x1,%ebx
  jne    L5
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: label: D1
datatype: string
name: parse_data
arguments: label: D39
datatype: f64
name: parse_data
arguments: label: D16
datatype: byte[8]
name: parse_data
arguments: label: D23
datatype: byte[4]

tool: D1: %3.0f %6.1f


tool: D39: 32

tool: {"D16": [0x72,0x1c,0xc7,0x71,0x1c,0xc7,0xe1,0x3f]}

tool: {"D23": [0x00,0x00,0xa0,0x41]}

assistant: ```c
int main(void) {
  float fahr, celsius;
  float lower, upper, step;
  lower = 0;
  upper = 300;
  step = 20;
  fahr = lower;
  while (fahr <= upper) {
    celsius = (5.0 / 9.0) * (fahr - 32.0);
    printf("%3.0f %6.1f\n", fahr, celsius);
    fahr = fahr + step;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<emit_set_gz32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <emit_cmpimm@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  mov    $0x1,%edi
  call   <emit_movimm@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  mov    $0x0,%edi
  call   <emit_cmovl_imm@plt>
  nop
  leave
  ret
```

assistant: ```c
void emit_set_gz32(int rs, int rt) {
  emit_cmpimm(rs, 1);
  emit_movimm(1, rt);
  emit_cmovl_imm(0, rt);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<emit_set_gz32>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  mov    $0x1,%esi
  call   <emit_cmpimm@plt>
  mov    %ebx,%esi
  mov    $0x1,%edi
  call   <emit_movimm@plt>
  mov    %ebx,%esi
  mov    $0x0,%edi
  call   <emit_cmovl_imm@plt>
  pop    %rbx
  ret
```

assistant: ```c
void emit_set_gz32(int rs, int rt) {
  emit_cmpimm(rs, 1);
  emit_movimm(1, rt);
  emit_cmovl_imm(0, rt);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<emit_set_gz32>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  mov    $0x1,%esi
  call   <emit_cmpimm@plt>
  mov    %ebp,%esi
  mov    $0x1,%edi
  call   <emit_movimm@plt>
  mov    %ebp,%esi
  xor    %edi,%edi
  pop    %rbp
  jmp    1080 <emit_cmovl_imm@plt>
```

assistant: ```c
void emit_set_gz32(int rs, int rt) {
  emit_cmpimm(rs, 1);
  emit_movimm(1, rt);
  emit_cmovl_imm(0, rt);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<emit_set_gz32>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  mov    $0x1,%esi
  call   <emit_cmpimm@plt>
  mov    %ebp,%esi
  mov    $0x1,%edi
  call   <emit_movimm@plt>
  mov    %ebp,%esi
  xor    %edi,%edi
  pop    %rbp
  jmp    1080 <emit_cmovl_imm@plt>
```

assistant: ```c
void emit_set_gz32(int rs, int rt) {
  emit_cmpimm(rs, 1);
  emit_movimm(1, rt);
  emit_cmovl_imm(0, rt);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x2,%edi
  call   <square_unsigned@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x4,-0x4(%rbp)
  je     L15
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  jmp    L13
L15:
  mov    $0x0,%eax
L13:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D2
type_infered: string

tool: D2: Got %u instead of 4


assistant: ```c
int main(void) {
  unsigned int ret = square_unsigned(2);
  if (ret != 4) {
    printf("Got %u instead of 4\n", ret);
    return 1;
  }
  return 0;
}
```