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

```asm
<gotcha_mmap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  mov    %r8d,-0x2c(%rbp)
  mov    %r9d,-0x30(%rbp)
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  mov    -0x2c(%rbp),%r9d
  mov    -0x28(%rbp),%r8d
  mov    -0x24(%rbp),%ecx
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rsi
  sub    $0x8,%rsp
  mov    -0x30(%rbp),%edi
  push   %rdi
  mov    %eax,%edi
  call   <syscall@plt>
  add    $0x10,%rsp
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *gotcha_mmap(void *addr, size_t length, int prot, int flags, int fd, off_t offset) {
  long result;
  result = syscall(SYS_mmap, addr, length, prot, flags, fd, offset);
  return (void *)result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<gotcha_mmap>:
  endbr64
  sub    $0x10,%rsp
  push   %r9
  mov    %r8d,%r9d
  mov    %ecx,%r8d
  mov    %edx,%ecx
  mov    %rsi,%rdx
  mov    %rdi,%rsi
  mov    D8(%rip),%rax
  mov    (%rax),%edi
  call   <syscall@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
void *gotcha_mmap(void *addr, size_t length, int prot, int flags, int fd, off_t offset) {
  long result;
  result = syscall(SYS_mmap, addr, length, prot, flags, fd, offset);
  return (void *)result;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<gotcha_mmap>:
  endbr64
  push   %r12
  mov    D3(%rip),%rax
  sub    $0x8,%rsp
  push   %r9
  mov    %r8d,%r9d
  mov    %ecx,%r8d
  mov    %edx,%ecx
  mov    %rsi,%rdx
  mov    %rdi,%rsi
  mov    (%rax),%edi
  call   <syscall@plt>
  pop    %rdx
  pop    %rcx
  pop    %r12
  ret
```

assistant: ```c
void *gotcha_mmap(void *addr, size_t length, int prot, int flags, int fd, off_t offset) {
  long result;
  result = syscall(SYS_mmap, addr, length, prot, flags, fd, offset);
  return (void *)result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<gotcha_mmap>:
  endbr64
  push   %r12
  mov    D9(%rip),%rax
  sub    $0x8,%rsp
  push   %r9
  mov    %r8d,%r9d
  mov    %ecx,%r8d
  mov    %edx,%ecx
  mov    %rsi,%rdx
  mov    %rdi,%rsi
  mov    (%rax),%edi
  call   <syscall@plt>
  pop    %rdx
  pop    %rcx
  pop    %r12
  ret
```

assistant: ```c
void *gotcha_mmap(void *addr, size_t length, int prot, int flags, int fd, off_t offset) {
  long result;
  result = syscall(SYS_mmap, addr, length, prot, flags, fd, offset);
  return (void *)result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L1
L10:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L1:
  cmpl   $0x13,-0x4(%rbp)
  jle    L10
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D1
datatype: string

tool: D1: i: %d rand(): %d


assistant: ```c
int main() {
  int i;
  srand(time(NULL));
  for (i = 0; i < 20; i++)
    printf("i: %d rand(): %d\n", i, rand());
  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    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%ebx
  lea    D2(%rip),%rbp
L9:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0x14,%ebx
  jne    L9
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D2
type_infered: string

tool: D2: i: %d rand(): %d


assistant: ```c
int main() {
  int i;
  srand(time(NULL));
  for (i = 0; i < 20; i++)
    printf("i: %d rand(): %d\n", i, rand());
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  xor    %edi,%edi
  lea    D7(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  nopl   0x0(%rax)
L8:
  xor    %eax,%eax
  call   <rand@plt>
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  mov    %eax,%edx
  call   <printf@plt>
  cmp    $0x14,%ebx
  jne    L8
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D7
datatype: string

tool: D7: i: %d rand(): %d


assistant: ```c
int main() {
  int i;
  srand(time(NULL));
  for (i = 0; i < 20; i++)
    printf("i: %d rand(): %d\n", i, rand());
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  xor    %edi,%edi
  lea    D4(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  nopl   0x0(%rax)
L7:
  xor    %eax,%eax
  call   <rand@plt>
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  mov    %eax,%edx
  call   <printf@plt>
  cmp    $0x14,%ebx
  jne    L7
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D4
datatype: string

tool: D4: i: %d rand(): %d


assistant: ```c
int main() {
  int i;
  srand(time(NULL));
  for (i = 0; i < 20; i++)
    printf("i: %d rand(): %d\n", i, rand());
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<solcon_number_of_standard_solutions>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    -0x8(%rbp),%rcx
  mov    -0x28(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    $0x0,%r8d
  mov    %rax,%rsi
  mov    $0x20,%edi
  call   <_ada_use_c2phc4c@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  leave
  ret
```

assistant: ```c
int solcon_number_of_standard_solutions(int *length) {
  int *a, fail;
  double *c;
  fail = _ada_use_c2phc4c(32, a, length, c, 0);
  return fail;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<solcon_number_of_standard_solutions>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rdx
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    $0x0,%esi
  mov    $0x20,%edi
  call   <_ada_use_c2phc4c@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int solcon_number_of_standard_solutions(int *length) {
  int *a, fail;
  double *c;
  fail = _ada_use_c2phc4c(32, a, length, c, 0);
  return fail;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<solcon_number_of_standard_solutions>:
  endbr64
  mov    %rdi,%rdx
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  xor    %esi,%esi
  mov    $0x20,%edi
  jmp    1050 <_ada_use_c2phc4c@plt>
```

assistant: ```c
int solcon_number_of_standard_solutions(int *length) {
  int *a, fail;
  double *c;
  fail = _ada_use_c2phc4c(32, a, length, c, 0);
  return fail;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<solcon_number_of_standard_solutions>:
  endbr64
  mov    %rdi,%rdx
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  xor    %esi,%esi
  mov    $0x20,%edi
  jmp    1050 <_ada_use_c2phc4c@plt>
```

assistant: ```c
int solcon_number_of_standard_solutions(int *length) {
  int *a, fail;
  double *c;
  fail = _ada_use_c2phc4c(32, a, length, c, 0);
  return fail;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<multiply_array_number>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movsd  %xmm0,-0x28(%rbp)
  mov    %edx,-0x2c(%rbp)
  mov    %ecx,-0x30(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L29
L16:
  movl   $0x0,-0x4(%rbp)
  jmp    L33
L19:
  mov    -0x8(%rbp),%eax
  imul   -0x30(%rbp),%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x8(%rbp),%eax
  imul   -0x30(%rbp),%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mulsd  -0x28(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L33:
  mov    -0x4(%rbp),%eax
  cmp    -0x30(%rbp),%eax
  jl     L19
  addl   $0x1,-0x8(%rbp)
L29:
  mov    -0x8(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L16
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void multiply_array_number(double *out, double *a, double f, int ncol, int nlns) {
  int i, j;
  for (i = 0; i < ncol; ++i)
    for (j = 0; j < nlns; ++j)
      (out[(i) * (nlns) + (j)]) = (a[(i) * (nlns) + (j)]) * f;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<multiply_array_number>:
  endbr64
  mov    %rsi,%r8
  test   %edx,%edx
  jle    L29
  mov    $0x0,%r10d
  mov    $0x0,%r9d
  mov    %ecx,%r11d
  jmp    L38
L8:
  movslq %r10d,%rsi
  lea    0x0(,%rsi,8),%rax
  add    %r11,%rsi
  shl    $0x3,%rsi
L56:
  movapd %xmm0,%xmm1
  mulsd  (%r8,%rax,1),%xmm1
  movsd  %xmm1,(%rdi,%rax,1)
  add    $0x8,%rax
  cmp    %rsi,%rax
  jne    L56
L37:
  add    $0x1,%r9d
  add    %ecx,%r10d
  cmp    %r9d,%edx
  je     L29
L38:
  test   %ecx,%ecx
  jg     L8
  jmp    L37
L29:
  ret
```

assistant: ```c
void multiply_array_number(double *out, double *a, double f, int ncol, int nlns) {
  int i, j;
  for (i = 0; i < ncol; ++i)
    for (j = 0; j < nlns; ++j)
      (out[(i) * (nlns) + (j)]) = (a[(i) * (nlns) + (j)]) * f;
}
```
user: What is the c source code of the assembly code below?

```asm
<multiply_array_number>:
  endbr64
  mov    %edx,%r10d
  test   %edx,%edx
  jle    L37
  lea    -0x1(%rcx),%r11d
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  add    $0x1,%r11
  nopl   0x0(%rax)
L3:
  test   %ecx,%ecx
  jle    L18
  movslq %r9d,%rdx
  lea    0x0(,%rdx,8),%rax
  add    %r11,%rdx
  shl    $0x3,%rdx
  cs nopw 0x0(%rax,%rax,1)
L17:
  movsd  (%rsi,%rax,1),%xmm1
  mulsd  %xmm0,%xmm1
  movsd  %xmm1,(%rdi,%rax,1)
  add    $0x8,%rax
  cmp    %rax,%rdx
  jne    L17
L18:
  add    $0x1,%r8d
  add    %ecx,%r9d
  cmp    %r8d,%r10d
  jne    L3
L37:
  ret
```

assistant: ```c
void multiply_array_number(double *out, double *a, double f, int ncol, int nlns) {
  int i, j;
  for (i = 0; i < ncol; ++i)
    for (j = 0; j < nlns; ++j)
      (out[(i) * (nlns) + (j)]) = (a[(i) * (nlns) + (j)]) * f;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<multiply_array_number>:
  endbr64
  test   %edx,%edx
  jle    L76
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %ecx,%ebp
  push   %rbx
  test   %ecx,%ecx
  jle    L28
  mov    %ecx,%eax
  movslq %ecx,%r12
  mov    %ecx,%r8d
  mov    %edx,%r14d
  shr    %eax
  movapd %xmm0,%xmm2
  shl    $0x3,%r12
  lea    0x8(%rsi),%r11
  lea    -0x1(%rax),%r10d
  lea    -0x1(%rcx),%eax
  mov    %rdi,%rdx
  and    $0xfffffffe,%r8d
  mov    %rax,%r15
  shl    $0x3,%rax
  add    $0x1,%r10
  xor    %ecx,%ecx
  sub    %rsi,%rax
  shl    $0x4,%r10
  xor    %ebx,%ebx
  unpcklpd %xmm2,%xmm2
  mov    %rax,-0x18(%rsp)
  mov    $0xfffffffffffffff8,%rax
  sub    %rsi,%rax
  mov    %rax,-0x10(%rsp)
  nopl   (%rax)
L88:
  cmp    %r11,%rdx
  je     L9
  cmp    $0x1,%r15d
  lea    -0x8(%r11),%r9
  seta   %r13b
  xor    %eax,%eax
  test   %r13b,%r13b
  je     L9
L0:
  movupd (%r9,%rax,1),%xmm1
  mulpd  %xmm2,%xmm1
  movups %xmm1,(%rdx,%rax,1)
  add    $0x10,%rax
  cmp    %r10,%rax
  jne    L0
  cmp    %ebp,%r8d
  je     L65
  lea    (%rcx,%r8,1),%eax
  cltq
  movsd  (%rsi,%rax,8),%xmm1
  mulsd  %xmm0,%xmm1
  movsd  %xmm1,(%rdi,%rax,8)
L65:
  add    $0x1,%ebx
  add    %r12,%r11
  add    %r12,%rdx
  add    %ebp,%ecx
  cmp    %ebx,%r14d
  jne    L88
L28:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  mov    -0x10(%rsp),%rax
  mov    -0x18(%rsp),%r9
  add    %r11,%rax
  add    %r11,%r9
L16:
  movsd  (%rsi,%rax,1),%xmm1
  mulsd  %xmm0,%xmm1
  movsd  %xmm1,(%rdi,%rax,1)
  add    $0x8,%rax
  cmp    %r9,%rax
  jne    L16
  jmp    L65
L76:
  ret
```

assistant: ```c
void multiply_array_number(double *out, double *a, double f, int ncol, int nlns) {
  int i, j;
  for (i = 0; i < ncol; ++i)
    for (j = 0; j < nlns; ++j)
      (out[(i) * (nlns) + (j)]) = (a[(i) * (nlns) + (j)]) * f;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<getbranchno>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <countnumflds@plt>
  mov    %eax,%r13d
  mov    %r13d,%eax
  sar    $0x1f,%eax
  shr    $0x1f,%eax
  mov    %eax,%edx
  lea    0x0(%r13,%rdx,1),%eax
  and    $0x1,%eax
  sub    %edx,%eax
  cmp    $0x1,%eax
  jne    L14
  mov    -0x28(%rbp),%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  jmp    L15
L14:
  mov    -0x28(%rbp),%rbx
  mov    -0x30(%rbp),%r12
  mov    $0x1,%r14d
  jmp    L33
L28:
  mov    %rbx,%rdx
  lea    0x1(%rdx),%rbx
  mov    %r12,%rax
  lea    0x1(%rax),%r12
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L22:
  movzbl (%rbx),%eax
  cmp    $0x2e,%al
  jne    L28
  mov    %rbx,%rdx
  lea    0x1(%rdx),%rbx
  mov    %r12,%rax
  lea    0x1(%rax),%r12
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  add    $0x1,%r14d
L33:
  cmp    %r13d,%r14d
  jl     L22
  lea    -0x1(%r12),%rax
  movb   $0x0,(%rax)
L15:
  nop
  add    $0x10,%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
```

assistant: ```c
getbranchno(revno, branchno) char *revno, *branchno;
{
  register int i, numflds;
  register char *tp, *sp;
  numflds = countnumflds(revno);
  if (numflds % 2 == 1)
    strcpy(branchno, revno);
  else {
    sp = revno;
    tp = branchno;
    for (i = 1; i < numflds; i++) {
      while (*sp != '.')
        *tp++ = *sp++;
      *tp++ = *sp++;
    }
    *(tp - 1) = '\0';
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<getbranchno>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    $0x0,%eax
  call   <countnumflds@plt>
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  lea    (%rax,%rcx,1),%edx
  and    $0x1,%edx
  sub    %ecx,%edx
  cmp    $0x1,%edx
  je     L24
  mov    $0x1,%ecx
  mov    %rbp,%rdx
  cmp    $0x1,%eax
  jg     L8
L5:
  movb   $0x0,-0x1(%rdx)
L42:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L24:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  jmp    L42
L61:
  add    $0x1,%rbx
  add    $0x1,%rbp
  mov    %dl,-0x1(%rbp)
  movzbl (%rbx),%edx
  cmp    $0x2e,%dl
  jne    L61
L69:
  add    $0x1,%rbx
  lea    0x1(%rbp),%rdx
  movb   $0x2e,0x0(%rbp)
  add    $0x1,%ecx
  cmp    %ecx,%eax
  je     L5
  mov    %rdx,%rbp
L8:
  movzbl (%rbx),%edx
  cmp    $0x2e,%dl
  jne    L61
  jmp    L69
```

assistant: ```c
getbranchno(revno, branchno) char *revno, *branchno;
{
  register int i, numflds;
  register char *tp, *sp;
  numflds = countnumflds(revno);
  if (numflds % 2 == 1)
    strcpy(branchno, revno);
  else {
    sp = revno;
    tp = branchno;
    for (i = 1; i < numflds; i++) {
      while (*sp != '.')
        *tp++ = *sp++;
      *tp++ = *sp++;
    }
    *(tp - 1) = '\0';
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getbranchno>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <countnumflds@plt>
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  lea    (%rax,%rcx,1),%edx
  and    $0x1,%edx
  sub    %ecx,%edx
  cmp    $0x1,%edx
  je     L19
  mov    $0x1,%ecx
  mov    %r12,%rdx
  cmp    $0x1,%eax
  jg     L18
  jmp    L7
  nopl   0x0(%rax)
L32:
  mov    %dl,(%r12)
  add    $0x1,%r12
L18:
  movzbl 0x0(%rbp),%edx
  add    $0x1,%rbp
  cmp    $0x2e,%dl
  jne    L32
  add    $0x1,%ecx
  movb   $0x2e,(%r12)
  lea    0x1(%r12),%rdx
  cmp    %ecx,%eax
  je     L7
  mov    %rdx,%r12
  jmp    L18
  nopl   0x0(%rax,%rax,1)
L7:
  movb   $0x0,-0x1(%rdx)
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
L19:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
getbranchno(revno, branchno) char *revno, *branchno;
{
  register int i, numflds;
  register char *tp, *sp;
  numflds = countnumflds(revno);
  if (numflds % 2 == 1)
    strcpy(branchno, revno);
  else {
    sp = revno;
    tp = branchno;
    for (i = 1; i < numflds; i++) {
      while (*sp != '.')
        *tp++ = *sp++;
      *tp++ = *sp++;
    }
    *(tp - 1) = '\0';
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<getbranchno>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <countnumflds@plt>
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  lea    (%rax,%rcx,1),%edx
  and    $0x1,%edx
  sub    %ecx,%edx
  cmp    $0x1,%edx
  je     L47
  mov    $0x1,%ecx
  mov    %r12,%rdx
  cmp    $0x1,%eax
  jg     L10
  jmp    L26
  nopl   0x0(%rax)
L29:
  mov    %dl,(%r12)
  add    $0x1,%r12
L10:
  movzbl 0x0(%rbp),%edx
  add    $0x1,%rbp
  cmp    $0x2e,%dl
  jne    L29
  add    $0x1,%ecx
  movb   $0x2e,(%r12)
  lea    0x1(%r12),%rdx
  cmp    %ecx,%eax
  je     L26
  mov    %rdx,%r12
  jmp    L10
  nopl   0x0(%rax,%rax,1)
L26:
  movb   $0x0,-0x1(%rdx)
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
L47:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
getbranchno(revno, branchno) char *revno, *branchno;
{
  register int i, numflds;
  register char *tp, *sp;
  numflds = countnumflds(revno);
  if (numflds % 2 == 1)
    strcpy(branchno, revno);
  else {
    sp = revno;
    tp = branchno;
    for (i = 1; i < numflds; i++) {
      while (*sp != '.')
        *tp++ = *sp++;
      *tp++ = *sp++;
    }
    *(tp - 1) = '\0';
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<primoPosterior>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L12
L32:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <primo@plt>
  test   %rax,%rax
  je     L27
  mov    -0x4(%rbp),%eax
  jmp    L4
L27:
  addl   $0x1,-0x4(%rbp)
L12:
  mov    -0x4(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L32
  mov    $0xffffffff,%eax
L4:
  leave
  ret
```

assistant: ```c
int primoPosterior(int vetor[], int ind_primo, int tam_vetor) {
  int i;
  for (i = ind_primo + 1; i < tam_vetor; i++) {
    if (primo(vetor[i]))
      return i;
  }
  return -1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<primoPosterior>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    0x1(%rsi),%ebx
  cmp    %edx,%ebx
  jge    L38
  mov    %rdi,%rbp
  movslq %ebx,%rbx
  sub    $0x2,%edx
  sub    %esi,%edx
  movslq %esi,%rsi
  lea    0x2(%rdx,%rsi,1),%r13
L9:
  mov    %ebx,%r12d
  mov    0x0(%rbp,%rbx,4),%edi
  call   <primo@plt>
  test   %rax,%rax
  jne    L27
  add    $0x1,%rbx
  cmp    %r13,%rbx
  jne    L9
  mov    $0xffffffff,%r12d
L27:
  mov    %r12d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L38:
  mov    $0xffffffff,%r12d
  jmp    L27
```

assistant: ```c
int primoPosterior(int vetor[], int ind_primo, int tam_vetor) {
  int i;
  for (i = ind_primo + 1; i < tam_vetor; i++) {
    if (primo(vetor[i]))
      return i;
  }
  return -1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<primoPosterior>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    0x1(%rsi),%ebx
  sub    $0x8,%rsp
  cmp    %edx,%ebx
  jge    L34
  sub    $0x2,%edx
  mov    %rdi,%rbp
  movslq %ebx,%rbx
  sub    %esi,%edx
  movslq %esi,%rsi
  lea    0x2(%rdx,%rsi,1),%r13
  jmp    L26
  nopw   0x0(%rax,%rax,1)
L1:
  add    $0x1,%rbx
  cmp    %r13,%rbx
  je     L34
L26:
  mov    0x0(%rbp,%rbx,4),%edi
  mov    %ebx,%r12d
  call   <primo@plt>
  test   %rax,%rax
  je     L1
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L34:
  add    $0x8,%rsp
  mov    $0xffffffff,%r12d
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int primoPosterior(int vetor[], int ind_primo, int tam_vetor) {
  int i;
  for (i = ind_primo + 1; i < tam_vetor; i++) {
    if (primo(vetor[i]))
      return i;
  }
  return -1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<primoPosterior>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    0x1(%rsi),%ebx
  sub    $0x8,%rsp
  cmp    %edx,%ebx
  jge    L31
  sub    $0x2,%edx
  mov    %rdi,%rbp
  movslq %ebx,%rbx
  sub    %esi,%edx
  movslq %esi,%rsi
  lea    0x2(%rdx,%rsi,1),%r13
  jmp    L34
  nopw   0x0(%rax,%rax,1)
L22:
  add    $0x1,%rbx
  cmp    %r13,%rbx
  je     L31
L34:
  mov    0x0(%rbp,%rbx,4),%edi
  mov    %ebx,%r12d
  call   <primo@plt>
  test   %rax,%rax
  je     L22
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L31:
  add    $0x8,%rsp
  mov    $0xffffffff,%r12d
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int primoPosterior(int vetor[], int ind_primo, int tam_vetor) {
  int i;
  for (i = ind_primo + 1; i < tam_vetor; i++) {
    if (primo(vetor[i]))
      return i;
  }
  return -1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<show_usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  lea    D21(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <fprintf@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rcx
  mov    $0x7,%edx
  mov    $0x1,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <fwrite@plt>
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  lea    D16(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <fprintf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D21
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D29
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D16
type_guessed: string

tool: D21: %s [INFO] 

tool: D29: usage:


tool: D16: %s -k k --res r [--margin M] --iter1 n --iter2 m --acc a --fasta f --hic H --kmer c --out o [--pri p] [--sec s] [--verbose V] --thread_num t 


assistant: ```c
int show_usage(FILE *fp, const char *prog_name) {
  fprintf(fp, "%s [INFO] ", prog_name);
  fprintf(fp, "usage:\n");
  fprintf(fp, "%s -k k --res r [--margin M] --iter1 n --iter2 m --acc a --fasta f --hic H --kmer c --out o [--pri p] [--sec s] [--verbose V] --thread_num t \n", prog_name);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<show_usage>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %rsi,%rdx
  lea    D24(%rip),%rsi
  mov    $0x0,%eax
  call   <fprintf@plt>
  mov    %rbx,%rcx
  mov    $0x7,%edx
  mov    $0x1,%esi
  lea    D8(%rip),%rdi
  call   <fwrite@plt>
  mov    %rbp,%rdx
  lea    D17(%rip),%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <fprintf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D24
type: string
name: extract_symbolic_data
arguments: data_tag: D8
type: string
name: extract_symbolic_data
arguments: data_tag: D17
type: string

tool: D24: %s [INFO] 

tool: D8: usage:


tool: D17: %s -k k --res r [--margin M] --iter1 n --iter2 m --acc a --fasta f --hic H --kmer c --out o [--pri p] [--sec s] [--verbose V] --thread_num t 


assistant: ```c
int show_usage(FILE *fp, const char *prog_name) {
  fprintf(fp, "%s [INFO] ", prog_name);
  fprintf(fp, "usage:\n");
  fprintf(fp, "%s -k k --res r [--margin M] --iter1 n --iter2 m --acc a --fasta f --hic H --kmer c --out o [--pri p] [--sec s] [--verbose V] --thread_num t \n", prog_name);
  return 0;
}
```