Question

我有一个int变量和一个int指针变量，我想将它们传递给一个可以将两个变量都设置为等于数字的函数。我的理解是，我只需要更改setint(b, 20);代码行即可解决此问题

我一直试图在&变量前添加*和b，但是这导致文件无法编译。

void setint(int* ip, int i)
{
  *ip = i;
}

int main(int argc, char** argv)
{
  int a = -1;
  int *b = NULL;

  setint(&a, 10);
  cout << a << endl;
  setint(b, 20);
  cout << b << endl;

此代码的结果应输出：

10
20

当前输出为：

10
segment fault

Answer 1

您的代码执行/尝试/失败的操作（请参阅添加的注释）：

int main(int argc, char** argv)
{
  int a = -1;    // define and init an int variable, fine
  int *b = NULL; // define and init a pointer to int, albeit to NULL, ...
                 // ... not immediatly a problem

  setint(&a, 10);    // give the address of variable to your function, fine
  cout << a << endl; // output, fine

  setint(b, 20);  // give the NULL pointer to the function, 
                  // which attempts to dereference it, segfault

  cout << b << endl;

什么可以实现您的预期（至少可以实现我认为的目标...）：

int main(int argc, char** argv)
{
  int a = -1;
  int *b = &a; // changed to init with address of existing variable

  setint(&a, 10);
  cout << a << endl;
  setint(b, 20);      // now gives dereferencable address of existing variable
  cout << *b << endl; // output what is pointed to by the non-NULL pointer

顺便说一句，如果之后再次输出a，它将显示通过指针设置的值，即20，它将覆盖先前写入的值10。

Answer 2

void setint(int* ip, int i)
{
  if(ip != NULL) //< we should check for null to avoid segfault
  {
    *ip = i;
  }
  else
  {
    cout << "!!attempt to set value via NULL pointer!!" << endl;
  }
}

int main(int argc, char** argv)
{
  int a = -1;
  int *b = NULL;

  setint(&a, 10);
  cout << a << endl;
  setint(b, 20);      //< should fail (b is currently NULL)
  // cout << *b << endl;

  b = &a; //< set pointer value to something other than NULL

  setint(b, 20);      //< should work
  cout << *b << endl;

  return 0;
}

Answer 3

您需要使用以下方式为b（编译器已在堆栈上分配a）分配内存：

#include <stdlib.h>
int *b = (int *)malloc(sizeof(int));

如何将int和int *传递到函数中以定义变量

3 个答案: