Question

我正在将C代码翻译成另一种语言。我对指针很困惑。说我有这个代码。函数foo，调用函数bob，它们都是指针。

double
*foo(double *x, double *init, double *a){
   double *y = (double*)malloc(5*sizeof(double));
   double *z = (double*)malloc(5*sizeof(double));
   double *sum, *update;

   sum = (double*)bob(y, z)    //<---Q1: why y and z don't need stars in front of them? 
                               //I thought they are pointers?

   for (i<0; i<5; i++){
       z[i]=y[i]               //Q2: howcome it's ok to assign y to z? 
   }                           //aren't they pointer?(i.e.hold memory address)
}

double
*bob(*arg1, *arg2){
   something...
}

所以，
1）为什么y和z不需要在它们前面的星星，是不是y和z只是地址？
2）为什么sum没有星星，我认为sum被声明为指针 3）为什么可以将y分配给z？

我已经学会了这些，但他们已经太久了，有人可以给我一个提示吗？

Answer 1

值y和z不需要星号，因为添加星号会取消引用它们，并且您希望传递指针而不是值。
同样，您需要指针而不是值。虽然从代码看来你实际上想要的是值，所以那里可能应该有一颗星。
您可以指定指向另一个指针的指针，这意味着更改指针指向的地址。因此，当y=z时，y现在指向z点。

Answer 2

他们不需要明星因为你没有取消引用它们，而是作为指针传递。
您可以比较指针，但这可能不是您想要的。
分配指针是可以的，但请注意，它与复制值不同。

Answer 3

Ans for Q1：您正在将y和z传递给另一个函数，该函数将指针作为参数，为什么要传递指针的值？

答案2：他们是指针，你正在分配给另一个

Answer 4

基本指针概念：

double a = 42.0;   // a is a double
double b;          // b is a double
double *x;         // x is a pointer to double

x = &a;  // x is the pointer, we store the address of a in pointer x
b = *x;  // *x is the pointee (a double), we store the value pointed by x in b

// now b value is 42.0

Answer 5

关于2）以及为什么z [i] = y [i]是可以接受的，我认为最好指向this页面，其中详细描述了数组和指针，尤其是2.1到2.8。在该特定表达式中发生的事情（不一定按顺序）是从位置y开始，获取指针，将i添加到指针，然后获取指向该位置的值。从z开始，获取指针，将i添加到指针，并将值（y [i]）分配给该位置。

Answer 6

double
*foo(double *x, double *init, double *a){
// x is a pointer to a double, init is a pointer to a double, a is a pointer to a double
// foo is a function returning a pointer to a double
//   and taking three pointers to doubles as arguments

   double *y = (double*)malloc(5*sizeof(double));
   // y is a pointer to a double. It is assigned a pointer returned by malloc.
   //   That pointer returned by malloc points to memory space for 5 doubles.

   double *z = (double*)malloc(5*sizeof(double));
   // z is a pointer to a double. It is assigned a pointer returned by malloc.
   //   That pointer returned by malloc points to memory space for 5 doubles.

   double *sum, *update;
   // sum and update are pointers to doubles    

   sum = (double*)bob(y, z)    //<---Q1: why y and z don't need stars in front of them? 
                               //I thought they are pointers?
   // sum (whose type is pointer to double) 
   //   is assigned a pointer to double returned by bob

   for (i<0; i<5; i++){
       y[i] = z[i]       //and are z and y pointers?!?!
       // y[i] === *(y+i)
       // (y + i) points to the i-th element in the space previously allocated by malloc
       // *(y + i) dereferences that pointer
       // equivalently, y[i] accesses the i-th element from an array
   }
   if(sum<0)
   z=y //Q2: howcome it's ok to assign y to z? 
       //aren't they pointer?(i.e.hold memory address)
   // after the assignment, z contains the same address as y (points to the same memory)
}

double
*bob(*arg1, *arg2){
   something...
}

指针：指的是第一个元素是否有助于定位整个数组？

6 个答案: