Question

int *c;
int d,noofentries;

struct A
{
unsigned int *ptr;
int entry;
}a;


a->ptr=memalloc(34,unsigned int);

a->ptr = (unsigned int*) entry


nofoentries = 8 ;
    d =56749;


for(i=0;i<noofentries;i++)
{
c[i] = d; // how is a pointer treated as array ?
}
for(i=0;i<34;i++)
{
a->ptr[i] = c[i]; //segmentation fault occurs
}

我要求将c[i]中填充的值分配给a->ptr[i]。因此，当a->ptr[i]被删除时，c[i]也会被释放。

请帮助!!

Answer 1

通常您不希望将pointer视为array，而是拥有array并使用其name as pointer至refer to any particular member数组

例如

int arr[5];
//the array name 'arr' points to the zeroth element

现在您可以使用*(arr+ indexNo) = value or arr[indexNo] = value为特定元素分配值

当您为其指定了数组时，您可能希望将指针用作数组。

例如

int arr[5];
int *ptr;

如果你这样做

ptr = arr;

您可以像访问arr

一样访问ptr

as

ptr[index]= value;

Answer 2

指向类型的指针与该类型的数组相同 *（c + x）= a ＆LT; =＆GT; c [x] = a;

c + x找到正确的指针位置，因为它将x * sizeof（type）添加到c指针。

您的代码在gcc下编译：

#include <stdlib.h>
#include <assert.h>
#include <stdio.h>

struct A { unsigned int *ptr; int entry; };

int main(int argc, char ** argv)
{  
  unsigned int * c;
  unsigned int d;
  int noofentries, i;
  struct A a;

  noofentries=34; 
  c=malloc(noofentries * sizeof(unsigned int));

  d =56749;


  for(i=0;i<noofentries;i++) { c[i] = d; }
  // no need to copy full array c, since ptr is a pointer over it...
  a.ptr = c;
  // warning if above line is not done then allocation of ptr is required:
  // a.ptr = malloc(noofentries * sizeof(unsigned int));
  // and content copy
  // for(i=0;i<noofentries;i++) { a.ptr[i] = c[i]; }

  for(i=0;i<noofentries;i++) { 
    assert( a.ptr[i] == *(c + i) );
    printf("a.ptr[%u]=%u\n",i,*(a.ptr + i));
  }

  free(c);
}

unsigned int * ptr如何被视为prt [i]？

2 个答案: