如何通过在C ++中的[]中传递变量来定义数组的大小？

Question

My code is given below. I want to declare an array of size n .

FILE *fp;
fp=fopen("myfile.b", "rb");
if(fp==NULL){ fputs("file error", stderr); exit(1); }
int* a;
fread(a, 0x1, 0x4, fp);
int n=*a;
int array[n];  // here is an error 

How can I declare an array of size n in this code?

Answer 1

That is a declaration of a variable-length array and it's not in C++ yet.

Instead I suggest you use std::vector instead:

std::vector<int> array(n);

---

You also have other problems, like declaring a pointer but not initializing it, and then using that pointer. When you declare a local variable (like a ) then its initial value is undefined , so using that pointer (except to assign to it) leads to undefined behavior . In this case what will probably happen is that your program will crash.

Answer 2

int *array = (int*)malloc( n * sizeof(int) );
//..
//your code
//..
//..
free(array);

Answer 3

You cannot declare an array of variable size in C++, but you can allocate memory once you know how much you need:

int* a = new int[n];

//do something with your array...

//after you are done:

delete [] a;

Answer 4

Since your code looks more like C...

FILE *fp = fopen("myfile.b", "rb");

if(fp==NULL)
{ 
  fputs("file error", stderr); 
  exit(1); 
}

//fseek( fp, 0, SEEK_END ); // position at end
//long filesize = ftell(fp);// get size of file
//fseek( fp, 0, SEEK_SET ); // pos at start

int numberOfInts = 0;
fread(&numberOfInts, 1, 4, fp); // you read 4 bytes sizeof(int)=4?
int* array = malloc( numberOfInts*sizeof(int) );

Answer 5

数组仅接受const对象或表达式，该值可以由编译器在编译期间确定，在这种情况下，c ++的vector更适合，否则我们需要为其动态分配内存。