C++ expected constant expression

#include <iostream>
#include <fstream>
#include <cmath>
#include <math.h>
#include <iomanip>
using std::ifstream;
using namespace std;

int main (void)

{
int count=0;
float sum=0;
float maximum=-1000000;
float sumOfX;
float sumOfY;
int size;
int negativeY=0;
int positiveX=0;
int negativeX=0;
ifstream points; //the points to be imported from file
//points.open( "data.dat");
//points>>size;
//cout<<size<<endl;

size=100;
float x[size][2];
while (count<size) {



points>>(x[count][0]);
//cout<<"x= "<<(x[count][0])<<"  ";//read in x value
points开发者_运维技巧>>(x[count][1]);
//cout<<"y= "<<(x[count][1])<<endl;//read in y value


count++;
}

This program is giving me expected constant expression error on the line where I declare float x[size][2]. Why?

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float x[size][2];

That doesn't work because declared arrays can't have runtime sizes. Try a vector:

std::vector< std::array<float, 2> > x(size);

Or use new

// identity<float[2]>::type *px = new float[size][2];
float (*px)[2] = new float[size][2];

// ... use and then delete
delete[] px;

---

If you don't have C++11 available, you can use boost::array instead of std::array.

If you don't have boost available, make your own array type you can stick into vector

template<typename T, size_t N>
struct array {
  T data[N];
  T &operator[](ptrdiff_t i) { return data[i]; }
  T const &operator[](ptrdiff_t i) const { return data[i]; }
};

For easing the syntax of new, you can use an identity template which effectively is an in-place typedef (also available in boost)

template<typename T> 
struct identity {
  typedef T type;
};

---

If you want, you can also use a vector of std::pair<float, float>

std::vector< std::pair<float, float> > x(size);
// syntax: x[i].first, x[i].second

---

The array will be allocated at compile time, and since size is not a constant, the compiler cannot accurately determine its value.

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You cannot have variable length arrays (as they are called in C99) in C++. You need to use dynamically allocated arrays (if the size varies) or a static integral constant expression for size.

---

The line float x[size][2] won't work, because arrays have to be allocated at compile time (with a few compiler-specific exceptions). If you want to be able to easily change the size of the array x at compile time, you can do this:

 #define SIZE 100
 float x[SIZE][2];

If you really want to allocate the array based on information you only have at runtime, you need to allocate the array dynamically with malloc or new.

---

The size of an automatic array must be a compile-time constant.

 const int size = 100;
 float x[size][2];

If the size weren't known at compile-time (e.g entered by the user, determined from the contents of the file), you'd need to use dynamic allocation, for example:

std::vector<std::pair<float, float> > x(somesize);

(Instead of a pair, a dedicated Point struct/class would make perfect sense.)

---

Because it expected a constant expression!

Array dimensions in C (ignoring C99's VLAs) and C++ must be quantities known at compile-time. That doesn't mean just labelled with const: they have to be hard-coded into the program.

Use dynamic allocation or std::vector (which is a wrapper around dynamic array allocation) to determine array sizes at run-time.

---

It is a restriction of the language. Array sizes must be constant expressions. Here's a partial jsutification from cplusplus.com

NOTE: The elements field within brackets [] which represents the number of elements the array is going to hold, must be a constant value, since arrays are blocks of non-dynamic memory whose size must be determined before execution. In order to create arrays with a variable length dynamic memory is needed, which is explained later in these tutorials.

---

You haven't assigned any value to size; thus the compiler cannot allocate the memory for the array. (An array of null size? What?)

Additionally, you'd need to make SIZE a constant, and not a variable.

EDIT: Unfortunately, this response no longer makes sense since the poster has changed their question.