Ошибка variable sized object may not be initialized

asked Jun 21, 2010 at 7:52

I am assuming that you are using a C99 compiler (with support for dynamically sized arrays). The problem in your code is that at the time when the compilers sees your variable declaration it cannot know how many elements there are in the array (I am also assuming here, from the compiler error that length is not a compile time constant).

You must manually initialize that array:

int boardAux[length][length];
memset( boardAux, 0, length*length*sizeof(int) );

answered Jun 21, 2010 at 8:06

You receive this error because in C language you are not allowed to use initializers with variable length arrays. The error message you are getting basically says it all.

6.7.8 Initialization

…

3 The type of the entity to be initialized shall be
an array of unknown size or an object
type that is not a variable length
array type.

answered Jun 21, 2010 at 8:09

Amitesh RanjanAmitesh Ranjan

1,1621 gold badge12 silver badges9 bronze badges

After declaring the array

int boardAux[length][length];

the simplest way to assign the initial values as zero is using for loop, even if it may be a bit lengthy

int i, j;
for (i = 0; i<length; i++)
{
    for (j = 0; j<length; j++)
        boardAux[i][j] = 0;
}

answered Mar 27, 2016 at 12:35

The array is not initialized with the memory specified anf throws an error
variable sized array may not be initialised
I prefer usual way of initialization,

for (i = 0; i < bins; i++)
        arr[i] = 0;

answered Sep 21, 2020 at 13:25

The question is already answered but I wanted to point out another solution which is fast and works if length is not meant to be changed at run-time. Use macro #define before main() to define length and in main() your initialization will work:

#define length 10

int main()
{
    int boardAux[length][length] = {{0}};
}

Macros are run before the actual compilation and length will be a compile-time constant (as referred by David Rodríguez in his answer). It will actually substitute length with 10 before compilation.

answered Apr 12, 2020 at 21:42

int size=5;
int ar[size ]={O};

/* This  operation gives an error -  
variable sized array may not be 
initialised.  Then just try this. 
*/
int size=5,i;
int ar[size];
for(i=0;i<size;i++)
{
    ar[i]=0;
}

answered May 22, 2020 at 6:57

answered Mar 2, 2016 at 20:26

asked Jun 21, 2010 at 7:52

I am assuming that you are using a C99 compiler (with support for dynamically sized arrays). The problem in your code is that at the time when the compilers sees your variable declaration it cannot know how many elements there are in the array (I am also assuming here, from the compiler error that length is not a compile time constant).

You must manually initialize that array:

int boardAux[length][length];
memset( boardAux, 0, length*length*sizeof(int) );

answered Jun 21, 2010 at 8:06

You receive this error because in C language you are not allowed to use initializers with variable length arrays. The error message you are getting basically says it all.

6.7.8 Initialization

…

3 The type of the entity to be initialized shall be
an array of unknown size or an object
type that is not a variable length
array type.

answered Jun 21, 2010 at 8:09

Amitesh RanjanAmitesh Ranjan

1,1621 gold badge12 silver badges9 bronze badges

After declaring the array

int boardAux[length][length];

the simplest way to assign the initial values as zero is using for loop, even if it may be a bit lengthy

int i, j;
for (i = 0; i<length; i++)
{
    for (j = 0; j<length; j++)
        boardAux[i][j] = 0;
}

answered Mar 27, 2016 at 12:35

The array is not initialized with the memory specified anf throws an error
variable sized array may not be initialised
I prefer usual way of initialization,

for (i = 0; i < bins; i++)
        arr[i] = 0;

answered Sep 21, 2020 at 13:25

The question is already answered but I wanted to point out another solution which is fast and works if length is not meant to be changed at run-time. Use macro #define before main() to define length and in main() your initialization will work:

#define length 10

int main()
{
    int boardAux[length][length] = {{0}};
}

Macros are run before the actual compilation and length will be a compile-time constant (as referred by David Rodríguez in his answer). It will actually substitute length with 10 before compilation.

answered Apr 12, 2020 at 21:42

int size=5;
int ar[size ]={O};

/* This  operation gives an error -  
variable sized array may not be 
initialised.  Then just try this. 
*/
int size=5,i;
int ar[size];
for(i=0;i<size;i++)
{
    ar[i]=0;
}

answered May 22, 2020 at 6:57

answered Mar 2, 2016 at 20:26

asked Nov 7, 2013 at 11:31

Compile-time, you compiler does not know how many elements are in your matrix. In C, you can dynamically allocate memory using malloc.

You could use a define to create a constant value:

#define N 3

int main()
{
    double InputM[N][N] = { { 2, 0, 1 }, { 3, 1, 2 }, { 5, 2, 5} };
}

Or malloc:

int main()
{
    int n = 3;
    int idx;
    int row;
    int col;

    double **inputM;
    inputM = malloc(n * sizeof(double *));
    for (idx = 0; idx != n; ++idx)
    {
        inputM[idx] = malloc(n * sizeof(double));
    }

    // initialise all entries on 0
    for (row = 0; row != n; ++row)
    {
        for (row = 0; row != n; ++row)
        {
            inputM[row][col] = 0;
        }
    }

    // add some entries
    inputM[0][0] = 2;
    inputM[1][1] = 1;
    inputM[2][0] = 5;
}

answered Nov 7, 2013 at 11:33

linkddlinkdd

9951 gold badge9 silver badges24 bronze badges

I’m trying to let the user choose between three functions for the program to run, but when I compile I get the error» variable-sized object `addnumber’ may not be initialized» on line 47. My code:

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#include <iostream>
#include <cmath>
using namespace std;

int main()
{
    //char b;
    cout << "Choose to find the average, square or add" << endl;
    string input;
    cin >> input;
    
    if (input== "Average")
    {
    
    double x;
    double y;
    double w; 
    double z;
	cout << "Enter numbers to be added" << endl;
	cin >> x; 
	cin >> y; 
	cin >> w;
	cin >> z;
	cout << "The average is: " << endl;
	cout << (x + y + w + z) / 4 << endl;
}
    else if (input== "Square")
    {
	double a;

	cout << "Enter number to be squared" << endl;
	cin >> a;
	cout << "Square is: " << a * (a) << endl;
}
    else if (input== "Add")
    {
         int c;
         double d;
         double e;
         double f;
         double g;
         double h;
         double i;
         double j;
         cout << "Enter how many numbers you want to add" << endl;
         cin >> c;
         double addnumber[c] = {d,e,f,g,h,i,j};
         cout << "Enter numbers to be added" << endl;
         
         if (c = 2)
         {
               cin >> d;
               cin >> e;
               cout << "Sum is: " << d + e << endl;
               }
         else if (c = 3)
         {
              cin >> d;
              cin >> e;
              cin >> f;
              cout << "Sum is: " << d + e + f << endl;
              }      
         else if (c = 4)
         {
              cin >> d;
              cin >> e;
              cin >> f;
              cin >> g;
              cout << "Sum is: " << d + e + f + g << endl;
              }     
         else if (c = 5)
         {
              cin >> d;
              cin >> e;
              cin >> f;
              cin >> g;
              cin >> h;
              cout << "Sum is: " << d + e + f + g + h << endl;
              } 
         else if (c = 6)
         {
              cin >> d;
              cin >> e;
              cin >> f;
              cin >> g;
              cin >> h;
              cin >> i;
              cout << "Sum is: " << d + e + f + g + h + i << endl;
              }    
         else if (c > 6)
         {
              cout << "Yo number is too big bro" << endl;
              }
         else 
         {
              cout << "Yo number is invalid dawg" << endl;
              }
         }
	
	
	system("PAUSE");
	return 0;
}

The length of an array must be known at compile-time. In other words, the length must be constant. Therefore, a value only known at run-time cannot be used to determine the array length.

If it’s possible, you can use dynamic memory allocation^[1]. This will allow you to create an array during run-time:

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int input(0);

// Get the length of the array:
std::cin >> input;

// Create an array dynamically:
double *dynamic_array(new double[input]);

// Delete the array (MUST DO THIS!):
delete [] dynamic_array;
dynamic_array = 0x0;

References:
^[1] http://www.cplusplus.com/doc/tutorial/dynamic/

_Wazzak

You could just remove that line, since «addnumber» is not used anywhere else in the program.

However, the explanation is that «c» is an integer with a value not known at compile time (since it is input by the user).

Hence this double addnumber[c] is attempting to define an array with a variable length. This is not usually valid in C++ (though some compilers may accept it).

If you did need to use addnumber[], the simplest solution would be to set it to a size at least as large as required, such as double addnumber[10], and make sure the program never tries to use any element beyond addnumber[9].

Note also the initialisation part = {d,e,f,g,h,i,j}; does not make sense in any case.

I should think the simplest way would be not to use array storage at all.

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#include <iostream>

int main()
{
    std::cout << "Enter the amount of numbers to sum: " ;

    unsigned numbers ;
    std::cin >> numbers ;

    std::cout << "Enter the numbersn" ;

    double total = 0.0 ;
    for ( unsigned i=0;  i<numbers; ++i )
    {
        double input ;
        std::cin >> input ;
        total += input ;
    }

    std::cout << "The total is " << total ;
    std::cout << "nThe average is " << total / numbers << 'n' ;
}