C compile error: "Variable-sized object may not be initialized"

CCompiler ErrorsInitializer ListVariable Length-Array

C Problem Overview

Why do I receive the error "Variable-sized object may not be initialized" with the following code?

int boardAux[length][length] = {{0}};

C Solutions

Solution 1 - C

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) );

Solution 2 - C

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.

Solution 3 - C

This gives error:

int len;
char str[len]="";

This also gives error:

int len=5;
char str[len]="";

But this works fine:

int len=5;
char str[len]; //so the problem lies with assignment not declaration

You need to put value in the following way:

str[1]='b'; //like that; and not like str="ab";

Solution 4 - C

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;

Solution 5 - C

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.

Solution 6 - C

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;

Solution 7 - C

Variable length arrays are arrays whose length is not known by the compiler at compile time. In your case length is a variable. I conclude this, because if length was a e.g. preprocessor macro defined as a literal integer your initialization would work. The first C language standard from 1989 did not allow variable length arrays, they were added in 1999. Still the C standard does not allow these to be initialized with an expression like yours (although one could argue that it could or should allow it).

The best way to initialize a variable array is like this:

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

memset is a very fast standard library function for initializing memory (to 0 in the above case). sizeof(boardAux) returns the number of bytes occupied by boardAux. sizeof is always available but memset requires #include <string.h>. And yes - sizeof allows a variable sized object as argument.

Note that if you have a normal array (not variable length) and just want to initialize the memory to zero you never need nested brackets, you can initialize it simply like this:

struct whatEver name[13][25] = {0};

Solution 8 - C

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];

Solution 9 - C

Simply declare length to be a cons, if it is not then you should be allocating memory dynamically


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QuestionhelloWorldView Question on Stackoverflow
Solution 1 - CDavid Rodríguez - dribeasView Answer on Stackoverflow
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Solution 3 - CAmitesh RanjanView Answer on Stackoverflow
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