Error array bound is not an integer constant before token

asked Mar 15, 2020 at 15:35

aRange is a constant integer, but not an integer constant. Isn’t English a fun language?

• C11 §6.4.4.1 Integer constants
• C11 §6.6 Constant expressions ¶6
  • An integer constant expression¹¹⁷⁾ shall have integer type and shall only have operands that are integer constants, enumeration constants, character constants, sizeof expressions whose results are integer constants, _Alignof expressions, and floating constants that are the immediate operands of casts. Cast operators in an integer constant expression shall only convert arithmetic types to integer types, except as part of an operand to the sizeof or _Alignof operator.
  • ¹¹⁷⁾ An integer constant expression is required in a number of contexts such as the size of a bit-field member of a structure, the value of an enumeration constant, and the size of a non-variable length array. Further constraints that apply to the integer constant expressions used in conditional-inclusion preprocessing directives are discussed in 6.10.1.
• C11 §6.7.6.2 Array declarators — one of the more inscrutable parts of the standard. (¶2 is a constraint; ¶4 and ¶5 specify the semantics of array declarators.)
  • ¶2 If an identifier is declared as having a variably modified type, it shall be an ordinary identifier (as defined in 6.2.3), have no linkage, and have either block scope or function prototype scope. If an identifier is declared to be an object with static or thread storage duration, it shall not have a variable length array type.
  • ¶4 If the size is not present, the array type is an incomplete type. If the size is * instead of being an expression, the array type is a variable length array type of unspecified size, which can only be used in declarations or type names with function prototype scope;¹⁴³⁾ such arrays are nonetheless complete types. If the size is an integer constant expression and the element type has a known constant size, the array type is not a variable length array type; otherwise, the array type is a variable length array type. (Variable length arrays are a conditional feature that implementations need not support; see 6.10.8.3.)
  • ¹⁴³⁾ Thus, * can be used only in function declarations that are not definitions (see 6.7.6.3).
  • ¶5 If the size is an expression that is not an integer constant expression: if it occurs in a declaration at function prototype scope, it is treated as if it were replaced by *; otherwise, each time it is evaluated it shall have a value greater than zero. The size of each instance of a variable length array type does not change during its lifetime. Where a size expression is part of the operand of a sizeof operator and changing the value of the size expression would not affect the result of the operator, it is unspecified whether or not the size expression is evaluated.

At file (global) scope, you must have an integer constant expression for the dimensions of an array. In a local variable in C99 or later, what you’ve written would be OK for a VLA (variable-length array).

You could work around this with:

enum { A_MIN = 0, A_MAX = 127 };
const uint8_t a[2] = { A_MIN, A_MAX };
const uint8_t aRange = a[1] - a[0];
double sums[A_MAX - A_MIN];

In C, you can’t write const uint8_t aRange = a[1] - a[0]; at file (global) scope, so your code should have been OK unless you’re using an antiquated C compiler that doesn’t recognize C99 or later (or it defines __STDC_NO_VLA__).

answered Mar 15, 2020 at 15:42

Jonathan’s answer is accepted. As a workaround I used macros though.

#define A_MIN 0
#define A_MAX 127
const uint8_t a[2] = {A_MIN, A_MAX};
const uint8_t aRange = A_MAX - A_MIN;
double sums[aRange];

answered Mar 15, 2020 at 15:56

Thomas LiThomas Li

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An assignment such as e=a(); cannot exist outside a function.

int d[e]; is also not valid C++ as e needs to be a compile time evaluable constant expression.

So compilation fails with an albeit cryptic error.

Note that the behaviour of d[4] will be undefined due to your attempting to access an element outside the bounds of an array.

answered May 4, 2017 at 12:41

Try this:

#include <iostream>
using namespace std;

int a(){
int b = 3;
return b;
}

int main(){
int e;
e = a();
int d[e] = {12,2,3};
cout << d[0] << endl;
return 0;
}

As you can see, the size of the vector is just 3 (remember that in c++ the 0 counts as a position), so in this case you just can print 3 values, d[0],d[1],d[2], remember that you can not assign a value outside of a function. Also if you plan to assign a global variable to array bound, it has to be constant. Like this: const int e = 3;, hope it helps.

answered Aug 4, 2018 at 18:06

Please Help me When ever
Getting error “array bound is not an integer constant before ‘]’ token”
‪#‎include‬ <iostream>
using namespace std;
int maxr;
int maxc;
void readmatrix(int m[maxr][maxc]); // size depend upon input of maxr and
// maxc
main()
{
cout<<«enter Number of Row :»;
cin>>maxr;
cout<<«enter Number of columb :»;
cin>>maxc;
}

you declare an array with index specified with non constant variable.
when you declare 2d array , the 2nd index must be specified only with a number or a constant variable.

Wait , is your array dynamic or static … most likely i’ll guess it’s static and if that’s the case you’re not allowed to do this

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cout<<"enter Number of Row :";
cin>>maxr;
cout<<"enter Number of columb :";
cin>>maxc;

That only works for dynamic arrays, static arrays dimensions must be const values, so your variables in main should be like

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const int maxr=/*some value*/; 
const int maxc=/*some value*/;

also, it is a good habit to call global var with scope operator